STAAR Algebra I Released Tests

The y-intercept is the point where the graph intersects the y-axis.
The graph intersects the y-axis at (0, 3)
Hence, the y-intercept is (0, 3)

$ \sqrt{50} \\[3ex] = \sqrt{25 \cdot 2} \\[3ex] = \sqrt{25} \cdot \sqrt{2} \\[3ex] = 5\sqrt{2} $

$ \text{slope} = m \\[5ex] \underline{\text{Line with the positive slope}} \\[3ex] \text{Point 1: } (-2, 0) \\[3ex] x_1 = -2 \\[3ex] y_1 = 0 \\[5ex] \text{Point 2: } (0, 2) \\[3ex] x_2 = 0 \\[3ex] y_2 = 2 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{2 - 0}{0 - (-2)} \\[5ex] m = \dfrac{2}{2} \\[5ex] m = 1 \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 0 = 1(x - (-2)) \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = x + 2 \\[5ex] \underline{\text{Line with the negative slope}} \\[3ex] \text{Point 1: } (3, 5) \\[3ex] x_1 = 3 \\[3ex] y_1 = 5 \\[5ex] \text{Point 2: } (5, -3) \\[3ex] x_2 = 5 \\[3ex] y_2 = -3 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-3 - 5}{5 - 3} \\[5ex] m = -\dfrac{8}{2} \\[5ex] m = -4 \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 5 = -4(x - 3) \\[3ex] y = -4x + 12 + 5 \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = -4x + 17 \\[5ex] \therefore \text{The system of equations shown on the graph is:} \\[3ex] y = x + 2 \\[3ex] y = -4x + 17 $

The functions are quadratic functions.
We can solve the question using at least two approaches.
Use any approach you prefer.

1st Approach: Vertex Method
The y-coordinate of the vertex is the maximum or the minimum value of the function.
This value helps us determine the range of the function.

Another important information we need for this question is the leading coefficient.
If the leading coefficient is negative, the quadratic function has a maximum value.
If the leading coefficient is positive, the quadratic function has a minimum value.

The set of all real numbers greater than or equal to –1 implies that the graph of the function has a minimum value.
The minimum value must be –1
This also means that the leading coefficient is positive.
Let us analyze each option.

$ A.\;\; f(x) = -x^2 - 1 \\[3ex] \text{The leading coefficient is negative} \\[3ex] \text{Discard}. \\[5ex] B.\;\; f(x) = x^2 + 4x + 3 \\[3ex] \text{Compare to the standard form: } ax^2 + bx + c \\[3ex] a = 1 \\[3ex] b = 4 \\[3ex] c = 3 \\[3ex] x-coordinate \text{ of the vertex} = -\dfrac{b}{2a} = -\dfrac{4}{2(1)} = -\dfrac{4}{2} = -2 \\[5ex] y-coordinate \text{ of the vertex} = f(-2) \\[3ex] = (-2)^2 + 4(-2) + 3 \\[3ex] = 4 - 8 + 3 \\[3ex] = -1 \\[3ex] $ This is the correct option.
But, we can try other options to convince you.

$ C.\;\; f(x) = (x + 3)^2 + 1 \\[3ex] \text{Compare to the vertex form: } a(x - h)^2 + k \\[3ex] k = 1 \\[3ex] y-coordinate \text{ of the vertex} = k = 1 \\[3ex] \text{Discard}. \\[5ex] D.\;\; f(x) = (x + 4)^2 - 17 \\[3ex] k = -17 \\[3ex] y-coordinate \text{ of the vertex} = -17 \\[3ex] \text{Discard}. \\[3ex] $ 2nd Approach: Let us analyze each option.
Because of time, it is better to use the process of elimination: eliminate any option that has any output less than –1.
Something important to note in this question is that the square of any integer besides zero is always positive.
The square of zero is zero.

$ \text{Option A} \\[3ex] f(x) = -x^2 - 1 \\[3ex] \text{Assume } x = -1 \\[3ex] f(-1) = -(-1)^2 - 1 \\[3ex] f(-1) = -1 - 1 \\[3ex] f(-1) = -2 \\[3ex] -2 \lt -1 \\[3ex] \text{Discard}. \\[5ex] \text{Option B} \\[3ex] f(x) = x^2 + 4x + 3 \\[3ex] \text{Assume } x \ge 0 \\[3ex] f(x) \text{ must be } \gt -1...okay \\[5ex] \text{Assume } x \lt 0 \\[3ex] \text{Let } x = -1 \\[3ex] f(-1) = (-1)^2 + 4(-1) + 3 \\[3ex] f(-1) = 1 - 4 + 3 \\[3ex] f(-1) = 0 \\[3ex] 0 \gt -1...okay \\[3ex] \text{Let } x = -2 \\[3ex] f(-2) = (-2)^2 + 4(-2) + 3 \\[3ex] f(-2) = 4 - 8 + 3 \\[3ex] f(-2) = -1 \\[3ex] -1 = -1 ...okay \\[3ex] \text{Let } x = -3 \\[3ex] f(-3) = (-3)^2 + 4(-3) + 3 \\[3ex] f(-3) = 9 - 12 + 3 \\[3ex] f(-3) = 0 \\[3ex] 0 \gt -1...okay \\[3ex] \text{Let } x = -4 \\[3ex] f(-4) = (-4)^2 + 4(-4) + 3 \\[3ex] f(-4) = 16 - 16 + 3 \\[3ex] f(-4) = 3 \\[3ex] 3 \gt -1...okay \\[3ex] \text{Let } x = -5 \\[3ex] f(-5) = (-5)^2 + 4(-5) + 3 \\[3ex] f(-5) = 25 - 20 + 3 \\[3ex] f(-5) = 8 \\[3ex] 8 \gt -1...okay \\[3ex] $ This is the correct option.
But, we can try other options to convince you.

$ \text{Option C} \\[3ex] f(x) = (x + 3)^2 + 1 \\[3ex] (x + 3)^2 \text{ must be positive} \\[3ex] (x + 3)^2 + 1 \text{ must be positive} \\[3ex] \text{So } (x + 3)^2 + 1 \gt -1 \\[3ex] \text{This satisfies the condition that the range is } \gt -1 \\[3ex] $ Let us see whether it satisfies the other condition.
Can we find x such that $f(x) = -1$?

$ \text{Assume } f(x) = -1 \\[3ex] (x + 3)^2 + 1 = -1 \\[3ex] (x + 3)^2 = -1 - 1 \\[3ex] (x + 3)^2 = -2 \\[3ex] x + 3 = \pm\sqrt{-2} ...x \text{ is not a real number} \\[3ex] \text{Discard}. \\[5ex] \text{Option D} \\[3ex] f(x) = (x + 4)^2 - 17 \\[3ex] \text{Assume } x = -1 \\[3ex] f(-1) = (-1 + 4)^2 - 17 \\[3ex] f(-1) = 3^2 - 17 \\[3ex] f(-1) = 9 - 17 \\[3ex] f(-1) = -8 \\[3ex] -8 \lt -1 \\[3ex] \text{Discard}. $

$ \text{Point 1: } (-6, -10) \\[3ex] x_1 = -6 \\[3ex] y_1 = -10 \\[5ex] \text{Point 2: } (10, 14) \\[3ex] x_2 = 10 \\[3ex] y_2 = 14 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{14 - (-10)}{10 - (-6)} \\[5ex] m = \dfrac{24}{16} \\[5ex] m = \dfrac{3}{2} \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - (-10) = \dfrac{3}{2}[x - (-6)] \\[5ex] y + 10 = \dfrac{3}{2}(x + 6) \\[5ex] y + 10 = \dfrac{3}{2}x + 9 \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = mx + b ...\text{where } b \text{ is the y-intercept} \\[3ex] y = \dfrac{3}{2}x + 9 - 10 \\[5ex] y = \dfrac{3}{2}x + (-1) $

Observing the data, as the number of months increases (from 2 to 23), the remaining balance decreases (from 7,592 to 2,854).
This consistent downward trend indicates a strong negative correlation between the number of months and the remaining balance.

$ \text{slope, } m = -\dfrac{2}{11} \\[5ex] \text{Point 1: } (22, 5) \\[3ex] x_1 = 22 \\[3ex] y_1 = 5 \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 5 = -\dfrac{2}{11}(x - 22) \\[5ex] y - 5 = -\dfrac{2}{11}x + 4 \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = -\dfrac{2}{11}x + 4 + 5 \\[5ex] y = -\dfrac{2}{11}x + 9 $

$ h(x) = 3(x - 5)(x + 7) \\[3ex] \text{zeros of } h(x) \implies \\[3ex] h(x) = 3(x - 5)(x + 7) = 0 \\[3ex] \underline{\text{Zero Product Property}} \\[3ex] 3(x - 5) = 0 \hspace{1em}\text{OR}\hspace{1em} x + 7 = 0 \\[3ex] x = 5 \hspace{1em}\text{OR}\hspace{1em} x = -7 \\[3ex] \implies \\[3ex] h(5) = 0 \hspace{1em}\text{OR}\hspace{1em} h(-7) = 0 \\[3ex] $ C. The zeros are 5 and –7 because $h(5) = 0$ and $h(-7) = 0$.

$ 3y - 2x \le -6 \\[3ex] \text{When } x = 0 \\[3ex] 3y - 2(0) \le -6 \\[3ex] 3y \le -6 \\[3ex] y \le -2 ...\text{Mark point } (0, -2) \\[3ex] \text{The shaded area is downwards from } y = -2 \\[5ex] \text{When } y = 0 \\[3ex] 3(0) - 2x \le -6 \\[3ex] -2x \le -6 \\[3ex] x \ge \dfrac{-6}{-2} \\[5ex] x \ge 3 ...\text{Mark point } (3, 0) \\[3ex] \text{The shaded area is to the right of } x = 3 \\[3ex] $ Draw a straight solid line (because of the equal sign) connecting the two points: (0, –2) and (3, 0)
Locate the intersection of the two areas.
The correct answer is Option B.

$ 3x - 5y = 12...eqn.(1) \\[3ex] y = -6x + 2...eqn.(2) \\[3ex] \text{Substitute eqn.(2) into eqn.(1) for } y \\[3ex] 3x - 5(-6x + 2) = 12 \\[3ex] 3x + 30x - 10 = 12 \\[3ex] 33x = 12 + 10 \\[3ex] x = \dfrac{22}{33} \\[5ex] x = \dfrac{2}{3} $

$ f(x) = x^2 \\[3ex] g(x) = f(x + 8.7) = (x + 8.7)^2 \\[3ex] $ Transformation from $f(x)$ to $g(x)$:
Horizontal Shift of f(x) 8.7 units left
This implies that:

$ (0, 0) \rightarrow (0 - 8.7, 0) \\[3ex] \rightarrow (-8.7, 0) $

The rate of change of y with respect to x is the slope of the function.
Let us represent the function in slope–intercept form so that we can determine the slope.

$ x + 7y = -14 \\[3ex] 7y = -x - 14 \\[3ex] y = \dfrac{-x - 14}{7} \\[5ex] y = -\dfrac{x}{7} - \dfrac{14}{7} \\[5ex] \text{Compare to slope–intercept form: } y = mx + b \\[3ex] \text{slope, } m = -\dfrac{1}{7} $

The number of pounds of apples, x @ $2.50 per pound = 2.50 * x = 2.5x
The number of pounds of blueberries, y @ $3.00 per pound = 3.00 * y = 3y
Total cost = 2.5x + 3y

The baker does not want to spend more than $75.00.
This means that the total cost should be less than or equal to $75.00.

$ 2.5x + 3y \le 75 $

The y-intercept is the point where the graph intersects the y-axis.
The correct location is:

The range of a function is the set of all output values that result from applying the function to its input values.
The range of the function is: $\{64.95, 109.90, 154.85, 199.80\}$

$ f(x) = 3(x - 4)^2 - 7 \\[3ex] f(-1) = 3(-1 - 4)^2 - 7 \\[3ex] = 3(-5)^2 - 7 \\[3ex] = 3(25) - 7 \\[3ex] = 75 - 7 \\[3ex] = 68 $

$ \underline{\text{Exponential Function}} \\[3ex] y = ab^x \\[3ex] a = \text{initial value} \\[3ex] b = \text{base} \\[5ex] \text{Point 1: } (0, 90) \\[3ex] x = 0 \\[3ex] y = 90 \\[3ex] 90 = a * b^{0} \\[3ex] 90 = a * 1 \\[3ex] a = 90 \\[5ex] \text{Point 2: } (1, 30) \\[3ex] x = 1 \\[3ex] y = 30 \\[3ex] 30 = a * b^{1} \\[3ex] 30 = 90 * b \\[3ex] b = \dfrac{30}{90} \\[5ex] b = \dfrac{1}{3} \\[5ex] \implies \\[3ex] y = 90\left(\dfrac{1}{3}\right)^x $

The function is the y
The zero of the function is the value of the x when y = 0
y = 0 when x = –5
Hence, the zero of the function is: x = –5

$ 5x^2 - 30x - 80 \\[3ex] 5(x^2 - 6x - 16) \\[3ex] 5(x + 2)(x - 8) \\[3ex] $ One of the factors is $x - 8$

The axis of symmetry of the graph of f is the vertical line through the vertex
It is the x-coordinate of the vertex
Axis of symmetry is: $x = -5$

$ \underline{\text{Line }n} \\[3ex] y = -\dfrac{3}{5}x + 6 \\[5ex] \text{Compare to slope–intercept form: } y = mx + b \\[3ex] \text{slope, } m = -\dfrac{3}{5} \\[5ex] \underline{\text{Line }p} \\[3ex] m = -\dfrac{3}{5} ...\text{parallel lines have the same slope} \\[5ex] \text{Passes through Point } (15, -6) \\[3ex] x_1 = 15 \\[3ex] y_1 = -6 \\[3ex] y - y_1 = m(x - x_1) ...\text{Point–Slope Form} \\[3ex] y - (-6) = -\dfrac{3}{5}(x - 15) \\[5ex] y + 6 = -\dfrac{3}{5}x + 9 \\[5ex] y = -\dfrac{3}{5}x + 9 - 6 \\[5ex] y = -\dfrac{3}{5}x + 3 ...\text{Slope–Intercept Form} $

$ (2k^2 - 3m)(3k - 4) \\[3ex] 6k^3 - 8k^2 - 9km + 12m $

$ y = -0.5x + 4...eqn.(1) \\[3ex] 3x + 6y = 24...eqn.(2) \\[3ex] \text{From } eqn.(2) \\[3ex] 6y = -3x + 24 \\[3ex] y = -\dfrac{3}{6}x + \dfrac{24}{6} \\[5ex] y = -0.5x + 4 ...\text{modified form of } eqn.(2) \\[3ex] $ We notice that eqn.(1) and eqn.(2) are the same.
So, we are saying the same thing.
Hence, the system has infinitely many solutions.

Linear relationship implies that we shall use the equation of a straight line
We shall use the equation of a straight line in point–slope form and slope intercept form.

$ \underline{\text{Linear Relationship}} \\[3ex] \text{Point 1: } (6.50, 107) \\[3ex] x_1 = 6.5 \\[3ex] y_1 = 107 \\[5ex] \text{Point 2: } (11.00, 65) \\[3ex] x_2 = 11 \\[3ex] y_2 = 65 \\[5ex] \text{slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{65 - 107}{11 - 6.5} \\[5ex] m = -\dfrac{42}{4.5} \\[5ex] m = -9.3\bar{3} \\[5ex] \underline{\text{Point–Slope Form}} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 107 = -9.33\bar{3}(x - 6.5) \\[3ex] y - 107 = -9.3\bar{3}x + 60.6\bar{6} \\[5ex] \underline{\text{Slope–Intercept Form}} \\[3ex] y = -9.3\bar{3}x + 60.6\bar{6} + 107 \\[3ex] y = -9.3\bar{3}x + 167.6\bar{6} \\[5ex] \text{Let us test each option and eliminate incorrect ones} \\[3ex] \text{1st option: } x = \$5.00 \\[3ex] y = -9.3\bar{3}(5) + 167.6\bar{6} \\[3ex] y = 121\text{ lunch specials} \\[3ex] 121 \ne 82 ...\text{Discard} \\[5ex] \text{2nd option: } x = \$6.00 \\[3ex] y = -9.3\bar{3}(6) + 167.6\bar{6} \\[3ex] y = 111.6\bar{6}\text{ lunch specials} \\[3ex] 111.6\bar{6} \approx 111 ...\text{rounded down...Keep} \\[5ex] \text{3rd option: } x = \$7.50 \\[3ex] y = -9.3\bar{3}(7.5) + 167.6\bar{6} \\[3ex] y = 97.6\bar{6}\text{ lunch specials} \\[3ex] 97.6\bar{6} \ne 102 ...\text{Discard} \\[5ex] \text{4th option: } x = \$12.50 \\[3ex] y = -9.3\bar{3}(12.5) + 167.6\bar{6} \\[3ex] y = 51\text{ lunch specials} \\[3ex] 51 \ne 74 ...\text{Discard} \\[5ex] \text{5th option: } x = \$15.00 \\[3ex] y = -9.3\bar{3}(15) + 167.6\bar{6} \\[3ex] y = 27.6\bar{6}\text{ lunch specials} \\[3ex] 27.6\bar{6} \approx 28 ...\text{rounded normal or rounded up...Keep} $

$ x(x + 3) = 108 \\[3ex] x^2 + 3x - 108 = 0 \\[3ex] (x + 12)(x - 9) = 0 \\[3ex] x + 12 = 0 \hspace{1em}\text{OR}\hspace{1em} x - 9 = 0 ...\text{Zero Product Property} \\[3ex] x = -12 ...\text{Discard because the width cannot be negative} \\[3ex] x = 9 ...\text{Keep because the width is positive} \\[3ex] x = 9\;ft $

$ \text{Point 1: } (-4, 23) \\[3ex] x_1 = -4 \\[3ex] y_1 = 23 \\[5ex] \text{Point 2: } (5, -1) \\[3ex] x_2 = 5 \\[3ex] y_2 = -1 \\[5ex] \text{slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-1 - 23}{5 - (-4)} \\[5ex] m = -\dfrac{24}{9} \\[5ex] m = -\dfrac{8}{3} $

$ \dfrac{2x^{12}}{6x^{-3}} \\[5ex] \dfrac{2}{6} * \dfrac{x^{12}}{x^{-3}} \\[5ex] \dfrac{2}{6} * x^{12 - (-3)}...\text{Law 2 Exp} \\[5ex] \dfrac{1}{3} * x^{12 + 3} \\[5ex] \dfrac{1}{3}x^{15} $

$ \dfrac{4}{3}z + 6 = -4\left(\dfrac{1}{6}z + 9\right) \\[5ex] \dfrac{4z}{3} + 6 = \dfrac{-4z}{6} - 36 \\[5ex] \dfrac{4z}{3} + \dfrac{4z}{6} = -36 - 6 \\[5ex] \dfrac{4z}{3} + \dfrac{2z}{3} = -42 \\[5ex] \dfrac{6z}{3} = -42 \\[5ex] 2z = -42 \\[3ex] z = -\dfrac{42}{2} \\[5ex] z = -21 $

Because the inequality does not contain an equal sign, the line will be a dashed line.

$ y \gt 5x - 3 \\[5ex] \text{When } x = 0 \\[3ex] y \gt 5(0) - 3 \\[3ex] y \gt -3 \\[5ex] \text{When } x = 1 \\[3ex] y \gt 5(1) - 3 \\[3ex] y \gt 5 - 3 \\[3ex] y \gt 2 \\[5ex] $ The graph of the inequality is



If you want to play with this some more, we can also use another approach.

$ \text{When } y = 2 \\[3ex] 2 \gt 5x - 3 \\[3ex] 5x - 3 \lt 2 \\[3ex] 5x \lt 2 + 3 \\[3ex] 5x \lt 5 \\[3ex] x \lt 1 \\[3ex] $ Student: Mr. C, wait a minute.
Why did you not do when $y = 0$?
Is there any reason for using 2?
Teacher: Good observation.
If we used 0, we would end up with a fraction.
Fractions are not allowed in the graph that we are given.
Hence, it is necessary that we use a value of $y$ that will give an integer value of $x$ within the domain [-7, 7]
What other values could we have used?
Student: 7
–2
Teacher: There you go...

The graph of the inequality is

$ 5x - 8y = 40 \\[3ex] 5x - 40 = 8y \\[3ex] 8y = 5x - 40 \\[3ex] y = \dfrac{5x - 40}{8} \\[5ex] y = \dfrac{5x}{8} - \dfrac{40}{8} \\[5ex] y = \dfrac{5}{8}x - 5 $

The domain is the set of all the input values, x for which the function has an output.
The closed circle represents a closed interval. It shows that the end point is included.
On the graph, it is on the right hand side at x = 5

The arrow represents infinity
On the graph, it is on the left hand side and represents negative infinity.

So, the interval notation is: $(-\infty, 5]$
This represents all real numbers less than or equal to 5.

The function is an exponential function because the exponent, x is a variable and the base, $\dfrac{1}{3}$ is a constant.
So, we select the Exponential tool.

$ y = f(x) = 6\left(\dfrac{1}{3}\right)^x \\[5ex] \text{ When } x = 0 \\[3ex] y = 6\left(\dfrac{1}{3}\right)^0 \\[5ex] y = 6 * 1 \\[3ex] y = 6 \\[3ex] \text{Point 1: } (0, 6) \\[5ex] \text{ When } x = 1 \\[3ex] y = 6\left(\dfrac{1}{3}\right)^1 \\[5ex] y = 6 * \dfrac{1}{3} \\[5ex] y = 2 \\[3ex] \text{Point 2: } (1, 2) \\[3ex] $ The graph of the exponential function is:

$ \text{The perimeter of a rectangular garden is 48 meters.} \\[3ex] Perimeter = 2(length + width) \\[3ex] 2(y + x) = 48 \\[3ex] 2y + 2x = 48 \\[3ex] 2x + 2y = 48 ...eqn.(1) \\[5ex] \text{The length is 6 meters more than twice the width} \\[3ex] y = 6 + 2x \\[3ex] y = 2x + 6 ...eqn.(2) $

The graph has a positive slope.
So, Options A. and B. are discarded.

The graph has a y-intercept of 4
This implies that the graph intersects the y-axis at 4.
Option C is eliminated.

Option D. is the correct answer.

$ (2x + 1)^2 = 25 \\[3ex] 2x + 1 = \pm\sqrt{25} ...\text{Square Root Property} \\[3ex] 2x + 1 = \pm 5 \\[3ex] 2x = -1 \pm 5 \\[3ex] 2x = -1 + 5 \hspace{2em}\text{OR}\hspace{2em} 2x = -1 - 5 \\[3ex] 2x = 4 \hspace{2em}\text{OR}\hspace{2em} 2x = -6 \\[3ex] x = \dfrac{4}{2} \hspace{2em}\text{OR}\hspace{2em} x = -\dfrac{6}{2} \\[5ex] x = 2 \hspace{2em}\text{OR}\hspace{2em} x = -3 $

The domain is the set of all the input values, x for which the function has an output.
The closed circle represents a closed interval. It shows that the endpoint is included.

On the graph:
The closed circle on the left hand side is at x = 0
The closed circle on the right hand side is at x = 10

The interval notation is: [0, 10]
The set notation is: Domain = $\{x: 0 \le x \le 10 \}$

$ \underline{\text{Inequality 1}} \\[3ex] \text{This is a dashed line because of the absence of the equality sign} \\[3ex] y \lt x + 4 \\[3ex] \text{When } x = 0 \\[3ex] y \lt 0 + 4 \\[3ex] y \lt 4 \\[3ex] \text{When } y = 0 \\[3ex] 0 \lt x + 4 \\[3ex] x + 4 \gt 0 \\[3ex] x \gt 0 - 4 \\[3ex] x \gt -4 \\[5ex] \underline{\text{Inequality 2}} \\[3ex] \text{This is a solid line because of the presence of the equality sign} \\[3ex] y \le -2x - 2 \\[3ex] \text{When } x = 0 \\[3ex] y \le -2(0) - 2 \\[3ex] y \le 0 - 2 \\[3ex] y \le -2 \\[3ex] \text{When } y = 0 \\[3ex] 0 \le -2x - 2 \\[3ex] -2x - 2 \ge 0 \\[3ex] -2x \ge 0 + 2 \\[3ex] -2x \ge 2 \\[3ex] x \le \dfrac{2}{-2} \\[5ex] x \le -1 \\[3ex] $ To graph a system of inequalities, each inequality is shaded and the solution is where the shaded regions overlap.

$ f(x) = x^2 \\[3ex] g(x) = f(x - 6) + 2 \\[3ex] = (x - 6)^2 ...\text{Horizontal Shift 6 units right} \\[3ex] + 2 ...\text{Vertical Shift 2 units up} \\[3ex] $

$ a_n = 1.1a_{(n - 1)} \\[3ex] a_1 = 100 \\[5ex] a_2 = 1.1 * a_{2 - 1} \\[3ex] a_2 = 1.1 * a_1 \\[3ex] a_2 = 1.1 * 100 \\[3ex] a_2 = 110 \\[3ex] $ We now know the correct answer is Option D.
But if we want to continue, we can.

$ a_3 = 1.1 * a_2 \\[3ex] a_3 = 1.1 * 110 \\[3ex] a_3 = 121 \\[5ex] a_4 = 1.1 * a_3 \\[3ex] a_4 = 1.1 * 121 \\[3ex] a_4 = 133.1 $

$ \underline{\text{Equation of a Quadratic Function in Vertex Form}} \\[3ex] f(x) = a(x - h)^2 + k ...\text{where } (h, k) \text{ is the vertex} \\[3ex] (h, k) = (5, -1) \\[3ex] h = 5 \\[3ex] k = -1 \\[3ex] \implies \\[3ex] f(x) = a(x - 5)^2 + (-1) \\[3ex] f(x) = a(x - 5)^2 - 1 \\[3ex] \text{Passes through the point } (3, -17) \\[3ex] x = 3 \\[3ex] y = f(x) = -17 \\[3ex] \implies \\[3ex] -17 = a(3 - 5)^2 - 1 \\[3ex] -17 = a(-2)^2 - 1 \\[3ex] -17 + 1 = 4a \\[3ex] 4a = -16 \\[3ex] a = -\dfrac{16}{4} \\[5ex] a = -4 \\[3ex] \implies \\[3ex] f(x) = -4(x - 5)^2 - 1 \\[5ex] \underline{\text{Equation of a Quadratic Function in Standard Form}} \\[3ex] f(x) = -4(x - 5)(x - 5) - 1 \\[3ex] f(x) = -4(x^2 - 5x - 5x + 25) - 1 \\[3ex] f(x) = -4(x^2 - 10x + 25) - 1 \\[3ex] f(x) = -4x^2 + 40x - 100 - 1 \\[3ex] f(x) = -4x^2 + 40x - 101 $

$ \text{Point 1: } (-6, 1) \\[3ex] x_1 = -6 \\[3ex] y_1 = 1 \\[5ex] \text{Point 2: } (9, 6) \\[3ex] x_2 = 9 \\[3ex] y_2 = 6 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{6 - 1}{9 - (-6)} \\[5ex] m = \dfrac{5}{15} \\[5ex] m = \dfrac{1}{3} \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 1 = \dfrac{1}{3}[x - (-6)] \\[5ex] y - 1 = \dfrac{1}{3}(x + 6) \\[5ex] \text{Simplify} \\[3ex] y - 1 = \dfrac{1}{3}x + 2 \\[5ex] y = \dfrac{1}{3}x + 2 + 1 \\[5ex] y = \dfrac{1}{3}x + 3 \\[5ex] y = \dfrac{1}{3}(x + 9) $

$ \underline{\text{Exponential Function}} \\[3ex] y = ab^x \\[3ex] a = \text{initial value} \\[3ex] b = \text{base} \\[5ex] \text{Point 1: } (0, 72) \\[3ex] x = 0 \\[3ex] y = 72 \\[3ex] 72 = a * b^{0} \\[3ex] 72 = a * 1 \\[3ex] a = 72 \\[5ex] \text{Point 2: } (1, 43.2) \\[3ex] x = 1 \\[3ex] y = 43.2 \\[3ex] 43.2 = a * b^{1} \\[3ex] 43.2 = 72 * b \\[3ex] b = \dfrac{43.2}{72} \\[5ex] b = 0.6 \\[5ex] \implies \\[3ex] y = 72(0.6)^x $

$ 6x^2 + 3x - 9 \\[3ex] 3(2x^2 + x - 3)...\text{Option 2} \\[3ex] ............................................. \\[3ex] 2x^2 + x - 3 \\[3ex] 2x^2 * - 3 = -6x^2 \\[3ex] \text{Factors are: } 3x \text{ and } -2x \\[3ex] \implies \\[3ex] 2x^2 - 2x + 3x - 3 \\[3ex] 2x(x - 1) + 3(x - 1) \\[3ex] (x - 1)(2x + 3) \\[3ex] ............................................. \\[3ex] 3(2x + 3)(x - 1)...\text{Option 4} $

$ x + y = -8...eqn.(1) \\[3ex] 2x + 2y = 10...eqn.(2) \\[3ex] \text{From } eqn.(2) \\[3ex] 2(x + y) = 10 \\[3ex] x + y = \dfrac{10}{2} \\[5ex] x + y = 5 ...\text{modified form of } eqn.(2) \\[3ex] $ We notice that eqn.(1) and eqn.(2) are the same on the Left Hand Side (LHS) but give different results on the Right Hand Side (RHS).
This is an inconsistent system.
Hence, the system has no solution.

$ \dfrac{v - 6}{5} = \dfrac{v + 10}{2} \\[5ex] \text{Cross Multiply} \\[3ex] 2(v - 6) = 5(v + 10) \\[3ex] 2v - 12 = 5v + 50 \\[3ex] -12 - 50 = 5v - 2v \\[3ex] -62 = 3v \\[3ex] 3v = -62 \\[3ex] v = -\dfrac{62}{3} $

Each year the bank will add interest to te account, which will increase the account's value by 0.5%.
This implies that the interest is being compounded.
We can solve the question using at least three approaches.
Use any approach you prefer.

$ 0.5\% = \dfrac{0.5}{100} = 0.005 \\[5ex] \underline{\text{1st Approach: Exponential Growth Function}} \\[3ex] A = a(1 + r)^t \\[3ex] a = \text{initial value} = \$2500 \\[3ex] r = \text{growth rate} = 0.5\% = 0.005 \\[5ex] \implies \\[3ex] A = 2500(1 + 0.005)^t \\[3ex] A = 2500(1.005)^t \\[5ex] \underline{\text{2nd Approach: Exponential Function}} \\[3ex] A = ab^t \\[3ex] a = \text{initial value} = \$2500 \\[3ex] b = \text{base} \\[5ex] \text{Point 1: } (0, 2500) \\[3ex] t = 0 \\[3ex] A = 2500 \\[3ex] 2500 = a * b^{0} \\[3ex] 2500 = a * 1 \\[3ex] a = 2500 \\[5ex] \text{After 1 year} \\[3ex] \text{Interest, } I = 0.005 * 2500 = 12.5 \\[3ex] A = 2500 + 12.5 = 2512.5 \\[5ex] \text{Point 2: } (1, 2512.5) \\[3ex] t = 1 \\[3ex] A = 2512.5 \\[3ex] 2512.5 = a * b^{1} \\[3ex] 2512.5 = 2500 * b \\[3ex] b = \dfrac{2512.5}{2500} \\[5ex] b = 1.005 \\[5ex] \implies \\[3ex] A = 2500(1.005)^t \\[5ex] \underline{\text{3rd Approach: Compound Interest Formula}} \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] \text{Where} \\[3ex] \text{Principal, } P = \$2500 \\[3ex] \text{Time} = t \\[3ex] \text{Amount} = A \\[3ex] \text{Interest rate, } r = 0.5\% = 0.005 \\[3ex] \text{Number of compounding periods per year, } m = 1 \text{ (Compounded annually)} \\[3ex] A = 2500\left(1 + \dfrac{0.005}{1}\right)^{1 * t} \\[5ex] A = 2500(1 + 0.005)^t \\[3ex] A = 2500(1.005)^t $

$ \dfrac{(6y^3)^{-2}}{y} \\[5ex] \dfrac{6^{-2} * y^{3 * -2}}{y}...Law\;5 Exp \\[5ex] 6^{-2} * y^{-6} * \dfrac{1}{y} \\[5ex] \dfrac{1}{6^2} * \dfrac{1}{y^6} * \dfrac{1}{y^1} ...Law\;6 Exp \\[5ex] \dfrac{1}{36} * \dfrac{1}{y^{6 + 1}} ...\text{Law 1 Exp} \\[5ex] \dfrac{1}{36 * y^7} \\[5ex] \dfrac{1}{36y^7} $

$ \underline{\text{Exponential Function}} \\[3ex] y = ab^x \\[3ex] a = \text{initial value} \\[3ex] b = \text{base} \\[5ex] \underline{\text{Exponential Decay Function}} \\[3ex] \text{This is because } 0.85 \lt 1 \\[3ex] y = a(1 - r)^x \\[3ex] r = \text{decay rate} \\[5ex] \underline{\text{Compare}} \\[3ex] P(t) = 26,080(0.85)^t \\[3ex] \text{initial population} = 26080 \text{ people} \\[3ex] 1 - r = 0.85 \\[3ex] 1 - 0.85 = r \\[3ex] r = 0.15 \\[3ex] r = 0.15 * 100 = 15\% \\[3ex] $ The population is decreasing at a rate of 15% per year.

The rate of change of the student's distance from home in miles with respect to the time in minutes is the slope of the graph.

$ \text{Point 1: } (0, 6) \\[3ex] x_1 = 0 \\[3ex] y_1 = 6 \\[5ex] \text{Point 2: } (8, 0) \\[3ex] x_2 = 8 \\[3ex] y_2 = 0 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{0 - 6}{8 - 0} \\[5ex] m = -\dfrac{6}{8} \\[5ex] m = -\dfrac{3}{4} \\[5ex] $ This is a decrease of $\dfrac{3}{4}$ miles per minute.

We can solve the question using at least two approaches.
Use any approach you prefer.

$ a = \text{leading coefficient} \\[5ex] \underline{\text{1st Approach: Equation of a Quadratic Function in Factored Form}} \\[3ex] f(x) = a(x - \alpha)(x - \beta) \\[3ex] \alpha, \beta = \text{zeros of the function} \\[3ex] \alpha = -5 \\[3ex] \beta = -3 \\[5ex] \implies \\[3ex] h(x) = a[x - (-5)][x - (-3)] \\[3ex] h(x) = a(x + 5)(x + 3) \\[5ex] \text{Passes through the point (vertex) } (-4, 1) \\[3ex] x = -4 \\[3ex] h(x) = 1 \\[3ex] \implies \\[3ex] 1 = a(-4 + 5)(-4 + 3) \\[3ex] 1 = a(1)(-1) \\[3ex] -a = 1 \\[3ex] a = \dfrac{1}{-1} = -1 \\[5ex] \implies \\[3ex] h(x) = -1(x + 5)(x + 3) \\[3ex] h(x) = -1(x^2 + 3x + 5x + 15) \\[3ex] h(x) = -1(x^2 + 8x + 15) \\[3ex] h(x) = -x^2 - 8x - 15 \\[5ex] \underline{\text{2nd Approach: Equation of a Quadratic Function in Vertex Form}} \\[3ex] h(x) = a(x - h)^2 + k ...\text{where } (h, k) \text{ is the vertex} \\[3ex] (h, k) = (-4, 1) \\[3ex] h = -4 \\[3ex] k = 1 \\[3ex] \implies \\[3ex] h(x) = a[x - (-4)]^2 + 1 \\[3ex] h(x) = a(x + 4)^2 + 1 \\[5ex] \text{Passes through the point: } (-3, 0) \\[3ex] x = -3 \\[3ex] h(x) = 0 \\[3ex] \implies \\[3ex] 0 = a(-3 + 4)^2 + 1 \\[3ex] 0 = a(1)^2 + 1 \\[3ex] 0 = a(1) + 1 \\[3ex] a + 1 = 0 \\[3ex] a = -1 \\[3ex] \implies \\[3ex] h(x) = -1(x + 4)^2 + 1 \\[3ex] h(x) = -1(x + 4)(x + 4) + 1 \\[3ex] h(x) = -1(x^2 + 4x + 4x + 16) + 1 \\[3ex] h(x) = -1(x^2 + 8x + 16) + 1 \\[3ex] h(x) = -x^2 - 8x - 16 + 1 \\[3ex] h(x) = -x^2 - 8x - 15 $

$ y \propto x \\[3ex] y = kx ...k\text{ is the constant of variation} \\[3ex] -12 = k * 4 \\[3ex] 4k = -12 \\[3ex] k = -\dfrac{12}{4} \\[5ex] k = -3 \\[5ex] \implies \\[3ex] y = -3x \\[3ex] x = -6 \\[3ex] y = -3(-6) \\[3ex] y = 18 $

The real zeros of a quadratic function are the x-values where the graph intersects (crosses or touches) the x-axis.
The only graph that does not intersect the x-axis is the graph in Option A.

$ \underline{\text{Exponential Function}} \\[3ex] A(t) = 1,550(1.02)^t \\[3ex] \text{When } t = 0 \\[3ex] A(0) = 1,550(1.02)^0 \\[3ex] A(0) = 1550 * 1 \\[3ex] A(0) = \$1550 \\[3ex] $ The initial deposit in the investment account was $1,550
The correct answer is Option C.

Student: Mr. C
Could you analyze the other options?
Teacher: Sure, let's do it.

$ \text{When } t = 1 \\[3ex] A(0) = 1,550(1.02)^1 \\[3ex] A(0) = 1550 * 1.02 \\[3ex] A(0) = \$1581 \\[3ex] $ The amount in the investment account after 1 year is $1,581

$ \underline{\text{Exponential Growth Function}} \\[3ex] \text{This is because } 1.02 \gt 1 \\[3ex] A(t) = P(1 + r)^t \\[3ex] P = \text{initial deposit} \\[3ex] r = \text{interest rate or the growth rate} \\[3ex] \text{Compare to } A(t) = 1,550(1.02)^t \\[3ex] 1 + r = 1.02 \\[3ex] r = 1.02 - 1 \\[3ex] r = 0.02 = 2\% \\[3ex] $ The amount of money in the investment account increases by 2% each year.

$ \dfrac{1}{3}(6x^2y)^2(2x^3y^4) \\[5ex] \text{DISSOCIATE} \\[3ex] \dfrac{1}{3} * 6^2 * x^{2 * 2} * y^2 * 2 * x^3 * y^4 \\[5ex] \text{SOLVE} \\[3ex] \dfrac{1}{3} * 36 * x^4 * y^2 * 2 * x^3 * y^4 \\[5ex] \dfrac{1}{3} * 36 * 2 * x^4 * x^3 * y^2 * y^4 \\[5ex] 24 * x^{4 + 3} * y^{2 + 4} \\[3ex] 24 * x^7 * y^6 \\[3ex] \text{ASSOCIATE} \\[3ex] 24x^7y^6 $

The rate of change of y with respect to x for the equation $5x - 8y = 40$ is the slope of the equation.
Express the equation in Slope–Intercept Form
Find the slope.

$ \underline{\text{Slope–Intercept Form of the Equation of a Straight Line}} \\[3ex] y = mx + b ...\text{where } m = slope \\[5ex] 5x - 8y = 40 \\[3ex] 5x - 40 = 8y \\[3ex] 8y = 5x - 40 \\[3ex] y = \dfrac{5x - 40}{8} \\[5ex] y = \dfrac{5}{8}x - \dfrac{40}{8} \\[5ex] m = \dfrac{5}{8} $

$ 47, 33, 19, 5, -9 \\[3ex] 33 - 47 = -14 \\[3ex] 19 - 33 = -14 \\[3ex] 5 - 19 = -14 \\[3ex] -9 - 5 = -14 \\[3ex] $ This is an Arithmetic Sequence because the common difference is –14
We can solve the question using at least two approaches.
Use any approach you prefer.

$ \underline{\text{1st Approach: nth term of an Arithmetic Sequence}} \\[3ex] AS_n = a + d(n - 1) \\[3ex] AS_n = \text{nth term} = a_n \\[3ex] a = \text{first term} = 47 \\[3ex] d = \text{common difference} = -14 \\[3ex] n = \text{number of terms} \\[3ex] \implies \\[3ex] a_n = 47 + -14(n - 1) \\[3ex] a_n = 47 - 14(n - 1) \\[3ex] a_n = 47 - 14n + 14 \\[3ex] a_n = -14n + 61 \\[3ex] $ For the 2nd Approach, let us represent the information as a linear function.
An Arithmetic Sequence is also known as a Linear Sequence.

Linear Function

x = $n$	y = $a_n$
1	47
2	33
3	19
4	5
5	–9

$ \underline{\text{2nd Approach: Linear Function in Slope–Intercept Form}} \\[3ex] \text{Point 1: } (1, 47) \\[3ex] x_1 = 1 \\[3ex] y_1 = 47 \\[5ex] \text{Point 2: } (2, 33) \\[3ex] x_2 = 2 \\[3ex] y_2 = 33 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{33 - 47}{2 - 1} \\[5ex] m = \dfrac{-14}{1} \\[5ex] m = -14 \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 47 = -14(x - 1) \\[5ex] \text{Slope–Intercept Form} \\[3ex] y - 47 = -14x + 14 \\[3ex] y = -14x + 14 + 47 \\[3ex] y = -14x + 61 \\[3ex] a_n = -14n + 61 $

The function is a quadratic function.
Let us use the Vertex Method.
The y-coordinate of the vertex is the maximum or the minimum value of the function.
This value helps us determine the range of the function.

Another important information we need for this question is the leading coefficient.
If the leading coefficient is negative, the quadratic function has a maximum value. It opens downwards.
If the leading coefficient is positive, the quadratic function has a minimum value. It opens upwards.

$ f(x) = -x^2 + 60x \\[3ex] \text{Compare to the standard form: } ax^2 + bx + c \\[3ex] \text{The leading coefficient, } a = -1 \\[3ex] \text{Parabola opens downwards, it has a maximum value.} \\[3ex] b = 60 \\[3ex] x-coordinate \text{ of the vertex} = -\dfrac{b}{2a} = -\dfrac{60}{2(-1)} = \dfrac{-60}{-2} = 30 \\[5ex] \text{maximum value of } y = y-coordinate \text{ of the vertex} = f(30) \\[3ex] = -(30)^2 + 60(30) \\[3ex] \\[3ex] = -900 + 1800 \\[3ex] = 900 \\[3ex] \implies \\[3ex] \text{Range} = \{y: y \le 900\} \\[3ex] $ The range is all real numbers that are less than or equal to 900.

Assume:
y = overall grade average
x = number of hours worked
Just as y depends on x,
The overall grade average depends on the number of hours worked.
The question we should ask ourselves is this:
How does y change when x changes?
How does the overall grade average change when the number of hours worked changes?
Let us observe the pattern by arranging the values of x in ascending order and including the corresponding values of y
The modified table is:

Weekly Hours and Grade Average

Hours Worked, x	Overall Grade Average, y
12	94
15	86
16	87
20	90
25	83
27	77
30	72

As we notice:
As the number of hours worked increases, the overall grade average decreases, then increases slightly, then decreases.
Generally speaking, as the number of hours worked increases, the overall grade average decreases.
This is a Strong Negative association.
We note that these points (16, 87) to (20, 90) is a small increase, however, the overall pattern is strongly negative.
Correlation looks at the overall trend, not whether every single point moves perfectly downward.
One or two slight deviations do not change the overall trend.

Let us analyze each option to convince you that the correct answer is Option C.

$ \underline{\text{Option A.}} \\[3ex] y = 16(0.5)^x \\[3ex] \text{When } x = 0 \\[3ex] y = 16(0.5)^0 \\[3ex] y = 16(1) \\[3ex] y = 16 \\[3ex] y-intercept = (0, 16) \\[3ex] TRUE \\[5ex] \underline{\text{Option B.}} \\[3ex] y = 16(0.5)^x \\[3ex] \text{Compare to: } y = ab^x \\[3ex] b = 0.5 \\[3ex] $ 0.5 < 1
This implies that the exponential function is an exponential decay function \\[3ex] The graph is decreasing for all values of x.
TRUE

$ \underline{\text{Option C.}} \\[3ex] y = 16(0.5)^x \\[3ex] \text{When } y = 0 \\[3ex] 0 = 16(0.5)^x \\[3ex] 16(0.5)^x = 0 \\[3ex] 0.5^x = \dfrac{0}{16} \\[5ex] 0.5^x = 0 \\[3ex] \log 0.5^x = \log 0 \\[3ex] x\log 0.5 = \log 0 \\[3ex] $ But, $\log 0$ does not exist.
Hence, we cannot find any value of x that will make y to be 0.
There is no x-intercept.
Generally speaking, a basic exponential function does not have an x-intercept.
Unless the function is shifted downward (transformed exponential function), a basic exponential function does not cross the x‑axis.
So, saying that the x-intercept is (0.5, 0) is NOT TRUE.

Option D.
The horizontal asymptote is: y = 0
This is because the graph approaches the x-axis as the input increases without bound.
The function keeps decreasing (approaches but does not touch the x-axis) as the input increases.
TRUE.

Because the inequality contains an equal sign, the line will be a solid line.

$ y \ge \dfrac{1}{2}x - 2 \\[5ex] \text{When } x = 0 \\[3ex] y \ge \dfrac{1}{2}(0) - 2 \\[5ex] y \ge 0 - 2 \\[3ex] y \ge -2 \\[5ex] \text{When } x = 2 \\[3ex] y \ge \dfrac{1}{2}(2) - 2 \\[5ex] y \ge 1 - 2 \\[3ex] y \ge -1 \\[5ex] $ Student: Mr. C, wait a minute.
Why did you not do when $x = 1$?
Is there any reason for using $x = 2$?
Teacher: Good observation.
If we used 1, we would end up with a fraction.
Fractions are not allowed in the graph that we are given.
Hence, it is necessary that we use a value of $x$ that will give an integer value of $y$ within the domain [-4, 4]
What other values could we have used?
Student: 4, 6, 8, 10, 12, –2, –4
–2
Teacher: There you go...

The graph of the inequality is

Let the:
cost of 1 cup of coffee = c
cost of 1 doughout = d

$4.90 for 1 cup of coffee and 2 doughouts implies
c + 2d = $4.90 ...equation (1)

$8.60 for 2 cups of coffee and 3 doughouts implies
2c + 3d = $8.60 ...equation (2)

Based on the question, we only need to find c
Let us use the Elimination Method.
To find c, eliminate d

$ c + 2d = 4.9...eqn.(1) \\[3ex] 2c + 3d = 8.6...eqn.(2) \\[5ex] eqn.(1) * 3 \implies 3(c + 2d) = 3(4.9) \\[3ex] 3c + 6d = 14.7...eqn.(3) \\[5ex] eqn.(2) * 2 \implies 2(2c + 3d) = 2(8.6) \\[3ex] 4c + 6d = 17.2 ...eqn.(4) \\[5ex] eqn.(4) - eqn.(3) \implies \\[3ex] (4c + 6d) - (3c + 6d) = 17.2 - 14.7 \\[3ex] 4c + 6d - 3c - 6d = 2.5 \\[3ex] c = 2.5 \\[3ex] $ The cost of a cup of coffee is $2.50

$ \text{Point 1: } (-3, 0) \\[3ex] x_1 = -3 \\[3ex] y_1 = 0 \\[5ex] \text{Point 2: } (0, -5) \\[3ex] x_2 = 0 \\[3ex] y_2 = -5 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-5 - 0}{0 - (-3)} \\[5ex] m = -\dfrac{5}{3} \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 0 = -\dfrac{5}{3}[x - (-3)] \\[5ex] y = -\dfrac{5}{3}(x + 3) \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = mx + b ...\text{where } b \text{ is the y-intercept} \\[3ex] y = -\dfrac{5}{3}x - 5 $

The number of hours the two hikers have been hiking when they are the same distance from the start of the trail is the point of intersection of the two graphs.
As is observed, the point of intersection is between 1 hour and 2 hours.
This is about 1.3 hours.

$ f(x) = 6x^2 - 23x + 21 \\[3ex] 6x^2 * 21 = 126x^2 \\[3ex] \text{Factors are: } -9x \text{ and } -14x \\[3ex] f(x) = 6x^2 - 9x - 14x + 21 \\[3ex] f(x) = 3x(2x - 3) - 7(2x - 3) \\[3ex] f(x) = (2x - 3)(3x - 7) $

Option 2: The domain is the set of all real numbers greater than or equal to 2 and less than or equal to 5.

Option 4: The range is the set of all real numbers greater than or equal to 1 and less than or equal to 8.

$ g(x) = 3x^2 - 2x - 5 \\[3ex] g(x) = 0 \\[3ex] \implies \\[3ex] 3x^2 - 2x - 5 = 0 \\[3ex] 3x^2 * -5 = -15x^2 \\[3ex] \text{Factors are: } 3x \text{ and } -5x \\[3ex] 3x^2 + 3x - 5x - 5 = 0 \\[3ex] 3x(x + 1) - 5(x + 1) = 0 \\[3ex] (x + 1)(3x - 5) = 0 \\[3ex] x + 1 = 0 \hspace{1em}OR\hspace{1em} 3x - 5 = 0...\text{Zero Product Property} \\[3ex] x = -1 \hspace{1em}OR\hspace{1em} 3x = 5 \\[3ex] x = -1 \hspace{1em}OR\hspace{1em} x = \dfrac{5}{3} $

$ \underline{\text{Exponential Function}} \\[3ex] y = ab^x \\[3ex] y = g(x) \\[3ex] a = \text{initial value} \\[3ex] b = \text{base} \\[5ex] \text{Point 1: } (1, 48) \\[3ex] x = 1 \\[3ex] g(x) = 48 \\[3ex] 48 = a * b^{1} \\[3ex] 48 = ab \\[3ex] ab = 48...eqn.(1) \\[5ex] \text{Point 2: } (2, 72) \\[3ex] x = 2 \\[3ex] g(x) = 72 \\[3ex] 72 = a * b^{2} \\[3ex] 72 = ab^2 \\[3ex] ab^2 = 72...eqn.(2) \\[5ex] eqn.(2) \div eqn.(1) \implies \\[3ex] \dfrac{ab^2}{ab} = \dfrac{72}{48} \\[5ex] b = \dfrac{3}{2} \\[5ex] \text{From } eqn.(1) \\[3ex] a = \dfrac{48}{b} \\[5ex] a = 48 \div \dfrac{3}{2} \\[5ex] a = 48 * \dfrac{2}{3} \\[5ex] a = 32 \\[5ex] \implies \\[3ex] g(x) = 32\left(\dfrac{3}{2}\right)^x $

Let us analyze each option.
A. When the number of spectators at a football game increases, the number of wins for the home team increases.
The number of spectators at a football game does not really cause win for the home team.
What if the majority of the spectators do not want the home team to win?
This is not a causation.

B. When more gallons of water are put into an aquarium, the weight of the aquarium increases.
More gallons of water in an aquarium does increases the weight of the aquarium because of the weight of the weight.
This is a causation.

C. When the amount of time an athlete runs increases, the distance she runs increases.
If I am asked to run for 5 minutes, I shall cover some meters.
If I am asked to run for 10 minutes, I shall cover more meters.
So, the extended time caused more distance to be covered.
This is a causation.

D. When the charge of a cell phone battery increases, the number of minutes the phone can be used increases.
Ceteris paribus:
When a cell phone battery is at 50%, the phone can be used for some number of minutes.
When the cell phone battery is fully charged (at 100%), the phone can be used for an extended number of minutes.
This implies that the percent of charge in a cell phone causes the phone to be used for a certain number of minutes.
This is a causation.

Observing the two graphs from the graph of $f(x)$ to the graph of $g(x)$, we notice a shift to the left, a stretch, and a shift downwards. $f(x) = x^2$
Vertex of $f(x) = (0, 0)$

Vertex of $g(x) = (-3, -2)$
1st transformation:
Horizontal Shift of 3 units to the left: $(0, 0) \rightarrow (0 - 3, 0) \rightarrow (-3, 0)$
Transformed function: $f(x + 3)$

2nd transformation:
We notice a vertical stretch: a narrowing of $f(x)$ to give $g(x)$
Based on the options, we see that the y-coordinate is multiplied by 2
Vertical Stretch by a factor of 2 units: $(-3, 0) \rightarrow (-3, 0 * 2) \rightarrow (-3, 0)$
Transformed function: $2f(x + 3)$

3rd transformation:
Vertical Shift of 2 units down: $(-3, 0) \rightarrow (-3, 0 - 2) \rightarrow (-3, -2)$
Transformed function: $2f(x + 3) - 2$

$g(x) = 2f(x + 3) - 2$

$ 2x + 3y = 6 \\[3ex] 3y = 6 - 2x \\[3ex] y = \dfrac{6 - 2x}{3} \\[5ex] \text{When } x = 0 \\[3ex] y = \dfrac{6 - 2(0)}{3} \\[5ex] y = \dfrac{6 - 0}{3} \\[5ex] y = \dfrac{6}{3} \\[5ex] y = 2 \\[3ex] \text{Point 1: } (0, 2) \\[5ex] \text{When } x = 3 \\[3ex] y = \dfrac{6 - 2(3)}{3} \\[5ex] y = \dfrac{6 - 6}{3} \\[5ex] y = \dfrac{0}{3} \\[5ex] y = 0 \\[3ex] \text{Point 2: } (3, 0) \\[3ex] $ The graph of the equation is:

The cost in dollars to rent the pool is a function of the number of hours the pool is rented.
This means that the:
cost in dollars to rent the pool is the dependent variable (output) = y
number of hours the pool is rented is the independent variable (input) = x

The pool can be rented for either 1 hour or 2 hours.
It can be rented for $110 per hour.
cost for 1 hour @ $110 per hour = 1(110) = $110
cost for 2 hours @ $110 per hour = 2(110) = $220

B. Domain: {1, 2}
  Range: {110, 220}

$ \dfrac{2}{3}(m - 9) = \dfrac{1}{3}(m - 27) \\[5ex] \dfrac{2(m - 9)}{3} = \dfrac{1(m - 27)}{3} \\[5ex] $ There is a single term on each side (one term on the Left Hand Side and one term on the Right Hand Side)
The denominators are the same.
So, we just equate the numerators. This is similar to multiplying each term by the denominator.

$ \text{single term on each side, same denominator; equate the numerators} \\[3ex] 2(m - 9) = 1(m - 27) \\[3ex] 2m - 18 = m - 27 \\[3ex] 2m - m = -27 + 18 \\[3ex] m = -9 $

The function has a minimum value of –4

One family paid $8.00 for 2 hot dogs and 3 bags of popcorn.

$ 8 = 2x + 3y \\[3ex] 2x + 3y = 8...eqn.(1) \\[3ex] $ Another family paid $16.50 for 5 hot dogs and 4 bags of popcorn.

$ 16.5 = 5x + 4y \\[3ex] 5x + 4y = 16.5...eqn.(2) $

The y-intercept is 2.
This implies that one of the points on the graph is (0, 2)
The graph must pass through (0, 2)
Options A. and B. are eliminated.
Options C. and D. remain.

The function increases at a rate of 50%.
This is an exponential growth function.
Option D. is an exponential decay function. It is eliminated.
Option C. is the correct answer.

$ \underline{\text{Line P}} \\[3ex] \text{Point 1: } (-5, -2) \\[3ex] x_1 = -5 \\[3ex] y_1 = -2 \\[5ex] \text{Point 2: } (5, 6) \\[3ex] x_2 = 5 \\[3ex] y_2 = 6 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{6 - (-2)}{5 - (-5)} \\[5ex] m = \dfrac{6 + 2}{5 + 5} \\[5ex] m = \dfrac{8}{10} \\[5ex] m = \dfrac{4}{5} \\[5ex] \underline{\text{Line W}} \\[3ex] \text{Line W } || \text{ Line P} \\[3ex] m = \dfrac{4}{5}...\text{parallel lines have the same slope} \\[5ex] \text{Passes through Point } (1, 5) \\[3ex] x_3 = 1 \\[3ex] y_3 = 5 \\[3ex] \text{Point–Slope Form} \\[3ex] y - y_3 = m(x - x_3) \\[3ex] y - 5 = \dfrac{4}{5}(x - 1) $

$ f(x) = x^2...\text{parent function} \\[3ex] \text{Vertex} = (0, 0) \\[5ex] g(x) = f(x + 7.2)...\text{child (transformed) function} \\[3ex] g(x) = (x + 7.2)^2 \\[3ex] $ Transformation from $f(x)$ to $g(x)$:
Horizontal Shift of $f(x)$ 7.2 units left
This implies that:

$ (0, 0) \rightarrow (0 - 7.2, 0) \\[3ex] \rightarrow (-7.2, 0) \\[3ex] \text{Vertex} = (-7.2, 0) $

The rate of change of the altitude of the airplane with respect to time is the slope.
Observing the pattern, as the time increases, the altitude decreases.
We should expect a negative slope.

$ \text{Point 1: } (1.5, 25500) \\[3ex] x_1 = 1.5 \\[3ex] y_1 = 25500 \\[5ex] \text{Point 2: } (6, 12000) \\[3ex] x_2 = 6 \\[3ex] y_2 = 12000 \\[5ex] \text{Slope} = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{12000 - 25500}{6 - 1.5} \\[5ex] = -\dfrac{13500}{4.5} \\[5ex] = -3000 \\[3ex] $ The altitude of the airplane decreases at a rate of 3000 feet per minute.

$ 5(x + 3)^2 = 75 \\[3ex] (x + 3)^2 = \dfrac{75}{5} \\[5ex] (x + 3)^2 = 15 \\[3ex] \underline{\text{Square Root Property}} \\[3ex] x + 3 = \pm\sqrt{15} \\[3ex] x = -3 \pm \sqrt{15} $

The domain is the set of all the input values, x for which the function has an output.
The closed circle represents a closed interval.
On the graph,
It is on the left hand side at x = –2
It is on the right hand side at x = 3

So, the interval notation is: $[-2, 3]$
This represents all real numbers between –2 and 3 (both ends included).
This means all real numbers greater than or equal to –2 and less than or equal to 3.

$ \dfrac{16k^6m^7}{4k^3m^{14}} \\[5ex] \text{DISSOCIATE} \\[3ex] \dfrac{16 * k^6 * m^7}{4 * k^3 * m^{14}} \\[5ex] \dfrac{16}{4} * \dfrac{k^9}{k^3} * \dfrac{m^7}{m^{14}} \\[5ex] \text{SOLVE} \\[3ex] 4 * k^{9 - 3} * \dfrac{1}{m^{14 - 7}}...\text{Laws 1 and 2 Exp} \\[5ex] 4 * k^6 * \dfrac{1}{m^7} \\[5ex] \text{ASSOCIATE} \\[3ex] \dfrac{4k^6}{m^7} $

$ \underline{\text{Zeros}} \\[3ex] 1st:\;\; x = -\dfrac{3}{2} \\[5ex] 2nd:\;\; x = 5 \\[5ex] \underline{\text{Equate to number zero}} \\[3ex] 1st:\;\; 2x = -3 \\[3ex] 2x + 3 = 0 \\[3ex] 2nd:\;\;x - 5 = 0 \\[5ex] \underline{\text{Factors}} \\[3ex] 1st:\;\; 2x + 3 \\[3ex] 2nd:\;\; x - 5 \\[5ex] \underline{\text{Factored Form}} \\[3ex] (2x + 3)(x - 5) $

$ \text{Perimeter} \\[3ex] = \left(\dfrac{3}{2}x + \dfrac{5}{6}\right) + \left(3x - \dfrac{1}{6}\right) + \left(2x + \dfrac{1}{3}\right) \\[5ex] = \dfrac{3}{2}x + \dfrac{5}{6} + 3x - \dfrac{1}{6} + 2x + \dfrac{1}{3} \\[5ex] = \dfrac{3}{2}x + 3x + 2x + \dfrac{5}{6} - \dfrac{1}{6} + \dfrac{1}{3} \\[5ex] = \dfrac{3x}{2} + 5x + \dfrac{4}{6} + \dfrac{1}{3} \\[5ex] = \dfrac{3x + 10x}{2} + \dfrac{4 + 2}{6} \\[5ex] = \dfrac{13x}{2} + \dfrac{6}{6} \\[5ex] = \dfrac{13x}{2} + 1 $

Let the:
flat fee = p
cost per ticket = c

$ y = p + cx \\[3ex] $ One customer purchased 4 tickets for a total cost of $160.

$ x = 4 \\[3ex] y = 160 \\[3ex] 160 = p + 4c ...eqn.(1) \\[3ex] $ Another customer purchased 8 tickets for a total cost of $300.

$ x = 8 \\[3ex] y = 300 \\[3ex] 300 = p + 8c ...eqn.(2) \\[5ex] eqn.(2) - eqn.(1) \implies \\[3ex] (300 - 160) = (p + 8c) - (p + 4c) \\[3ex] 140 = p + 8c - p - 4c \\[3ex] 4c = 140 \\[3ex] c = \dfrac{140}{4} \\[5ex] c = 35 \\[3ex] \text{Substitute the value of c in eqn.(1)} \\[3ex] 160 = p + 4(35) \\[3ex] p + 140 = 160 \\[3ex] p = 160 - 140 \\[3ex] p = 20 \\[3ex] \implies \\[3ex] y = 20 + 35x \\[3ex] y = 35x + 20 \\[3ex] $ Student: SamDom For Peace
Teacher: That's me.
What's good?
Student: What if we decide to check each option and eliminate the incorrect ones until we get the correct one?
Teacher: Sure, we can do that.

$ \underline{\text{Testing: Trial and Error Approach}} \\[3ex] \text{Option A} \\[3ex] y = 40x \\[3ex] \text{1st Customer} \\[3ex] x = 4 \\[3ex] y = 160 \\[3ex] 160 = 40(4) \checkmark \\[5ex] \text{2nd Customer} \\[3ex] x = 8 \\[3ex] y = 300 \\[3ex] 300 \ne 40(8)...\text{Discard} \\[5ex] \text{NEXT} \\[3ex] \text{Option B} \\[3ex] y = 37.5x \\[3ex] \text{1st Customer} \\[3ex] x = 4 \\[3ex] y = 160 \\[3ex] 160 \ne 37.5(4) ...\text{Discard} \\[5ex] \text{NEXT} \\[3ex] \text{Option C} \\[3ex] y = 35x + 20 \\[3ex] \text{1st Customer} \\[3ex] x = 4 \\[3ex] y = 160 \\[3ex] 160 = 35(4) + 20 \checkmark \\[5ex] \text{2nd Customer} \\[3ex] x = 8 \\[3ex] y = 300 \\[3ex] 300 = 35(8) + 20 \checkmark \\[5ex] \text{STOP} $

$ x^2 + 7x - 30 \\[3ex] (x + 10)(x - 3) $

$ f(x) = x \\[3ex] g(x) = \dfrac{1}{2}f(x) + 3 \\[5ex] $ Transformation of $f(x)$ to give $g(x)$

1st: Vertical Compression of $f(x)$ by a factor of $\dfrac{1}{2}$ unit
This makes the graph of $g(x)$ to be less than steep than that of $f(x)$

2nd: Vertical Shift of $f(x)$ 3 units up
This moves up the modified graph of $f(x)$ by 3 units.
In other words, all y-coordinates increase by 3 units.

The graph of g is less steep than the graph of f and has a y-intercept that is greater than that of f.

$ \text{slope} = m \\[3ex] \text{y-intercept} = b \\[5ex] \underline{\text{Line } n} \\[3ex] \text{Point 1: } (0, -2) \\[3ex] x_1 = 0 \\[3ex] y_1 = -2 \\[5ex] \text{Point 2: } (1, 1) \\[3ex] x_2 = 1 \\[3ex] y_2 = 1 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{1 - (-2)}{1 - 0} \\[5ex] m = \dfrac{3}{1} \\[5ex] m = 3 \\[5ex] b = -2 \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = mx + b \\[3ex] y = 3x - 2 \\[5ex] \text{Standard Form} \\[3ex] 2 = 3x - y \\[3ex] 3x - y = 2 \\[5ex] \underline{\text{Line } p} \\[3ex] \text{Point 1: } (-1, 6) \\[3ex] x_1 = -1 \\[3ex] y_1 = 6 \\[5ex] \text{Point 2: } (2, 0) \\[3ex] x_2 = 2 \\[3ex] y_2 = 0 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{0 - 6}{2 - (-1)} \\[5ex] m = \dfrac{-6}{3} \\[5ex] m = -2 \\[5ex] b = 4 \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = mx + b \\[3ex] y = -2x + 4 \\[5ex] \text{Standard Form} \\[3ex] y + 2x = 4 \\[3ex] 2x + y = 4 $

$ f(x) = 4(x - 3)^2 + 5 \\[3ex] .............................................. \\[3ex] (x - 3)^2 \\[3ex] (x - 3)(x - 3) \\[3ex] x^2 - 3x - 3x + 9 \\[3ex] x^2 - 6x + 9 \\[3ex] .............................................. \\[3ex] = 4(x^2 - 6x + 9) + 5 \\[3ex] = 4x^2 - 24x + 36 + 5 \\[3ex] = 4x^2 - 24x + 41 $

$ x + 2y \ge 2 \\[3ex] \text{When } x = 0 \\[3ex] 0 + 2y \ge 2 \\[3ex] 2y \ge 2 \\[3ex] y \ge \dfrac{2}{2} \\[5ex] y \ge 1 ...\text{Mark point } (0, 1) \\[3ex] \text{The shaded area is upwards from } y = 1 \\[5ex] \text{When } y = 0 \\[3ex] x + 2(0) \ge 2 \\[3ex] x + 0 \ge 2 \\[3ex] x \ge 2 ...\text{Mark point } (2, 0) \\[3ex] \text{The shaded area is to the right of } x = 2 \\[3ex] $ Draw a straight solid line (because of the equal sign) connecting the two points: (0, 1) and (2, 0)
Locate the intersection of the two areas.
The correct answer is Option A.

$ 6x - 2y = 10...eqn.(1) \\[3ex] y = 3x - 5...eqn.(2) \\[3ex] \text{Substitute eqn.(2) into eqn.(1) for } y \\[3ex] 6x - 2(3x - 5) = 10 \\[3ex] 6x - 6x + 10 = 10 \\[3ex] 10 = 10 \\[3ex] $ The system has infinitely many solutions.

$ \underline{\text{Exponential Growth Function}} \\[3ex] y = a(1 + r)^x \\[3ex] a = \text{initial value} = 500 \\[3ex] r = \text{growth rate} = 2\% = \dfrac{2}{100} = 0.02 \\[5ex] y = 500(1 + 0.02)^x \\[3ex] y = 500(1.02)^x \\[3ex] y = 500 \cdot 1.02^x $

A function is a relation that has a unique output.
This means that an input can only one output, two inputs can have the same output, however, an inout cannot have more than one output.

Let us analyze each option and discard the incorrect ones.
Option A.
The input, –2 has two different outputs, –4 and 4.
This is not a function.

Option B.
Vertical Line Test: A vertical line through the graph intersects it at more than one point.
This implies that at least one input, say –3 has two different outputs, 4 and –4
This is not a function.

Option C.
Each input has a different output.
This is a function.
It is a one-to-one (injective) function.

Option D.
Vertical Line Test: A vertical line through the graph intersects it at more than one point.
This implies that at least one input, say –3 has two different outputs, 2 and –2
This is not a function.

Because the second equation has a different slope than the first equation, this implies that the two equations will intersect at a point.
Hence, the system has exactly one solution.

$ 7w - 2(w - 9) = 4 - 8(w + 2) \\[3ex] 7w - 2w + 18 = 4 - 8w - 16 \\[3ex] 7w - 2w + 8w = 4 - 16 - 18 \\[3ex] 13w = -30 \\[3ex] w = -\dfrac{30}{13} $

$ 4x^2 - 81 \\[3ex] 2^2x^2 - 9^2 \\[3ex] (2x)^2 - 9^2 \\[3ex] (2x + 9)(2x - 9) ...\text{Difference of Two Squares} $

$ f(x) = 3x^2 - 7 \\[3ex] f(4) = 3(4)^2 - 7 \\[3ex] = 3(16) - 7 \\[3ex] = 48 - 7 \\[3ex] = 41 $

The domain is the set of all the input values, x for which the function has an output.
The closed circle represents a closed interval. It shows that the endpoint is included.

For the graph:
The graph starts from the origin at x = 0
The circle on the right hand side is at x = 12

The interval notation is: [0, 12]
The set notation is: Domain = $\{x: 0 \le x \le 12 \}$

The y-intercept is the point where the graph intersects the y-axis.
The graph intersects the y-axis at (0, –2)
Hence, the y-intercept is (0, –2)

The rate of change of tof the water level with respect to time is the slope of the graph.

$ \text{Point 1: } (0, 9) \\[3ex] x_1 = 0 \\[3ex] y_1 = 9 \\[5ex] \text{Point 2: } (4, 8) \\[3ex] x_2 = 4 \\[3ex] y_2 = 8 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{8 - 9}{4 - 0} \\[5ex] m = -\dfrac{1}{4} \\[5ex] m = -0.25...\text{This is a decrease because of the negative sign.} \\[3ex] $ The water level decreases at a rate of 0.25 m/h.

Let:
α, β be the zeros of the function
a = leading coefficient

$ \alpha = -3 \\[3ex] \beta = 1 \\[3ex] h(x) = a(x - \alpha)(x - \beta) \\[3ex] h(x) = a[x - (-3)](x - 1) \\[3ex] h(x) = a(x + 3)(x - 1) \\[3ex] \text{Passes through the vertex: } (-1, -4) \\[3ex] x = -1 \\[3ex] h(x) = -4 \\[3ex] \implies \\[3ex] -4 = a(-1 + 3)(-1 - 1) \\[3ex] -4 = a(2)(-2) \\[3ex] -4 = -4a \\[3ex] -4a = -4 \\[3ex] a = \dfrac{-4}{-4} \\[5ex] a = 1 \\[3ex] \implies \\[3ex] h(x) = 1(x + 3)(x - 1) \\[3ex] h(x) = x^2 - x + 3x - 3 \\[3ex] h(x) = x^2 + 2x - 3 $

$ \underline{\text{Rectangle}} \\[3ex] Area = Length * Width \\[3ex] = (5 + 2x) * 10 \\[3ex] = 10(5 + 2x)\text{ square inches} \\[5ex] \underline{\text{Triangle}} \\[3ex] Area = \dfrac{1}{2} * base * \perp height \\[5ex] = \dfrac{1}{2} * 30 * (4x - 10) \\[5ex] = 15(4x - 10) \\[3ex] = 15 * 2 * (2x - 5) \\[3ex] = 30 * (2x - 5) \text{ square inches} \\[3ex] $ The area in square inches of the rectangle is equal to the area in square inches of the triangle.

$ 10(5 + 2x) = 30(2x - 5) \\[3ex] \text{Divide both sides by 10} \\[3ex] 5 + 2x = 3(2x - 5) \\[3ex] 5 + 2x = 6x - 15 \\[3ex] 5 + 15 = 6x - 2x \\[3ex] 20 = 4x \\[3ex] 4x = 20 \\[3ex] x = \dfrac{20}{4} \\[5ex] x = 5 $

$ 30x^2 - 4x - 16 \\[3ex] 2(15x^2 - 2x - 8) \\[3ex] ....................................... \\[3ex] 15x^2 - 2x - 8 \\[3ex] 15x^2 * -8 = -120x^2 \\[3ex] \text{Factors are: } 10x \text{ and } -12x \\[3ex] \implies \\[3ex] 15x^2 + 10x - 12x - 8 \\[3ex] 5x(3x + 2) -4(3x + 2) \\[3ex] (3x + 2)(5x - 4) \\[3ex] ....................................... \\[3ex] 2(3x + 2)(5x - 4) $

Observing the data, as the year increases (from 2011 to 2017), the number of species added fluctuates: it increases, then decreases, then increase, and finally decrease.
There is no consistent upward or downward trend.
Hence, the association is a Weak Correlation.
A weak negative correlation has a correlation coefficient close to 0, while a strong negative correlation has a correlation coefficient close to –1.
This implies that the association is a weak negative correlation, because r ≈ –0.09.

The range is the set of all the output values, y produced by the input values of the function.

The closed circle represents a closed interval.
On the graph, the closed circle appears on the left-hand side at y = 2
This value is included.

The open circle represents an open interval.
On the graph, the open circle appears on the right-hand side at y = -4
This value is not included.

So, the interval notation is: $(-4, 2]$
This represents all real numbers between –4 (excluded) and 2 (included).
The range is the set of all real numbers greater than –4 and less than or equal to 2.

$ f(x) = 4x^2 - 36x + 81 \\[3ex] 4x^2 * 81 = 324x^2 \\[3ex] \text{Factors are: } -18x \text{ and } -18x \\[3ex] \implies \\[3ex] f(x) = 4x^2 - 18x - 18x + 81 \\[3ex] f(x) = 2x(2x - 9) - 9(2x - 9) \\[3ex] f(x) = (2x - 9)(2x - 9) \\[3ex] f(x) = (2x - 9)^2 \\[5ex] \underline{\text{Zeros}} \\[3ex] f(x) = (2x - 9)^2 = 0 \\[3ex] 2x - 9 = \pm\sqrt{0} \\[3ex] 2x - 9 = \pm 0 \\[3ex] 2x = 9 \pm 0 \\[3ex] 2x = 9 \\[3ex] x = \dfrac{9}{2} ...\text{twice} \\[5ex] $ The only zero is $\dfrac{9}{2}$ because $f(x) = (2x - 9)^2$

The only asymptote that applies to standard exponential functions is the horizontal asymptote.
A horizontal asymptote is the horizontal line say y = k where the graph/output approaches the specific value, k as the input, x increases or decreases without bound.

$ 0.75 = \dfrac{75}{100} = \dfrac{3}{4} \\[5ex] y = 16(0.75)^x \\[3ex] y = 16\left(\dfrac{3}{4}\right)^x \\[5ex] \text{As } x \to \infty, \\[3ex] y = 16\left(\dfrac{3}{4}\right)^\infty \\[5ex] y = 16(0) \\[3ex] y = 0 \\[5ex] \text{As } x \to -\infty, \\[3ex] y = 16\left(\dfrac{3}{4}\right)^{-\infty} \\[5ex] y = 16\left(\dfrac{4}{3}\right)^\infty ...\text{Law 6 Exp} \\[5ex] y = 16(\infty) \\[3ex] y = \infty \\[3ex] $ Because ∞ is not a value, the horizontal asymptote is: $y = 0$

$ \left(x^{\dfrac{3}{7}}\right)^2 \\[5ex] x^{\dfrac{3}{7} * 2} ...\text{Law 5 Exp} \\[5ex] x^{\dfrac{6}{7}} $

Let us analyze each option and discard the incorrect ones.
For the function: $n(x) = 5x^2 - 20x + 12$,

The domain is all real numbers.
Options A, and C. are discarded.
Now, onto the range.

The leading coefficient is 5.
5 is a positive value. It is greater than 0.
Hence, the parabola has a minimum value. It is concave up.
The vertex is the minimum value. No other value is lower than the y-value of the vertex.
Let's find it.

$ n(x) = 5x^2 - 20x + 12 \\[3ex] a = 5 \\[3ex] b = -20 \\[3ex] c = 12 \\[3ex] \text{x-coordinate of the vertex} = -\dfrac{b}{2a} \\[5ex] = -\dfrac{-20}{2(5)} \\[5ex] = 2 \\[5ex] \text{x-coordinate of the vertex} = n(2) \\[3ex] = 5(2)^2 - 20(2) + 12 \\[3ex] = 20 - 40 + 12 \\[3ex] = -8 \\[3ex] $ The range of the function is $n(x) \ge -8$.

Each hat costs $7, and each T-shirt costs $10.
... hats, x, and T-shirts, y

x hats @ $7 per hat = $7x$
y T-shirts @ $10 per T-shirt = $10y$
Total cost = $7x + 10y$

The customer receives free shipping if the amount of the order is over $100.
This implies that:
$7x + 10y \gt 100$

For a quadratic function:
Observing the trend, as x increases,
y increases, reaches the vertex, then begins to decrease;
This is a quadratic function that has a maximum point.
Let's say the vertex hovers around 30 before y-value begins to decrease
When x = 65, y should be lower than 29.1
Options B. and C. are discarded.

For x-values from 30 to 35, there is a gradual increase in the y-values from 28.3 to 29.1
Similarly, for the x-values from 60 to 65, there should be a gradual decrease in the y-value from 29.1
Option D: 24.9 is too distant. Discard it.
Option A.: 27.9 is a reasonable estimate.

$ \underline{\text{Exponential Function}} \\[3ex] p(x) = ab^x \\[3ex] a = \text{initial value} = 48000 \\[3ex] b = \text{base} \\[3ex] $ After the first year, the population of the town was 50,400.

$ \text{Point: } (1, 50400) \\[3ex] x = 1 \\[3ex] p(x) = 50400 \\[3ex] 50400 = 48000 * b^{1} \\[3ex] 48000 * b = 50400 \\[3ex] b = \dfrac{50400}{48000} \\[5ex] \implies \\[3ex] p(x) = 48,000\left(\dfrac{50,400}{48,000}\right)^x $

$ f(x) = x \\[3ex] g(x) = 3f(x) = 3x \\[3ex] \text{This means that when: } x = 1 \\[3ex] f(1) = 1 \\[3ex] g(1) = 3(1) = 3 \\[3ex] $ Let us analyze each option and discard the incorrect ones.

$ \text{When } x = 1 \\[5ex] \text{Option A.} \\[3ex] f(1) = 1 \\[3ex] g(1) = -2 \ne 3 ...\text{Discard} \\[3ex] \text{NEXT} \\[3ex] \text{Option B.} \\[3ex] f(1) = 1 \\[3ex] g(1) = 4 \ne 3 ...\text{Discard} \\[3ex] \text{NEXT} \\[3ex] \text{Option C.} \\[3ex] f(1) = 1 \\[3ex] g(1) \ne 3 ...\text{Discard} \\[3ex] \text{NEXT} \\[3ex] \text{Option D.} \\[3ex] f(1) = 1 \\[3ex] g(1) = 3 ...\text{This is the correct option.} $

$ \dfrac{8w^2 - 20w - 12}{2w + 1} \\[5ex] ................................................... \\[3ex] \underline{\text{Numerator}} \\[3ex] 8w^2 - 20w - 12 \\[3ex] 4(2w^2 - 5w - 3) \\[3ex] \text{For } 2w^2 - 5w - 3 \\[3ex] 2w^2 * -3 = -6w^2 \\[3ex] \text{Factors are } w \text{ and } -6w \\[3ex] \implies \\[3ex] 2w^2 + w - 6w - 3 \\[3ex] w(2w + 1) -3(2w + 1) \\[3ex] (w - 3)(2w + 1) \\[3ex] \implies \\[3ex] 4(w - 3)(2w + 1) \\[3ex] ................................................... \\[3ex] \implies \\[3ex] \dfrac{4(w - 3)(2w + 1)}{2w + 1} \\[5ex] 4(w - 3) \\[3ex] 4w - 12 $

There are at least two approaches to solving this question.
Use any approach you prefer.

1st Approach: Algebraic Solution
This involves checking each option to see the one that satisfies the inequality.

$ y \le \dfrac{3}{5}x - 6 \\[5ex] \text{Option A.} \\[3ex] (5, -4) \\[3ex] x = 5 \\[3ex] y = -4 \\[3ex] -4 \stackrel{?}{\le} \dfrac{3}{5} * 5 - 6 \\[5ex] -4 \stackrel{?}{\le} 3 - 6 \\[3ex] -4 \le -3...\text{Yes, this is the correct answer} \\[5ex] \text{Option B.} \\[3ex] (-2, -5) \\[3ex] x = -2 \\[3ex] y = -5 \\[3ex] -5 \stackrel{?}{\le} \dfrac{3}{5} * -2 - 6 \\[5ex] -5 \stackrel{?}{\le} -\dfrac{6}{5} - \dfrac{30}{5} \\[5ex] -\dfrac{25}{5} \stackrel{?}{\le} -\dfrac{36}{5} \\[5ex] -\dfrac{25}{5} \ge -\dfrac{36}{5} \\[5ex] \text{Discard}. \\[5ex] \text{Option C.} \\[3ex] (9, 1) \\[3ex] x = 9 \\[3ex] y = 1 \\[3ex] 1 \stackrel{?}{\le} \dfrac{3}{5} * 9 - 6 \\[5ex] 1 \stackrel{?}{\le} \dfrac{27}{5} - \dfrac{30}{5} \\[5ex] \dfrac{5}{5} \stackrel{?}{\le} -\dfrac{3}{5} \\[5ex] \dfrac{5}{5} \ge -\dfrac{3}{5} \\[5ex] \text{Discard}. \\[5ex] \text{Option D.} \\[3ex] (-8, 3) \\[3ex] x = -8 \\[3ex] y = 3 \\[3ex] 3 \stackrel{?}{\le} \dfrac{3}{5} * -8 - 6 \\[5ex] 3 \stackrel{?}{\le} -\dfrac{24}{5} - \dfrac{30}{5} \\[5ex] \dfrac{15}{5} \stackrel{?}{\le} -\dfrac{54}{5} \\[5ex] \dfrac{15}{5} \ge -\dfrac{54}{5} \\[5ex] \text{Discard}. \\[3ex] $ 2nd Approach: Graphical Solution
They gave us a graph, so apparently, they want us to graph the inequality and use that graph in determining the solution.
Because the inequality contains an equal sign, the line will be a solid line.

$ y \le \dfrac{3}{5}x - 6 \\[5ex] \text{When } x = 0 \\[3ex] y \le \dfrac{3}{5} * 0 - 6 \\[5ex] y \le 0 - 6 \\[3ex] y \le -6 \\[5ex] \text{When } x = 5 \\[3ex] y \le \dfrac{3}{5} * 5 - 6 \\[5ex] y \le 3(1) - 6 \\[3ex] y \le 3 - 6 \\[3ex] y \le -3 \\[5ex] \text{When } y = 0 \\[3ex] 0 \le \dfrac{3}{5}x - 6 \\[5ex] \dfrac{3}{5}x - 6 \ge 0 \\[5ex] \dfrac{3}{5}x \ge 6 \\[5ex] 3x \ge 5(6) \\[3ex] x \ge \dfrac{5(6)}{3} \\[5ex] x \ge 10 \\[3ex] $ The graph of the inequality is:



An ordered pair in the shaded region is in the solution set of the inequality.
Let us analyze each option.

Option A.
$(5, -4)$ is inside the solution set.
This is the correct answer.
But, let us analyze the remaining options if you wish.

Option B.
$(-2, -5)$ is outside the solution set.


Option C.
$(9, 1)$ is outside the solution set.

Option D.
$(-8, 3)$ is outside the solution set.

For $p(x) = -7(4)^x$,
Any real number input for x, will have an output.
Hence, the domain is the set of all real numbers.

4 raised to a positive exponent gives a positive result greater than 1.
4 raised to exponent 0 gives a positive result, 1.
4 raised to a negative exponent gives a positive result less than 1.
So, 4 raised to any input, x gives a positive result.

Any positive result multiplied by –7 gives a negative output.
Therefore, the range is the set of all negative real numbers.
The range is all real numbers less than 0.

Let us analyze each option.
Option A.
A. The function has 3 zeros.
No, the function has 2 zeros.

B. The maximum value of the function is 3.
No, the maximum value of the function is 4.

C. The vertex of the graph of the function is (4, 1)
No, the vertex of the graph is (1, 4)

D. The equation of the axis of symmetry of the graph of the function is x = 1.
Yes, the equation of the axis of symmetry of the graph of the function is x = 1.
This is the correct answer.

$ \sqrt{600} \\[3ex] \sqrt{100 * 6} \\[3ex] \sqrt{100} * \sqrt{6} \\[3ex] 10 * \sqrt{6} \\[3ex] 10\sqrt{6} $

The slope of a vertical line is undefined.
Discard Options B. and D.
Options A. and C. remain.

The equation of the line is x = –4
Discard Option C.
The correct answer is Option A.

$y = f(x)$
y is a function of x
The zero of the function is the value of x for which the function, y is zero.
$y = 0$ when $x = 2$
Hence, the zero of f is $x = 2$.

$ -3x + 5y = 21 ...eqn.(1) \\[3ex] 6x - y = -15 ...eqn.(2) \\[5ex] \underline{\text{Substitution Method}} \\[3ex] \text{From } eqn.(2) \\[3ex] 6x + 15 = y \\[3ex] y = 6x + 15...eqn.(3) \\[5ex] \text{Substitute for y in eqn.(1)} \\[3ex] -3x + 5(6x + 15) = 21 \\[3ex] -3x + 30x + 75 = 21 \\[3ex] 27x = 21 - 75 \\[3ex] 27x = -54 \\[3ex] x = -\dfrac{54}{27} \\[5ex] x = -2 \\[5ex] \text{Substitute for x in eqn.(3)} \\[3ex] y = 6(-2) + 15 \\[3ex] y = -12 + 15 \\[3ex] y = 3 \\[5ex] \text{Solution set is: } (-2, 3) \\[3ex] $ Check

$x = -2, y = 3$

LHS	RHS
$ -3x + 5y \\[3ex] -3(-2) + 5(3) \\[3ex] 6 + 15 \\[3ex] 21 $ $ 6x - y \\[3ex] 6(-2) - 3 \\[3ex] -12 - 3 \\[3ex] -15 $	$21$ $-15$

$ 12x^2 - 48x + 48 \\[3ex] \text{GCF} = 12 \\[3ex] \implies \\[3ex] 12(x^2 - 4x + 4) \\[3ex] \text{Factors are: } -2 \text{ and } -2 \\[3ex] 12(x - 2)(x - 2) \\[3ex] 12(x - 2)^2 $

$ \underline{\text{Quadratic Function in Vertex Form}} \\[3ex] y = a(x - h)^2 + k \\[3ex] \text{Where} \\[3ex] \text{Leading Coefficient} = a \\[3ex] \text{Vertex} = (h, k) = (1, 46) \\[3ex] h = 1 \\[3ex] k = 46 \\[3ex] \implies \\[3ex] y = a(x - 1)^2 + 46 \\[5ex] \text{Passes through the point } (3, 10) \\[3ex] x = 3 \\[3ex] y = 10 \\[3ex] \implies \\[3ex] 10 = a(3 - 1)^2 + 46 \\[3ex] 10 - 46 = a(2)^2 \\[3ex] -36 = 4a \\[3ex] 4a = -36 \\[3ex] a = -\dfrac{36}{4} \\[5ex] a = -9 \\[5ex] \implies \\[3ex] y = -9(x - 1)^2 + 46 $

$ \text{Point 1: } (1, 296) \\[3ex] x_1 = 1 \\[3ex] y_1 = 296 \\[5ex] \text{Point 2: } (7, 51) \\[3ex] x_2 = 7 \\[3ex] y_2 = 51 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{51 - 296}{7 - 1} \\[5ex] m = \dfrac{-245}{6} \\[5ex] m = -40.8\bar{3} \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 296 = -40.8\bar{3}(x - 1) \\[3ex] y - 296 = -40.8\bar{3}x + 40.8\bar{3} \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = mx + b ...\text{where } b \text{ is the y-intercept} \\[3ex] y = -40.8\bar{3}x + 40.8\bar{3} + 296 \\[3ex] y = -40.8\bar{3}x + 336.8\bar{3} \\[3ex] y \approx 41x + 337 \\[3ex] y \approx 337 - 41x $

$ \underline{\text{Exponential Function}} \\[3ex] y = ab^x \\[3ex] a = \text{initial value} \\[3ex] b = \text{base} \\[3ex] $ If b < 1,
The exponential function is an exponential decay function.
The base is a decay factor.

If b > 1,
The exponential function is an exponential growth function.
The base is a growth factor.

$ w(t) = 270(1.029)^t \\[3ex] 1.029 \gt 1 \\[3ex] \text{growth factor} = 1.029 $

$ \text{slope} = m \\[3ex] \text{Point 1: } (-2, -2) \\[3ex] x_1 = -2 \\[3ex] y_1 = -2 \\[5ex] \text{Point 2: } (4, 2) \\[3ex] x_2 = 4 \\[3ex] y_2 = 2 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{2 - (-2)}{4 - (-2)} \\[5ex] m = \dfrac{4}{6} \\[5ex] m = \dfrac{2}{3} $

$ 5(2w + 4) = 4(2w + 9) \\[3ex] 10w + 20 = 8w + 36 \\[3ex] 10w - 8w = 36 - 20 \\[3ex] 2w = 16 \\[3ex] w = \dfrac{16}{2} \\[5ex] w = 8 $

The price of a discounted ticket was 80% of the price of a standard ticket.
80% = 0.8
If the second equation in the system is $y = 0.80x$
This implies that:
y = price of a discounted ticket
x = price of a standard ticket

There were 153 standard tickets sold at the baseball game.
Total cost for the standard tickets = 153x

There were 47 discounted tickets sold at the baseball game.
Total cost for the discounted tickets = 47y

The total amount of ticket sales from the standard and discounted tickets was $2,649.34.
This implies that:
$153x + 47y = 2649.34$

$ y = 0.8x...eqn.(1) \\[3ex] 153x + 47y = 2649.34...eqn.(2) \\[5ex] \underline{\text{Substitution Method}} \\[3ex] \text{Substitute for y in eqn.(2)} \\[3ex] 153x + 47(0.8x) = 2649.34 \\[3ex] 153x + 37.6x = 2649.34 \\[3ex] 190.6x = 2649.34 \\[3ex] x = \dfrac{2649.34}{190.6} \\[5ex] x = 13.9 \\[5ex] \text{Substitute for x in eqn.(1)} \\[3ex] y = 0.8(13.9) \\[3ex] y = 11.12 \\[3ex] $ The price of a discounted ticket is $11.12
The best estimate of the price of a discounted ticket is $11.15

The first term, $f(1) = 27$.
Options C. and D. are discarded.
Options A. and B. remain.

$ f(1) = 27 \\[3ex] f(n) = \dfrac{1}{3}f(n - 1) \\[5ex] f(2) = \dfrac{1}{3} * f(2 - 1) \\[5ex] f(2) = \dfrac{1}{3} * f(1) \\[5ex] f(2) = \dfrac{1}{3} * 27 \\[5ex] f(2) = 9 \\[3ex] $ Option B. is discarded.
Option A. is the answer.

But, if you would like us to find the remaining terms:

$ f(3) = \dfrac{1}{3} * 9 \\[5ex] f(3) = 3 \\[5ex] f(4) = \dfrac{1}{3} * 3 \\[5ex] f(4) = 1 $

The question basically wants us to compare fractions.
In that regard, let us modify the table:

g(x)	$\dfrac{7}{4}$	$4$	$\dfrac{19}{4}$	$4$	$\dfrac{7}{4}$	$-2$
g(x)	$\dfrac{7}{4}$	$\dfrac{16}{4}$	$\dfrac{19}{4}$	$\dfrac{16}{4}$	$\dfrac{7}{4}$	$-\dfrac{8}{4}$

$\dfrac{19}{4}$ is the maximum value of the range.

This implies that the range is: All real numbers less than or equal to $\dfrac{19}{4}$

$ (2a + 5)(3a - 2) \\[3ex] 6a^2 - 4a + 15a - 10 \\[3ex] 6a^2 + 11a - 10 $

Printer ink costs $18.99 per cartridge.
For c cartridges
This implies: Total cost of printer ink = $18.99c

A flat rate shipping fee of $7.95, no matter the number of cartridges purchased in a single order.
This implies: $7.95

∴ The total cost, t, for a single order of c cartridges is: $t = 18.99c + 7.95$

$ x^6y^3 \div \left(\dfrac{x^2}{y}\right) \\[5ex] x^6y^3 \cdot \left(\dfrac{y}{x^2}\right) \\[5ex] \text{DISSOCIATE} \\[3ex] x^6 \cdot y^3 \cdot y \cdot \dfrac{1}{x^2} \\[5ex] \dfrac{x^6}{x^2} \cdot y^3 \cdot y \\[5ex] \text{SOLVE} \\[3ex] x^{6 - 2} * y^{3 + 1} ...\text{Laws 1 and 2 Exp} \\[3ex] x^4 \cdot y^4 \\[3ex] \text{ASSOCIATE} \\[3ex] x^4y^4 $

$ \underline{\text{Exponential Growth Function}} \\[3ex] f(x) = a(1 + r)^x \\[3ex] a = \text{initial value} = \$317 \\[3ex] r = \text{growth rate} = 4\% = \dfrac{4}{100} = 0.04 \\[5ex] \implies \\[3ex] f(x) = 2500(1 + 0.04)^x \\[3ex] f(x) = 317(1.04)^x $

$ \text{slope} = m \\[3ex] \text{y-intercept} = b \\[3ex] \text{constants} = A, B, C \\[3ex] A \gt 0 \\[5ex] \underline{\text{Line a}} \\[3ex] \text{Point 1: } (-11, 81) \\[3ex] x_1 = -11 \\[3ex] y_1 = 81 \\[5ex] \text{Point 2: } (4, -9) \\[3ex] x_2 = 4 \\[3ex] y_2 = -9 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-9 - 81}{4 - (-11)} \\[5ex] m = \dfrac{-90}{15} \\[5ex] m = -6 \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 81 = -6(x - (-11)) \\[3ex] y - 81 = -6(x + 11) \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = mx + b \\[3ex] y = -6x - 66 + 81 \\[3ex] y = -6x + 15 \\[5ex] \text{Standard Form} \\[3ex] Ax + By = C \\[3ex] y + 6x = 15 \\[3ex] 6x + y = 15 ...eqn.(1) \\[5ex] \underline{\text{Line b}} \\[3ex] \text{Point 1: } (-9, 18) \\[3ex] x_1 = -9 \\[3ex] y_1 = 18 \\[5ex] \text{Point 2: } (6, -27) \\[3ex] x_2 = 6 \\[3ex] y_2 = -27 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-27 - 18}{6 - (-9)} \\[5ex] m = \dfrac{-45}{15} \\[5ex] m = -3 \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 18 = -3(x - (-9)) \\[3ex] y - 18 = -3(x + 9) \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = mx + b \\[3ex] y = -3x - 27 + 18 \\[3ex] y = -3x - 9 \\[5ex] \text{Standard Form} \\[3ex] Ax + By = C \\[3ex] y + 3x = -9 \\[3ex] 3x + y = -9 ...eqn.(2) \\[5ex] \therefore \text{The system of equations that represents the graph of the two lines is:} \\[3ex] 6x + y = 15 \\[3ex] 3x + y = -9 $

Parent Function: $f(x) = x^2$
Vertex = (0, 0)

Child Function: $g(x) = f(x - 3) - 9$

1st Transformation: Horizontal Shift 3 units to the right

$ (0, 0) \rightarrow (0 + 3, 0) \\[3ex] \hspace{2em} \rightarrow (3, 0) \\[3ex] $ 2nd Transformation: Vertical Shift 9 units down

$ (3, 0) \rightarrow (3, 0 - 9) \\[3ex] \hspace{2em} \rightarrow (3, -9) $

$ 24gh - 12g^2 + 18g \\[3ex] \text{GCF} = 6g \\[3ex] \implies \\[3ex] 6g(4h - 2g + 3) $

Let the:
cost per click on Website 1 = c
cost per click on Website 2 = d

$ \underline{\text{Monday}} \\[3ex] 15c + 29d = 94.15 \\[3ex] 29d = 94.15 - 15c...eqn.(1) \\[5ex] \underline{\text{Tuesday}} \\[3ex] 25c + 29d = 121.15 \\[3ex] 29d = 121.15 - 25c...eqn.(2) \\[5ex] 29d = 29d \implies \\[3ex] eqn.(1) = eqn.(2) \\[3ex] 94.15 - 15c = 121.15 - 25c \\[3ex] 25c - 15c = 121.15 - 94.15 \\[3ex] 10c = 27 \\[3ex] c = \dfrac{27}{10} \\[5ex] c = \$2.70 $

$ k(x) = x^2 + 32x + 248 \\[3ex] \text{Compare to the standard form: } k(x) = ax^2 + bx + c \\[3ex] a = 1 \\[3ex] b = 32 \\[3ex] c = 248 \\[5ex] \underline{\text{Quadratic Formula}} \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] x = \dfrac{-32 \pm \sqrt{32^2 - 4(1)(248)}}{2(1)} \\[5ex] x = \dfrac{-32 \pm \sqrt{1024 - 992}}{2} \\[5ex] x = \dfrac{-32 \pm \sqrt{32}}{2} \\[5ex] x = \dfrac{-32 \pm \sqrt{16 \cdot 2}}{2} \\[5ex] x = \dfrac{-32 \pm (\sqrt{16} * \sqrt{2})}{2} \\[5ex] x = \dfrac{-32 \pm 4\sqrt{2}}{2} \\[5ex] x = \dfrac{4(-8 \pm \sqrt{2})}{2} \\[5ex] x = 2(-8 \pm \sqrt{2}) \\[3ex] OR \\[3ex] x = -16 \pm 2\sqrt{2} $

The question is an exponential decay function.

$ \underline{\text{Exponential Decay Function}} \\[3ex] y = a(1 - r)^x \\[3ex] a = \text{initial value} = \$650 \\[3ex] r = \text{decay rate} = 30\% = \dfrac{30}{100} = 0.3 \\[5ex] \implies \\[3ex] y = 650(1 - 0.3)^x \\[3ex] y = 650(0.7)^x \\[5ex] \text{When } x = 0 \\[3ex] y = 650(0.7)^0 \\[3ex] y = 650(1) \\[3ex] y = 650 \\[3ex] $ The y-intercept of the graph is $650.
Option A. is the correct answer.

Student: Could you please give explanations for the other options?
Teacher: Sure, let's do it.

Option B. is incorrect because the y-intercept is 650.

The only asymptote that applies to standard exponential functions is the horizontal asymptote.
Option C. is incorrect.

A horizontal asymptote is the horizontal line say y = k where the graph/output approaches the specific value, k as the input, x increases or decreases without bound.

$ y = 650(0.7)^x \\[3ex] y = 650\left(\dfrac{7}{10}\right)^x \\[5ex] \text{As } x \to \infty, \\[3ex] y = 650\left(\dfrac{7}{10}\right)^\infty \\[5ex] y = 650(0) \\[3ex] y = 0 \\[5ex] \text{As } x \to -\infty, \\[3ex] y = 650\left(\dfrac{7}{10}\right)^{-\infty} \\[5ex] y = 650\left(\dfrac{10}{7}\right)^\infty ...\text{Law 6 Exp} \\[5ex] y = 650(\infty) \\[3ex] y = \infty \\[3ex] $ Because ∞ is not a value, the horizontal asymptote is: $y = 0$
Option D. is incorrect.

Because the inequality contains an equal sign, the line will be a solid line.

$ y \ge -x + 2 \\[5ex] \text{When } x = 0 \\[3ex] y \ge -0 + 2 \\[3ex] y \ge 2 \\[5ex] \text{When } x = 1 \\[3ex] y \ge -1 + 2 \\[3ex] y \ge 1 \\[3ex] $ The graph of the inequality is

$ \underline{\text{Line m}} \\[3ex] \text{Point 1: } (0, -3) \\[3ex] x_1 = 0 \\[3ex] y_1 = -3 \\[5ex] \text{Point 2: } (2, 0) \\[3ex] x_2 = 2 \\[3ex] y_2 = 0 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{0 - (-3)}{2 - 0} \\[5ex] m = \dfrac{3}{2} \\[5ex] \underline{\text{Line P}} \\[3ex] \text{Line P } || \text{ Line m} \\[3ex] m = \dfrac{3}{2}...\text{parallel lines have the same slope} \\[5ex] \text{Passes through Point } (-2, 2) \\[3ex] x_3 = -2 \\[3ex] y_3 = 2 \\[3ex] \text{Point–Slope Form} \\[3ex] y - y_3 = m(x - x_3) \\[3ex] y - 2 = \dfrac{3}{2}[x - (-2)] \\[5ex] y - 2 = \dfrac{3}{2}(x + 2) \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = \dfrac{3}{2}(x + 2) + 2 \\[5ex] y = \dfrac{3}{2}x + 3 + 2 \\[5ex] y = \dfrac{3}{2}x + 5 $

$ \underline{\text{Literal Equation}} \\[3ex] S = Ph + 2B \\[3ex] Ph + 2B = S \\[3ex] 2B = S - Ph \\[3ex] B = \dfrac{S - Ph}{2} $

Let us analyze each option, and discard the incorrect ones.

$ \underline{\text{Option A.}} \\[3ex] x \lt 5 \\[3ex] y \gt \approx 3.2 \\[3ex] \text{Incorrect option} \\[5ex] \underline{\text{Option B.}} \\[3ex] x \gt 5 \\[3ex] y \lt 2 \\[3ex] \text{Correct option} \\[5ex] \underline{\text{Option C.}} \\[3ex] x \lt 5 \\[3ex] y \lt 2 \\[3ex] \text{Incorrect option} \\[5ex] \underline{\text{Option D.}} \\[3ex] x \gt 5 \\[3ex] y \gt 2 \\[3ex] \text{Incorrect option} $

$ g(x) = 6(2x + 7) \\[3ex] g(20) = 6[2(20) + 7] \\[3ex] = 6(40 + 7) \\[3ex] = 6(47) \\[3ex] = 282 $

The x-intercept of the line is $(-2, 0)$
The y-intercept of the line is $(0, 8)$

The zeros of the function are the x-values where y is zero.
The zeros of the function are: $x = -1$ and $x = 3$

The rate of change of height off the ground with respect to the number of steps is the slope of the linear function.

$ \text{Point 1: } (3, 21) \\[3ex] x_1 = 3 \\[3ex] y_1 = 21 \\[5ex] \text{Point 2: } (11, 77) \\[3ex] x_2 = 11 \\[3ex] y_2 = 77 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{77 - 21}{11 - 3} \\[5ex] m = \dfrac{56}{8} \\[5ex] m = 7 \\[3ex] $ The rate of change of height off the ground with respect to the number of steps is 7 inches per step.

The rate of change of the number of items the worker packed in boxes with respect to the number of minutes the worker has been packing items in boxes is the slope
y = Number of items packed
x = Number of minutes

$ \text{Point 1: } (5, 20) \\[3ex] x_1 = 5 \\[3ex] y_1 = 20 \\[5ex] \text{Point 2: } (14, 56) \\[3ex] x_2 = 14 \\[3ex] y_2 = 56 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{56 - 20}{14 - 5} \\[5ex] m = \dfrac{36}{9} \\[5ex] m = 4 \\[3ex] $ The rate of change of the number of items the worker packed in boxes with respect to the number of minutes the worker has been packing items in boxes is 4 items in boxes per minute.

$ \text{slope} = m \\[3ex] \text{y-intercept} = b \\[5ex] \underline{\text{Line with the positive slope}} \\[3ex] \text{Point 1: } (0, -3) \\[3ex] x_1 = 0 \\[3ex] y_1 = -3 \\[5ex] \text{Point 2: } (5, -1) \\[3ex] x_2 = 5 \\[3ex] y_2 = -1 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-1 - (-3)}{5 - 0} \\[5ex] m = \dfrac{-1 + 3}{5} \\[5ex] m = \dfrac{2}{5} \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = mx + b \\[3ex] b = -3 \\[3ex] \implies \\[3ex] y = \dfrac{2}{5}x - 3...eqn.(1) \\[5ex] \underline{\text{Line with the negative slope}} \\[3ex] \text{Point 1: } (-5, -5) \\[3ex] x_1 = -5 \\[3ex] y_1 = -5 \\[5ex] \text{Point 2: } (0, -8) \\[3ex] x_2 = 0 \\[3ex] y_2 = -8 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-8 - (-5)}{0 - (-5)} \\[5ex] m = \dfrac{-8 + 5}{0 + 5} \\[5ex] m = -\dfrac{3}{5} \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = mx + b \\[3ex] b = -8 \\[3ex] \implies \\[3ex] y = -\dfrac{3}{5}x - 8...eqn.(2) \\[5ex] $ The system of equations best represented by the graph is:

$ B.\;\; y = \dfrac{2}{5}x - 3 \\[5ex] \hspace{1.6em} y = -\dfrac{3}{5}x - 8 $

The Vertical Line test states that if a vertical line is used to cut through the graph of a relation; the relation is a function if the vertical line intersects the graph at only one point, and the relation is not a function if the vertical line intersects the graph at more than one point.

Using the Vertical Line Test:
The only graph that passes the Vertical Line Test is the graph in Option D.
Option D. is the correct answer.

$ 2x + y = 40...eqn.(1) \\[3ex] x - 2y = -20...eqn.(2) \\[3ex] \underline{\text{Substitution Method}} \\[3ex] \text{From } eqn.(2) \\[3ex] x = -20 + 2y...eqn.(3) \\[3ex] \text{Substitute for } x \text{ in } eqn.(1) \\[3ex] 2(-20 + 2y) + y = 40 \\[3ex] -40 + 4y + y = 40 \\[3ex] 5y = 40 + 40 \\[3ex] 5y = 80 \\[3ex] y = \dfrac{80}{5} \\[5ex] y = 16 \\[3ex] \text{Substitute for } y \text{ in } eqn.(3) \\[3ex] x = -20 + 2(16) \\[3ex] x = -20 + 32 \\[3ex] x = 12 \\[3ex] (x , y) = (12, 16) $

Let us analyze each option.

Option A.
The range is the set of all real numbers less than or equal to –3
Incorrect answer.

Option B.
The range is the set of all real numbers less than or equal to 3
Incorrect answer.

Option C.
The range is the set of all real numbers greater than or equal to –3
Incorrect answer.

Option D.
The range is the set of all real numbers greater than or equal to 3
Correct answer.

$ y \propto x...\text{Direct Proportion} \\[3ex] y = kx ...k\text{ is the proportionality constant} \\[3ex] x = 512, y = 128, k = ? \\[3ex] 128 = k * 512 \\[3ex] k = \dfrac{128}{512} \\[5ex] k = \dfrac{1}{4} \\[5ex] x = 64, y = ? \\[3ex] y = \dfrac{1}{4} * 64 \\[5ex] y = 16 $

$ 8x + 5y \le 100 \\[3ex] \text{When } x = 0 \\[3ex] 8(0) + 5y \le 100 \\[3ex] 5y \le 100 \\[3ex] y \le \dfrac{100}{5} \\[5ex] y \le 20 ...\text{Mark point } (0, 20) \\[3ex] \text{The shaded area is downwards from } y = 20 \\[5ex] \text{When } y = 0 \\[3ex] 8x + 5(0) \le 100 \\[3ex] 8x \le 100 \\[3ex] x \le \dfrac{100}{8} \\[5ex] x \le 12.5 ...\text{Mark point } (12.5, 0) \\[3ex] \text{The shaded area is to the left of } x = 12.5 \\[3ex] $ Draw a straight solid line (because of the equal sign) connecting the two points: (0, 20) and (12.5, 0)
Locate the intersection of the two areas.
The correct answer is Option B.

$ \underline{\text{Exponential Function}} \\[3ex] y = ab^x \\[3ex] a = \text{initial value} = \text{y-intercept} = 1 \\[3ex] b = \text{base} \\[3ex] \implies \\[3ex] y = b^x \\[5ex] \text{Passes through the point } (-2, 16) \\[3ex] y = 16 \\[3ex] x = -2 \\[3ex] 16 = b^{-2} \\[3ex] b^{-2 * -\dfrac{1}{2}} = 16^{-\dfrac{1}{2}} \\[7ex] b = \dfrac{1}{16^{\dfrac{1}{2}}}...\text{Law 6 Exp} \\[9ex] b = \dfrac{1}{\sqrt{16}} ...\text{Law 7 Exp} \\[5ex] b = \dfrac{1}{4} \\[5ex] b = 0.25 \\[3ex] \implies \\[3ex] y = (0.25)^x $

$ (n - 4)(2n + 7) \\[3ex] 2n^2 + 7n - 8n - 28 \\[3ex] 2n^2 - n - 28 $

The correct answer is Option B.
B. When the number of hours worked each week by an hourly employee decreases, the amount of money earned by the employee also decreases.
This is because an hourly employee’s weekly pay depends directly on the number of hours worked.
If the employee works more hours, they earn more pay.
If they work fewer hours, they earn less pay.
In this case, the number of hours worked each week causes the change in weekly pay.

If we analyze the other options:
Options A and D show no meaningful relationship between the events.
There is no association and therefore no causation.

For Option C.,
When the discount offered during a sale increases, we would generally expect the number of items sold to increase, not decrease.
A larger discount means a lower price, which typically leads to higher sales.
Therefore, Option C is incorrect.

$ \text{slope} = m \\[3ex] \text{y-intercept} = b \\[5ex] \underline{\text{Table}} \\[3ex] \text{Point 1: } (-16, 7) \\[3ex] x_1 = -16 \\[3ex] y_1 = 7 \\[5ex] \text{Point 2: } (12, -14) \\[3ex] x_2 = 12 \\[3ex] y_2 = -14 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-14 - 7}{12 - (-16)} \\[5ex] m = \dfrac{-21}{28} \\[5ex] m = -\dfrac{3}{4} \\[5ex] \text{Point – Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 7 = -\dfrac{3}{4}[x - (-16)] \\[5ex] y - 7 = -\dfrac{3}{4}(x + 16) \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = -\dfrac{3}{4}(x + 16) + 7 \\[5ex] y = -\dfrac{3}{4}x - 12 + 7 \\[5ex] y = -\dfrac{3}{4}x - 5...eqn.(1) \\[5ex] \underline{\text{Graph}} \\[3ex] \text{Point 1: } (-4, 6) \\[3ex] x_1 = -4 \\[3ex] y_1 = 6 \\[5ex] \text{Point 2: } (4, -4) \\[3ex] x_2 = 4 \\[3ex] y_2 = -4 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-4 - 6}{4 - (-4)} \\[5ex] m = \dfrac{-10}{8} \\[5ex] m = -\dfrac{5}{4} \\[5ex] \text{Slope – Intercept Form} \\[3ex] y = mx + b \\[3ex] b = 1 \\[3ex] \implies \\[3ex] y = -\dfrac{5}{4}x + 1...eqn.(2) \\[5ex] $ The system of equations best represented by the graph is:

$ D.\;\; y = -\dfrac{3}{4}x - 5 \\[5ex] \hspace{1.6em} y = -\dfrac{5}{4}x + 1 $

Parent Function
$f(x) = x^2$

Child Function
$n(x) = x^2 - 1$

Transformation done to Parent to Give Child
Vertical Shift 1 unit down.

$ d^2 - d - 6 \\[3ex] \text{Factors are: } 2 \text{ and } -3 \\[3ex] (d + 2)(d - 3) \\[3ex] \implies \\[3ex] k = -3 $

$ \text{slope} = m \\[3ex] \text{y-intercept} = b \\[5ex] \text{Point 1: } (4, 0) \\[3ex] x_1 = 4 \\[3ex] y_1 = 0 \\[5ex] \text{Point 2: } (8, -7) \\[3ex] x_2 = 8 \\[3ex] y_2 = -7 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-7 - 0}{8 - 4} \\[5ex] m = -\dfrac{7}{4} \\[5ex] b = 7 \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = mx + b \\[3ex] y = -\dfrac{7}{4}x + 7 $

$ \dfrac{c^8(d^6)^3}{c^2} \\[5ex] \dfrac{c^8 \cdot d^{6 \cdot 3}}{c^2} ...\text{Law 5 Exp} \\[5ex] c^{8 - 2} \cdot d^{18} ...\text{Law 2 Exp} \\[5ex] c^6 \cdot d^{18} \\[3ex] c^6d^{18} $

$ 5x^2 = 30x - 45 \\[3ex] 5x^2 - 30x + 45 = 0 \\[3ex] 5(x^2 - 6x + 9) = 0 \\[3ex] x^2 - 6x + 9 = \dfrac{0}{5} \\[5ex] x^2 - 6x + 9 = 0 \\[3ex] \text{Factors are } -3 \text{ and } -3 \\[3ex] (x - 3)(x - 3) = 0 \\[3ex] x - 3 = 0 ...\text{twice}...\text{Zero Product Property} \\[3ex] x = 3 ...\text{twice} $

The domain is the set of all the input values, x for which the function has an output.
Based on the graph, the domain is: $\{0, 1, 2, 3, 4, 5, 6, 7, 8\}$

The rate of change of y with respect to x for this line is the slope of the line.

$ \text{Point 1: } (-3, 5) \\[3ex] x_1 = -3 \\[3ex] y_1 = 5 \\[5ex] \text{Point 2: } (9, -5) \\[3ex] x_2 = 9 \\[3ex] y_2 = -5 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-5 - 5}{9 - (-3)} \\[5ex] m = \dfrac{-10}{12} \\[5ex] m = -\dfrac{5}{6} $

$ h(x) = 12.3(4.9)^x \\[3ex] y-\text{intercept} = h(0) \\[3ex] h(0) = 12.3(4.9)^0 \\[3ex] h(0) = 12.3(1)...\text{Law 3 Exp} \\[3ex] h(0) = 12.3 \\[3ex] h(0) = +12.3 $

$ \dfrac{8.8 \times 10^9}{2.2 \times 10^{-3}} \\[5ex] \dfrac{8.8}{2.2} \times 10^{9 - (-3)}...\text{Law 2 Exp} \\[5ex] 4 \times 10^{9 + 3} \\[4ex] 4 \times 10^{12} $

There are at least two approaches to solve the question.
We can find the answer by analyzing the options.
We can also write a 3 by 3 linear system using the information given to us.
I recommend that we analyze the options to get the answer, rather than solving a 3 by 3 linear system to get the function.
Analyzing the options:

$ \text{Option A.} \\[3ex] d(x) = 0.05x^2 + 0.74x \\[3ex] d(1) = 0.05(1)^2 + 0.74(1) \ne -2.85 \\[3ex] \text{Discard} \\[5ex] \text{Option B.} \\[3ex] d(x) = 0.05x^2 + 0.74x + 9.17 \\[3ex] d(1) = 0.05(1)^2 + 0.74(1) + 9.17 \ne -2.85 \\[3ex] \text{Discard} \\[5ex] \text{Option C.} \\[3ex] d(x) = 0.26x^2 - 3.11x \\[3ex] d(1) = 0.26(1)^2 - 3.11(1) = -2.85 \\[3ex] \text{Before concluding, test the final option} \\[5ex] \text{Option D.} \\[3ex] d(x) = 0.26x^2 - 3.11x + 1 \\[3ex] d(1) = 0.26(1)^2 - 3.11(1) + 1 \ne -2.85 \\[3ex] \text{Discard} \\[5ex] \text{Just to confirm, test another value for Option C.} \\[3ex] d(11.5) = 0.26(11.5)^2 - 3.11(11.5) = -1.38 \\[3ex] \text{Option C. is the correct answer} $

$ (5rt - 3rw - 8tw) + (6rt - 4rw + 2tw) \\[3ex] 5rt - 3rw - 8tw + 6rt - 4rw + 2tw \\[3ex] 5rt + 6rt - 3rw - 4rw - 8tw + 2tw \\[3ex] 11rt - 7rw - 6tw $

$ \underline{\text{Zeros and Factors}} \\[3ex] x = -7 \implies x + 7 \\[3ex] x = 7 \implies x - 7 \\[5ex] \underline{\text{Factored Form}} \\[3ex] p(x) = (x + 7)(x - 7) \\[5ex] \underline{\text{Standard Form}} \\[3ex] p(x) = x^2 - 49 ...\text{Difference of Two Squares} $

$ \text{Point 1: } (-2, 6) \\[3ex] x_1 = -2 \\[3ex] y_1 = 6 \\[5ex] \text{Point 2: } (1, -4) \\[3ex] x_2 = 1 \\[3ex] y_2 = -4 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-4 - 6}{1 - (-2)} \\[5ex] m = -\dfrac{10}{3} \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 6 = -\dfrac{10}{3}[x - (-2)] \\[5ex] y - 6 = -\dfrac{10}{3}(x + 2) \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = -\dfrac{10}{3}(x + 2) + 6 \\[5ex] y = -\dfrac{10}{3}x - \dfrac{20}{3} + \dfrac{18}{3} \\[5ex] y = -\dfrac{10}{3}x - \dfrac{2}{3} $

$ \underline{\text{Exponential Function}} \\[3ex] v(x) = ab^x \\[3ex] a = \text{initial value} \\[3ex] b = \text{base} \\[5ex] \text{Point 1: } (0, 9000) \\[3ex] x = 0 \\[3ex] v(x) = 9000 \\[3ex] 9000 = a * b^{0} \\[3ex] 9000 = a * 1 \\[3ex] a = 9000 \\[5ex] \text{Point 2: } (1, 8100) \\[3ex] x = 1 \\[3ex] v(x) = 8100 \\[3ex] 8100 = a * b^{1} \\[3ex] 8100 = 9000 * b \\[3ex] b = \dfrac{8100}{9000} \\[5ex] b = 0.9 \\[5ex] \implies \\[3ex] v(x) = 9,000(0.9)^x $

$ x^2 + 9x - 22 = 0 \\[3ex] \text{Factors are } 11 \text{ and } -2 \\[3ex] (x + 11)(x - 2) = 0 \\[3ex] x + 11 \text{ or } x - 2 = 0 ...\text{Zero Product Property} \\[3ex] x = -11 \text{ or } x = 2 \\[3ex] \text{Positive Solution} \\[3ex] x = 2 \\[3ex] x = +2 $

x = the number of monitors
y = the number of keyboards

Each monitor will cost $250 ⇒
Cost of x monitors @ $250 per x = 250x

Each keyboard will cost $50 ⇒
Cost of y monitors @ $50 per y = 50y

At most $4,500 to buy monitors and keyboards for a computer lab ⇒
250x + 50y ≤ $4,500

$ \underline{\text{Linear Function}} \\[3ex] \text{slope} = m \\[3ex] \text{y-intercept} = b = 100 \\[5ex] \text{Point 1: } (0, 100) \\[3ex] x_1 = 0 \\[3ex] y_1 = 100 \\[5ex] \text{Point 2: } (5, 34) \\[3ex] x_2 = 5 \\[3ex] y_2 = 34 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{34 - 100}{5 - 0} \\[5ex] m = \dfrac{-66}{5} \\[5ex] m = -13.5 \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = mx + b \\[3ex] y = -13.5x + 100 \\[3ex] $ The closest answer (the function that best models the data) is: $y = -13.5x + 97.8$

$ f(x) = x^2 - 36 \\[3ex] \text{Zeros of } f(x) \\[3ex] x^2 - 36 = 0 \\[3ex] x^2 - 6^2 = 0 \\[3ex] (x + 6)(x - 6) = 0...\text{Difference of Two Squares} \\[3ex] x + 6 = 0 \text{ or } x - 6 = 0 ...\text{Zero Product Property} \\[3ex] x = -6 \text{ or } x = 6 \\[3ex] $ The zeros, –6 and 6, can be found when $0 = (x + 6)(x - 6)$

$ f(x) = 7 - 4x \\[3ex] f(-5) = 7 - 4(-5) \\[3ex] = 7 + 20 \\[3ex] = 27 $

We can solve this question using at least two approaches.
Use any approach you prefer.

$ \text{1st Approach: Slope–Intercept Form} \\[3ex] y = mx + b \\[3ex] \text{slope} = m \\[3ex] \text{y-intercept} = b \\[5ex] y = -4(x + 3) - 2 \\[3ex] y = -4x - 12 - 2 \\[3ex] y = -4x - 14 \\[3ex] m = -4 \\[3ex] b = -14 \\[3ex] $ The slope is negative.
Options A. and B. are discarded.

Options C. and D. remain.
We cannot verify the y-intercept.
So, let us verify a point.

$ y = -4x - 14 \\[3ex] \text{When } y = 2 \\[3ex] 2 = -4x - 14 \\[3ex] 4x = -14 - 2 \\[3ex] 4x = -16 \\[3ex] x = -\dfrac{16}{4} \\[5ex] x = -4 \\[3ex] $ Option D. is discarded.
Option C. is the answer.

$ \text{2nd Approach: Tranformation of Functions} \\[3ex] \underline{\text{Parent Function}} \\[3ex] y = x \\[3ex] \text{Coordinate} = (0, 0) \\[5ex] \underline{\text{Child Function}} \\[3ex] y = -4(x + 3) - 2 \\[5ex] \underline{\text{Transformations}} \\[3ex] x + 3 \implies \text{Horizontal Shift 3 units to the left} \\[3ex] (0, 0) \rightarrow (0 - 3, 0) \\[3ex] \hspace{2.5em} \rightarrow (-3, 0) \\[5ex] 4(x + 3) \implies \text{Vertical Shift by a factor of 4 units} \\[3ex] (-3, 0) \rightarrow (-3, 4 * 0) \\[3ex] \hspace{3em} \rightarrow (-3, 0) \\[5ex] -4(x + 3) \implies \text{Vertical Reflection} \\[3ex] (-3, 0) \rightarrow (-3, -1 * 0) \\[3ex] \hspace{3em} \rightarrow (-3, 0) \\[5ex] -4(x + 3) - 2 \implies \text{Vertical Shift 2 units down} \\[3ex] (-3, 0) \rightarrow (-3, 0 - 2) \\[3ex] \hspace{3em} \rightarrow (-3, -2) \\[3ex] $ Which of the graphs have a line that passes through $(-3, -2)$?
Options A. and D. are discarded because the lines do not pass through that point.
Options B. and C. remain.
Let us test another coordinate.

$ \text{Coordinate on Parent Function} = (1, 1) \\[5ex] \text{Horizontal Shift 3 units to the left} \\[3ex] (1, 1) \rightarrow (1 - 3, 1) \\[3ex] \hspace{2.5em} \rightarrow (-2, 1) \\[5ex] \text{Vertical Shift by a factor of 4 units} \\[3ex] (-2, 1) \rightarrow (-2, 4 * 1) \\[3ex] \hspace{3em} \rightarrow (-2, 4) \\[5ex] \text{Vertical Reflection} \\[3ex] (-2, 4) \rightarrow (-2, -1 * 4) \\[3ex] \hspace{3em} \rightarrow (-2, -4) \\[5ex] \implies \text{Vertical Shift 2 units down} \\[3ex] (-2, -4) \rightarrow (-2, -4 - 2) \\[3ex] \hspace{3em} \rightarrow (-2, -6) \\[3ex] $ Which of the graphs in Options B. and C. have a line that passes through $(-3, -2)$?
Option B. is discarded.
Option C. is the correct answer.

$ 10x^2 - 19x + 6 \\[3ex] 10x^2 * 6 = 60x^2 \\[3ex] \text{Factors are } -4x \text{ and } -15x \\[3ex] 10x^2 - 4x - 15x + 6 \\[3ex] 2x(5x - 2) - 3(5x - 2) \\[3ex] (5x - 2)(2x - 3) $

x = Time (seconds)
y = Distance Below Sea Level (feet)

The rate of change of of the distance in feet below sea level with respect to time is the slope of the graph.

$ \text{Point 1: } (0, 460) \\[3ex] x_1 = 0 \\[3ex] y_1 = 460 \\[5ex] \text{Point 2: } (70, 1020) \\[3ex] x_2 = 70 \\[3ex] y_2 = 1020 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{1020 - 460}{70 - 0} \\[5ex] m = -\dfrac{560}{70} \\[5ex] m = 8 \\[3ex] m = +8 \\[3ex] $ The rate of change of of the distance in feet below sea level with respect to time that the submarine traveled is 8 feet per seconds.

The equation of the line is: $y = -6$

$ g(x) = 18(1.3)^x \\[3ex] \text{This is an Exponential Growth Function because } 1.3 \gt 1 \\[5ex] g(x) = 18(1 + 0.3)^x \\[5ex] \underline{\text{General Form of an Exponential Growth Function}} \\[3ex] g(x) = a(1 + r)^x \\[3ex] \text{Compare} \\[3ex] a = \text{initial value} = 18 \\[3ex] r = \text{growth rate} = 0.3 = 30\% \\[3ex] $ The insect population increased at a rate of 30% each month.

The line is a dashed line (because there is no equality sign).
The area above the line is shaded. (because of the greater than sign)
An ordered pair in the shaded region (the region above the line) is in the solution set of the inequality.
Let us analyze each option.

Option A.
$(-8, 8)$ is in the region above the line.
This is the correct answer.
But, let us analyze the remaining options if you wish.

Option B.
$(6, -5)$ is in the line.
This is incorrect because the line is a dashed line.

Option C.
$(4, -6)$ is in the region below the line.

Option D.
$(-2, -7)$ is in the region below the line.

$ \underline{\text{Linear Function}} \\[3ex] \text{slope} = m \\[3ex] \text{y-intercept} = b \\[5ex] \text{Point 1: } (0, 32) \\[3ex] x_1 = 0 \\[3ex] y_1 = 32 \\[5ex] \text{Point 2: } (40, 104) \\[3ex] x_2 = 40 \\[3ex] y_2 = 104 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{104 - 32}{40 - 0} \\[5ex] m = \dfrac{72}{40} \\[5ex] m = \dfrac{9}{5} \\[5ex] b = 32 \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = mx + b \\[3ex] y = \dfrac{9}{5}x + 32 \\[3ex] $

$ \underline{\text{Inequality 1}} \\[3ex] \text{This is a dashed line because of the absence of the equality sign} \\[3ex] x + 2y \lt -2 \\[5ex] \text{When } x = 0 \\[3ex] 0 + 2y \lt -2 \\[3ex] 2y \lt -2 \\[3ex] y \lt -\dfrac{2}{2} \\[5ex] y \lt -1 \\[5ex] \text{When } y = 0 \\[3ex] x + 2(0) \lt -2 \\[3ex] x + 0 \lt -2 \\[3ex] x \lt -2 \\[5ex] \underline{\text{Inequality 2}} \\[3ex] \text{This is a dashed line because of the absence of the equality sign} \\[3ex] y - x \lt 3 \\[5ex] \text{When } x = 0 \\[3ex] y - 0 \lt 3 \\[3ex] y \lt 3 \\[5ex] \text{When } y = 0 \\[3ex] 0 - x \lt 3 \\[3ex] -x \lt 3 \\[3ex] x \gt \dfrac{3}{-1} \\[5ex] x \gt -3 \\[3ex] $ To graph a system of inequalities, each inequality is shaded and the solution is where the shaded regions overlap.
The correct answer is Option C.

$ \text{Zeros of } h(x) \\[3ex] h(x) = (x + 1)(x - 3) = 0 \\[3ex] x + 1 = 0 \text{ or } x - 3 = 0 \\[3ex] x = -1 \text{ or } x = 3 \\[3ex] $ Options B. and D. are discarded.
Options A. and C. remain.

$ \text{Vertex of } h(x) \\[3ex] h(x) = (x + 1)(x - 3) \\[3ex] h(x) = x^2 - 3x + x - 3 \\[3ex] h(x) = x^2 - 2x - 3 \\[3ex] \text{Compare to the standard form: } ax^2 + bx + c = 0 \\[3ex] a = 1 \\[3ex] b = -2 \\[3ex] x-\text{coordinate of the Vertex} = -\dfrac{b}{2a} \\[5ex] = \dfrac{-(-2)}{2(1)} \\[5ex] = \dfrac{2}{2} \\[5ex] = 1 \\[3ex] y-\text{coordinate of the Vertex} = h(1) \\[3ex] = 1^2 - 2(1) - 3 \\[3ex] = 1 - 2 - 3 \\[3ex] = -4 \\[3ex] \text{Vertex} = (1, -4) \\[3ex] $ Option C. is discarded.
Option A. is the correct answer.

There are at least two approaches to solving this question.
Use any approach you prefer.

$ \underline{\text{1st Approach: Geometric Sequence}} \\[3ex] a_n = ar^{n - 1} \\[4ex] \text{where} \\[3ex] a = \text{first term} = -4 \\[3ex] r = \text{common ratio} = \dfrac{-16}{-4} = 4 \\[5ex] n = \text{number of terms} \\[3ex] a_n = \text{nth term} \\[3ex] \implies \\[3ex] a_n = -4 * 4^{n - 1} \\[4ex] = -1 * 4^1 * 4^{n - 1} ...\text{Law 4 Exp} \\[4ex] = -1 * 4^{1 + n - 1} ...\text{Law 1 Exp} \\[4ex] = -1 * 4^n \\[3ex] = -(4^n) \\[3ex] = -(4)^n \\[3ex] $ For the 2nd Approach, we shall test the options and discard the incorrect ones.
The question tells us that that the sequence is a geometric sequence (also known as an exponential sequence).
In that regard,
Option A. is incorrect because it is a linear function.
Option C. is incorrect because it is a quadratic function.
Discard Options A. and C.

$ \text{Testing Option B.} \\[3ex] a_n = -(4)^n \[3ex] a_1 = -(4)^1 \\[3ex] a_1 = -4...\text{okay} \\[5ex] a_2 = -(4)^2 \\[3ex] a_2 = -16 ...\text{okay} \\[5ex] a_3 = -(4)^3 \\[3ex] a_3 = -64 ...\text{okay} \\[3ex] $ Option B. is the correct answer.
However, let us test Option D. just to be sure it is incorrect.

$ \text{Testing Option D.} \\[3ex] a_n = (-4)^n \[3ex] a_1 = (-4)^1 \\[3ex] a_1 = -4...\text{okay} \\[5ex] a_2 = (-4)^2 \\[3ex] a_2 = 16 \ne -16 \\[3ex] $ This is incorrect.
The 2nd term should be –16 rather than 16.

$ 4(q + 56.5) = 30q - 112 \\[3ex] 4q + 226 = 30q - 112 \\[3ex] 226 + 112 = 30q - 4q \\[3ex] 338 = 26q \\[3ex] 26q = 338 \\[3ex] q = \dfrac{338}{26} \\[5ex] q = 13 \\[3ex] q = +13 $

$ 36m^2 - 100 \\[4ex] 4(9m^2 - 25) \\[4ex] 4(3^2m^2 - 5^2) \\[4ex] 4[(3m)^2 - 5^2] \\[4ex] 4(3m + 5)(3m - 5) ...\text{Difference of Two Squares} $

If the data follow a smooth exponential pattern and the initial value is included, then using the initial value and one other point (such as the last point) can produce a reasonable exponential function model.
However, if the data fluctuate or do not follow a consistent exponential pattern, an exponential regression should be used to obtain the best-fit model.
Just so you know, the exponential function model uses only two points, whereas the exponential regression uses all the points.
In that regard, the exponential regression gives the best model.

For this question, we are not given an initial value.
Hence, it is necessary to use an exponential regression.
But, we are not given the exponential regression formulas.
So, we shall use the exponential function model and find the option that most closely resembles the model.

$ \underline{\text{Exponential Function}} \\[3ex] r(x) = ab^x \\[3ex] a = \text{initial value} \\[3ex] b = \text{base} \\[5ex] \text{Point 1: } (3, 274) \\[3ex] x = 3 \\[3ex] r(x) = 274 \\[3ex] 274 = a * b^{3} \\[3ex] ab^3 = 274...eqn.(1) \\[5ex] \text{Point 2: } (24, 1584) \\[3ex] x = 24 \\[3ex] r(x) = 1584 \\[3ex] 1584 = a * b^{24} \\[3ex] ab^{24} = 1584...eqn.(2) \\[5ex] eqn.(2) \div eqn.(1) \implies \\[3ex] \dfrac{ab^{24}}{ab^3} = \dfrac{1584}{274} \\[5ex] b^{24 - 3} = \dfrac{1584}{274} \\[5ex] b^{21} = \dfrac{1584}{274} \\[5ex] b = \sqrt[21]{\dfrac{1584}{274}} \\[5ex] b = 1.087141148 \\[3ex] b \approx 1.09...\text{to 2 decimal places} \\[5ex] \text{Substitute for } b \text{ in } eqn. (1) \\[3ex] a = \dfrac{274}{b^3} \\[5ex] a = \dfrac{274}{1.087141148^3} \\[5ex] a = 213.2518271 \\[3ex] a \approx 213.25 ...\text{to 2 decimal places} \\[5ex] \implies \\[3ex] r(x) \approx 223.06(1.09)^x \\[3ex] $ The function that closely resembles it (the function that best models the data) is the function in Option A.

$ \text{Equation 1} \\[3ex] 2x = 6 - y \\[3ex] \text{When } x = 0 \\[3ex] 2(0) = 6 - y \\[3ex] 0 = 6 - y \\[3ex] y = 6 - 0 \\[3ex] y = 6 \\[3ex] (x, y) = (0, 6) \\[5ex] \text{When } y = 0 \\[3ex] 2x = 6 - 0 \\[3ex] 2x = 6 \\[3ex] x = \dfrac{6}{2} \\[5ex] x = 3 \\[3ex] (x, y) = (3, 0) \\[5ex] \text{Equation 2} \\[3ex] 5x - 4y = 28 \\[3ex] \text{When } x = 0 \\[3ex] 5(0) - 4y = 28 \\[3ex] -4y = 28 \\[3ex] y = \dfrac{28}{-4} \\[5ex] y = -7 \\[3ex] (x, y) = (0, -7) \\[5ex] \text{When } y = 3 \\[3ex] 5x - 4(3) = 28 \\[3ex] 5x - 12 = 28 \\[3ex] 5x = 28 + 12 \\[3ex] 5x = 40 \\[3ex] x = \dfrac{40}{5} \\[5ex] x = 8 \\[3ex] (x, y) = (8, 3) \\[3ex] $ The correct answer is Option B.

$ k(x) = x^2 + 2x - 15 \\[3ex] \text{Factors are: } 5 \text{ and } -3 \\[3ex] k(x) = (x + 5)(x - 3) $

The domain is the set of all the input values, x for which the function has an output.
Let us analyze each option and discard the incorrect ones.

Option A.
The domain is the set of all real numbers less than 0
Discard.

Option B.
The domain is the set of all real numbers greater than –4.
Discard.

Option C.
The domain is the set of all real numbers less than –4
This is the correct option.

Option D.
The domain is the set of all real numbers greater than 0
Discard.

$ \text{Point 1: } (-3, 1) \\[3ex] x_1 = -3 \\[3ex] y_1 = 1 \\[5ex] \text{Point 2: } (5, 8) \\[3ex] x_2 = 5 \\[3ex] y_2 = 8 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{8 - 1}{5 - (-3)} \\[5ex] m = \dfrac{7}{8} $

This is an example of a linear system.
Let the distance:
on the morning route = c
on the afternoon route = d

$ \underline{\text{Last Month Distance}} \\[3ex] \text{Morning Route: } 16 \times c = 16c \\[3ex] \text{Afternoon Route: } 12 \times d = 12d \\[3ex] \text{Total Distance} = 141 \text{ miles} \\[3ex] \implies \\[3ex] 16c + 12d = 141 ...eqn.(1) \\[5ex] \underline{\text{This Month Distance}} \\[3ex] \text{Morning Route: } 10 \times c = 10c \\[3ex] \text{Afternoon Route: } 15 \times d = 15d \\[3ex] \text{Total Distance} = 123.75\text{ miles} \\[3ex] \implies \\[3ex] 10c + 15d = 123.75 ...eqn.(2) \\[5ex] \underline{\text{Elimination Method}} \\[3ex] \text{To find } c, \text{ eliminate } d \\[3ex] 15 * eqn.(1) \implies \\[3ex] 240c + 180d = 2115 ...eqn.(3) \\[5ex] 12 * eqn.(2) \implies \\[3ex] 120c + 180d = 1485 ...eqn.(4) \\[5ex] eqn.(3) - eqn.(4) \implies \\[3ex] 120c = 630 \\[3ex] c = \dfrac{630}{120} \\[5ex] c = 5.25 \\[3ex] $ The distance of the morning route is 5.25 miles.

$ \text{Graph of } p \\[3ex] \text{Parent Function: } p(x) = x^2 \\[3ex] \text{Vertex} = (0, 0) \\[5ex] \text{Graph of } q \\[3ex] \text{Transformations of Parent Function to give Child Function} \\[3ex] \text{1st: Vertical Reflection} \\[3ex] q(x) = -x^2 \\[3ex] (0, 0) \rightarrow (0, -1 * 0) \\[3ex] \hspace{2em} \rightarrow (0, 0) \\[5ex] \text{2nd: Vertical Shift 2 units up} \\[3ex] q(x) = -x^2 + 2 \\[3ex] (0, 0) \rightarrow (0, 0 + 2) \\[3ex] \hspace{2em} \rightarrow (0, 2) \\[3ex] $ The function that best represents the graph of q is $q(x) = -x^2 + 2$

$ 2(40 - 5y) = 10y + 5(1 - y) \\[3ex] 80 - 10y = 10y + 5 - 5y \\[3ex] 80 - 5 = 10y - 5y + 10y \\[3ex] 75 = 15y \\[3ex] 15y = 75 \\[3ex] y = \dfrac{75}{15} \\[5ex] y = 5 $

The initial value is $200,000
Options B. and D. are discarded because the initial value for those graphs is $100,000
Options A. and C. remain.

$ \underline{\text{Exponential Growth Function}} \\[3ex] y = a(1 + r)^x \\[3ex] a = \text{initial value} = \$200,000 \\[3ex] r = \text{growth rate} = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] \implies \\[3ex] y = 200000(1 + 0.06)^x \\[3ex] y = 200000(1.06)^x \\[5ex] \text{When } x = 5 \text{ years} \\[3ex] y = 200000(1.06)^5 \\[3ex] y = \$267645.1155 \\[3ex] $ Option C. is discarded.
Option A. is the correct answer.

The range is the set of all the output values, y produced by the input values of the function.
From the graph, the range is $\{96, 88, 80, 72, 64, 56, 48\}$

Let us analyze each option.

Option A.
Zero: x = 3
y-intercept = 6 Discard.

Option B.
Zero: x = –2
y-intercept = 6 This is the correct answer.

Option C.
Zero: x = 6
y-intercept = –2 Discard.

Option D.
Zero: x = –6
y-intercept = –2 Discard.

$ m = \text{slope} \\[3ex] b = y-\text{intercept} = 4 \\[5ex] \text{Point 1: } (0, 4) \\[3ex] x_1 = 0 \\[3ex] y_1 = 4 \\[5ex] \text{Point 2: } (6, 0) \\[3ex] x_2 = 6 \\[3ex] y_2 = 0 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{0 - 4}{6 - 0} \\[5ex] m = \dfrac{-4}{6} \\[5ex] m = -\dfrac{2}{3} \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = mx + b \\[3ex] y = -\dfrac{2}{3}x + 4 \\[5ex] g(x) = -\dfrac{2}{3}x + 4 $

The rate of change of y with respect to x is the slope of teh function.

$ \text{Point 1: } (-8.5, -92) \\[3ex] x_1 = -8.5 \\[3ex] y_1 = -92 \\[5ex] \text{Point 2: } (-1, -17) \\[3ex] x_2 = -1 \\[3ex] y_2 = -17 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-17 - (-92)}{-1 - (-8.5)} \\[5ex] m = \dfrac{75}{7.5} \\[5ex] m = 10 $

$ g(x) = x^2 - 6x - 16 \\[3ex] \text{Factors are } 2 \text{ and } -8 \\[3ex] g(x) = (x + 2)(x - 8) ...\text{Factored Form} \\[5ex] \underline{\text{Zeros}} \\[3ex] g(x) = (x + 2)(x - 8) = 0 \\[3ex] x + 2 = 0 \text{ or } x - 8 = 0 \\[3ex] x = -2 \text{ or } x = 8 $

There are at least two approaches to solving this question.
Use any approach you prefer.

1st Approach: Algebraic Solution
This involves checking each option to see the one that satisfies the inequality.

$ 8x + 16y \gt 32 \\[5ex] \text{Option A.} \\[3ex] (0, 2) \\[3ex] x = 0 \\[3ex] y = 2 \\[3ex] 8(0) + 16(2) \stackrel{?}{\gt} 32 \\[3ex] 0 + 32 \stackrel{?}{\gt} 32 \\[3ex] 32 = 32 \\[3ex] 32 \ngtr 32 \\[3ex] \text{Discard}. \\[5ex] \text{Option B.} \\[3ex] (-3, 5) \\[3ex] x = -3 \\[3ex] y = 5 \\[3ex] 8(-3) + 16(5) \stackrel{?}{\gt} 32 \\[3ex] -24 + 80 \stackrel{?}{\gt} 32 \\[3ex] 56 \gt 32 \\[3ex] \text{This is the correct answer}. \\[5ex] \text{Option C.} \\[3ex] (-1, 1) \\[3ex] x = -1 \\[3ex] y = 1 \\[3ex] 8(-1) + 16(1) \stackrel{?}{\gt} 32 \\[3ex] -8 + 16 \stackrel{?}{\gt} 32 \\[3ex] 8 \ngtr 32 \\[3ex] \text{Discard}. \\[5ex] \text{Option D.} \\[3ex] (4, 0) \\[3ex] x = 4 \\[3ex] y = 0 \\[3ex] 8(4) + 16(0) \stackrel{?}{\gt} 32 \\[3ex] 32 + 0 \stackrel{?}{\gt} 32 \\[3ex] 32 = 32 \\[3ex] 32 \ngtr 32 \\[3ex] \text{Discard}. \\[5ex] $ 2nd Approach: Graphical Solution
They gave us a graph, so apparently, they want us to graph the inequality and use that graph in determining the solution.
Because the inequality does not contain an equal sign, the line will be a dashed line.

$ 8x + 16y \gt 32 \\[5ex] \text{When } x = 0 \\[3ex] 8(0) + 16y \gt 32 \\[3ex] 0 + 16y \gt 32 \\[3ex] 16y \gt 32 \\[3ex] y \gt \dfrac{32}{16} \\[5ex] y \gt 2 \\[5ex] \text{When } x = -2 \\[3ex] 8(-2) + 16y \gt 32 \\[3ex] -16 + 16y \gt 32 \\[3ex] 16y \gt 32 + 16 \\[3ex] 16y \gt 48 \\[3ex] y \gt \dfrac{48}{16} \\[5ex] y \gt 3 \\[5ex] \text{When } y = 0 \\[3ex] 8x + 16y \gt 32 \\[3ex] 8x + 16(0) \gt 32 \\[3ex] 8x \gt 32 \\[3ex] x \gt \dfrac{32}{8} \\[5ex] x \gt 4 \\[3ex] $ The graph of the inequality is:



An ordered pair in the shaded region is in the solution set of the inequality.
Let us analyze each option.

Option A.
$(0, 2)$ is on the dashed line.
Because it is a dashed line (not a solid line because there is no equal sign), this is outside the solution set.

Option B.
$(-3, 5)$ is inside the solution set.
But, let us analyze the remaining options if you wish.

Option C.
$(-1, 1)$ is outside the solution set.

Option D.
$(4, 0)$ is on the dashed line.
Because it is a dashed line (not a solid line because there is no equal sign), this is outside the solution set.

$ \text{Point 1: } (-26, -11) \\[3ex] x_1 = -26 \\[3ex] y_1 = -11 \\[5ex] \text{Point 2: } (39, 34) \\[3ex] x_2 = 39 \\[3ex] y_2 = 34 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{34 - (-11)}{39 - (-26)} \\[5ex] m = \dfrac{45}{65} \\[5ex] m = \dfrac{9}{13} \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - (-11) = \dfrac{9}{13}[x - (-26)] \\[5ex] y + 11 = \dfrac{9}{13}(x + 26) \\[5ex] y + 11 = \dfrac{9}{13}x + 18 \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = mx + b ...\text{where } b \text{ is the y-intercept} \\[3ex] y = \dfrac{9}{13}x + 18 - 11 \\[5ex] y = \dfrac{9}{13}x + 7 $

Let us analyze each option.

Option A.
Zero: x = –3
Axis of Symmetry: x = –3
This is the correct answer.

Option B.
Zeros: x = –6 and x = 0
Axis of Symmetry: x = –3
Discard.

Option C.
Zero: x = 0 and x = 6
Axis of Symmetry: x = 3
Discard.

Option D.
Zero: x = 3
Axis of Symmetry: x = 3
Discard.

$ (x^9yz^4)^5 \\[4ex] \text{DISSOCIATE} \\[3ex] (x^9 \cdot y \cdot z^4)^5 \\[4ex] \text{SOLVE} \\[3ex] x^{9 \cdot 5} \cdot y^{1 \cdot 5} \cdot z^{4 \cdot 5} ...\text{Law 5 Exp} \\[4ex] x^{45} \cdot y^5 \cdot x^{20} \\[3ex] \text{ASSOCIATE} \\[3ex] x^{45}y^5z^{20} $

$ m = \text{slope} \\[3ex] b = y-\text{intercept} = 1 \\[5ex] \text{Point 1: } (-7, -4) \\[3ex] x_1 = -7 \\[3ex] y_1 = -4 \\[5ex] \text{Point 2: } (7, 6) \\[3ex] x_2 = 7 \\[3ex] y_2 = 6 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{6 - (-4)}{7 - (-7)} \\[5ex] m = \dfrac{10}{14} \\[5ex] m = \dfrac{5}{7} $

$ 4(y - 3) + 19 = 8(2y + 3) + 7 \\[3ex] 4y - 12 + 19 = 16y + 24 + 7 \\[3ex] 4y + 7 = 16y + 31 \\[3ex] 7 - 31 = 16y - 4y \\[3ex] -24 = 12y \\[3ex] 12y = -24 \\[3ex] y = -\dfrac{24}{12} \\[5ex] y = -2 $

The correct answer is Option B.

Student: Just like that...
How do you know?
From the options?
Teacher: Yes, I know the answer from the options.
Recall: If the base of a basic exponential function is less than 1, then the exponential function is an
Student: Exponential Decay function
Yes, I see it now.
The other bases are greater than 1, hence they are Exponential Growth functions.
Teacher: That is correct.
Let us show our work anyway.

$ \underline{\text{Exponential Function}} \\[3ex] g(x) = ab^x \\[3ex] a = \text{initial value} \\[3ex] b = \text{base} \\[5ex] \text{Point 1: } (0, 25.5) \\[3ex] x = 0 \\[3ex] g(x) = 25.5 \\[3ex] 25.5 = a * b^{0} \\[3ex] 25.5 = a * 1 ...\text{Law 3 Exp} \\[3ex] a = 25.5 \\[5ex] \text{Point 2: } (2, 9.18) \\[3ex] x = 2 \\[3ex] g(x) = 9.18 \\[3ex] 9.18 = 25.5 * b^{2} \\[3ex] 25.5 * b^2 = 9.18 \\[3ex] b^2 = \dfrac{9.18}{25.5} \\[5ex] b^2 = 0.36 \\[3ex] b = \sqrt{0.36} \\[3ex] b = 0.6 \\[5ex] \implies \\[3ex] g(x) = 25.5(0.6)^x $

Range = Distance from Ground (meters)
Minimum value = 0
Maximum value = 7.66
The range is: C. All real numbers greater than or equal to 0 and less than or equal to 7.66.

$ \underline{\text{Inequality 1}} \\[3ex] \text{This is a dashed line because of the absence of the equality sign} \\[3ex] x + y \lt 1 \\[5ex] \text{When } x = 0 \\[3ex] 0 + y \lt 1 \\[3ex] y \lt 1 \\[5ex] \text{When } y = 0 \\[3ex] x + 0 \lt 1 \\[3ex] x \lt 1 \\[5ex] \underline{\text{Inequality 2}} \\[3ex] \text{This is a solid line because of the presence of the equality sign} \\[3ex] x - y \le 2 \\[5ex] \text{When } x = 0 \\[3ex] 0 - y \le 2 \\[3ex] -y \le 2 \\[3ex] y \ge \dfrac{2}{-1} \\[5ex] y \ge -2 \\[5ex] \text{When } y = 0 \\[3ex] x - 0 \le 2 \\[3ex] x \le 2 \\[3ex] $ To graph a system of inequalities, each inequality is shaded and the solution is where the shaded regions overlap.
The correct answer is Option A.

$ p(x) = -4(x - 15)^2 + 2 \\[3ex] p(7) = -4(7 - 15)^2 + 2 \\[3ex] = -4(-8)^2 + 2 \\[3ex] = -4(64) + 2 \\[3ex] = -256 + 2 \\[3ex] = -254 $

b = the number of black-and-white copies
c = the number of color copies

A customer paid a total of $6.00 for 68 copies at a print shop.
Some of the copies were black-and-white copies, and the rest were color copies.
Number of copies:
$ b + c = 68 \\[3ex] $ Each black-and-white copy cost $0.08.
b copies @ $0.08 per copy = 0.08b

Each color copy cost $0.15.
c copies @ $0.15 per copy = 0.15c

Cost of copies:
$ 0.08b + 0.15c = 6 $

We can solve the question using at least two approaches.
Use any approach you prefer.

$ a = \text{leading coefficient} \\[5ex] \underline{\text{1st Approach: Equation of a Quadratic Function in Factored Form}} \\[3ex] f(x) = a(x - \alpha)(x - \beta) \\[3ex] \alpha, \beta = \text{zeros of the function} \\[3ex] \alpha = -4 \\[3ex] \beta = 4 \\[3ex] \implies \\[3ex] f(x) = a[x - (-4)](x - 4) \\[3ex] f(x) = a(x + 4)(x - 4) \\[3ex] f(x) = a(x^2 - 4^2)...\text{Difference of Two Squares} \\[4ex] f(x) = a(x^2 - 16) \\[5ex] \text{Passes through the point (vertex) } (0, 8) \\[3ex] x = 0 \\[3ex] f(x) = 8 \\[3ex] \implies \\[3ex] 8 = a(0^2 - 16) \\[3ex] 8 = a(0 - 16) \\[3ex] 8 = -16a \\[3ex] -16a = 8 \\[3ex] a = \dfrac{8}{-16} \\[5ex] a = -\dfrac{1}{2} \\[5ex] \implies \\[3ex] f(x) = -\dfrac{1}{2}(x^2 - 16) \\[5ex] f(x) = -\dfrac{1}{2}x^2 + 8 \\[5ex] \underline{\text{2nd Approach: Equation of a Quadratic Function in Vertex Form}} \\[3ex] f(x) = a(x - h)^2 + k ...\text{where } (h, k) \text{ is the vertex} \\[3ex] (h, k) = (0, 8) \\[3ex] h = 0 \\[3ex] k = 8 \\[3ex] \implies \\[3ex] f(x) = a(x - 0)^2 + 8 \\[3ex] f(x) = ax^2 + 8 \\[5ex] \text{Passes through the point: } (4, 0) \\[3ex] x = 4 \\[3ex] f(x) = 0 \\[3ex] \implies \\[3ex] 0 = a(4)^2 + 8 \\[3ex] 0 = 16a + 8 \\[3ex] 16a + 8 = 0 \\[3ex] 16a = -8 \\[3ex] a = \dfrac{-8}{16} \\[5ex] a = -\dfrac{1}{2} \\[5ex] \implies \\[3ex] f(x) = -\dfrac{1}{2}x^2 + 8 $

$ \text{Point 1: } (-7, 9) \\[3ex] x_1 = -7 \\[3ex] y_1 = 9 \\[5ex] \text{Point 2: } (8, -15) \\[3ex] x_2 = 8 \\[3ex] y_2 = -15 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-15 - 9}{8 - (-7)} \\[5ex] m = \dfrac{-24}{15} \\[5ex] m = -\dfrac{8}{5} $

$ 2x^2 - 25x + 63 \\[3ex] 2x^2 * 63 = 126x^2 \\[3ex] \text{Factors are: } -18x \text{ and } -7x \\[3ex] \implies \\[3ex] 2x^2 - 18x - 7x + 63 \\[3ex] 2x(x - 9) - 7(x - 9) \\[3ex] (x - 9)(2x - 7) $

$ m = \text{slope} \\[3ex] b = y-\text{intercept} \\[3ex] A, B, C = \text{constants} \\[3ex] \text{Standard Form:} Ax + By = C ...\text{where } A \gt 0 \\[3ex] \text{Slope–Intercept Form: } y = mx + b \\[3ex] \text{Point–Slope Form: } y - y_1 = m(x - x_1) \\[5ex] \underline{\text{Equation 1}} \\[3ex] 8x + 3y = 15 \\[3ex] 3y = -8x + 15 \\[3ex] y = -\dfrac{8}{3}x + \dfrac{15}{3} \\[5ex] y = -\dfrac{8}{3}x + 5...\text{Slope–Intercept Form} \\[5ex] m_1 = -\dfrac{8}{3} \\[5ex] \underline{\text{Equation 2}} \\[3ex] \text{Line 2 } || \text{ Line 1} \\[3ex] m_2 = m_1 = -\dfrac{8}{3} ...\text{Parallel lines have the same slope} \\[5ex] \text{Passes through the point } (6, -1) \\[3ex] x_1 = 6 \\[3ex] y_1 = -1 \\[5ex] \text{Point–Slope Form} \\[3ex] y - (-1) = -\dfrac{8}{3}(x - 6) \\[5ex] y + 1 = -\dfrac{8}{3}x + 16 \\[5ex] y = -\dfrac{8}{3}x + 16 - 1 \\[5ex] y = -\dfrac{8}{3}x + 15 ...\text{Slope–Intercept Form} \\[5ex] \text{LCD} = 3 \\[3ex] \text{Multiply each term by the LCD} \\[3ex] 3(y) = 3\left(-\dfrac{8}{3}x\right) + 3(15) \\[5ex] 3y = -8x + 45 \\[3ex] 3y + 8x = 45 \\[3ex] 8x + 3y = 45 ...\text{Standard Form} $

$ \underline{\text{Linear Function}} \\[3ex] \text{slope} = m \\[3ex] \text{y-intercept} = b = 0 \\[5ex] \text{Point 1: } (-5, -4) \\[3ex] x_1 = -5 \\[3ex] y_1 = -4 \\[5ex] \text{Point 2: } (10, 8) \\[3ex] x_2 = 10 \\[3ex] y_2 = 8 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{8 - (-4)}{10 - (-5)} \\[5ex] m = \dfrac{12}{15} \\[5ex] m = \dfrac{4}{5} \\[5ex] \text{Slope–Intercept Form} \\[3ex] h(x) = mx + b \\[3ex] h(x) = \dfrac{4}{5}x + 0 \\[5ex] h(x) = \dfrac{4}{5}x \\[5ex] f(x) = x \\[3ex] h(x) = af(x) \\[3ex] h(x) = a * f(x) \\[3ex] \implies \\[3ex] \dfrac{4}{5} * x = a * x \\[5ex] a = \dfrac{4}{5} \\[5ex] a = 0.8 \\[3ex] a = +0.8 $

If the data follow a smooth exponential pattern and the initial value is included, then using the initial value and one other point (such as the last point) can produce a reasonable exponential function model.
However, if the data fluctuate or do not follow a consistent exponential pattern, an exponential regression should be used to obtain the best-fit model.
Just so you know, the exponential function model uses only two points, whereas the exponential regression uses all the points.
In that regard, the exponential regression gives the best model.

For this question, we are not given an initial value.
Hence, it is necessary to use an exponential regression.
But, we are not given the exponential regression formulas.
So, we shall use the exponential function model and find the option that most closely resembles the model.

$ \underline{\text{Exponential Function}} \\[3ex] n(x) = ab^x \\[3ex] a = \text{initial value} \\[3ex] b = \text{base} \\[5ex] \text{Point 1: } (1, 46125) \\[3ex] x = 1 \\[3ex] n(x) = 46125 \\[3ex] 46125 = a * b^{1} \\[3ex] ab = 46125...eqn.(1) \\[5ex] \text{Point 2: } (7, 27691) \\[3ex] x = 7 \\[3ex] n(x) = 27691 \\[3ex] 27691 = a * b^{7} \\[4ex] ab^7 = 27691...eqn.(2) \\[5ex] eqn.(2) \div eqn.(1) \implies \\[3ex] \dfrac{ab^7}{ab} = \dfrac{27691}{46125} \\[5ex] b^{7 - 1} = 0.6003468835...\text{Law 2 Exp} \\[4ex] b^6 = 0.6003468835 \\[4ex] b = \sqrt[6]{0.6003468835} \\[3ex] b = 0.9184743733 \\[5ex] \text{Substitute for } b \text{ in } eqn.(1) \\[3ex] a * 0.9184743733 = 46125 \\[3ex] a = \dfrac{46125}{0.9184743733} \\[5ex] a = 50219.14747 \\[5ex] \implies \\[3ex] n(x) = 50219.14747(0.9184743733)^x \\[4ex] $ The function that closely resembles it (the function that best models the data) is the function in Option B.

The closed circle represents a closed interval.

The domain is the set of all the input values, x for which the function has an output.
On the graph, it is from –7 to 6
$\text{Domain: } -7 \le x \le 6$

The range of a function is the set of all output values that result from applying the function to its input values.
On the graph, it is from –4 to 5
$\text{Range: } -4 \le g(x) \le 5$

$ 3x^2 - 30x - 72 = 0 \\[3ex] 3(x^2 - 10x - 24) = 0 \\[3ex] \text{Divide both sides by 3} \\[3ex] x^2 - 10x - 24 = 0 \\[3ex] \text{Factors are: } 2 \text{ and } -12 \\[3ex] \implies \\[3ex] (x + 2)(x - 12) = 0 \\[3ex] \underline{\text{Apply the Zero Product Property}} \\[3ex] x + 2 = 0 \text{ or } x - 12 = 0 \\[3ex] x = -2 \text{ or } x = 12 $

$ \text{slope} = m \\[5ex] \underline{\text{Line 1}} \\[3ex] \text{Point 1: } (2, 7) \\[3ex] x_1 = 2 \\[3ex] y_1 = 7 \\[5ex] \text{Point 2: } (11, 22.75) \\[3ex] x_2 = 11 \\[3ex] y_2 = 22.75 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{22.75 - 7}{11 - 2} \\[5ex] m = \dfrac{15.75}{9} \\[5ex] m = 1.75 \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 7 = 1.75(x - 2) \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = 1,75x - 3.5 + 7 \\[3ex] y = 1.75x + 3.5 \\[5ex] \underline{\text{Line 2}} \\[3ex] \text{Point 1: } (-3, 4) \\[3ex] x_1 = -3 \\[3ex] y_1 = 4 \\[5ex] \text{Point 2: } (4, -24) \\[3ex] x_2 = 4 \\[3ex] y_2 = -24 \\[5ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{-24 - 4}{4 - (-3)} \\[5ex] m = \dfrac{-28}{7} \\[5ex] m = -4 \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 4 = -4[x - (-3)] \\[3ex] y - 4 = -4(x + 3) \\[5ex] \text{Slope–Intercept Form} \\[3ex] y = -4x - 12 + 4 \\[3ex] y = -4x - 8 \\[5ex] \therefore \text{The system of equations shown on the graph is:} \\[3ex] y = 1.75x + 3.5 \\[3ex] y = -4x - 8 $

The rate of change of y with respect to x for this function is the slope of the function.

$ \text{Point 1: } (-4, 7) \\[3ex] x_1 = -4 \\[3ex] y_1 = 7 \\[5ex] \text{Point 2: } (4, 1) \\[3ex] x_2 = 4 \\[3ex] y_2 = 1 \\[5ex] \text{slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{1 - 7}{4 - (-4)} \\[5ex] m = \dfrac{-6}{8} \\[5ex] m = -\dfrac{3}{4} $

$ \dfrac{36x^4y^5}{(3xy)^2} \\[5ex] \text{DISSOCIATE and SOLVE} \\[3ex] \dfrac{36 \cdot x^4 \cdot y^5}{(3 \cdot x \cdot y)^2} \\[5ex] \dfrac{36 \cdot x^4 \cdot y^5}{3^2 \cdot x^2 \cdot y^2}...\text{Law 5 Exp} \\[5ex] \dfrac{36}{3^2} \cdot \dfrac{x^4}{x^2} \cdot \dfrac{y^5}{y^2} \\[5ex] \dfrac{36}{9} \cdot x^{4 - 2} \cdot y^{5 - 2} ...\text{Law 2 Exp} \\[5ex] 4 \cdot x^2 \cdot y^3 \\[3ex] \text{ASSOCIATE} \\[3ex] 4x^2y^3 $

$ g(x) = 73\left(\dfrac{4}{5}\right)^x \\[5ex] \text{When } x = 0 \\[3ex] y-\text{intercept} = g(0) = 73 \cdot \left(\dfrac{4}{5}\right)^0 \\[5ex] g(0) = 73 \cdot 1 ...\text{Law 3 Exp} \\[3ex] g(0) = 73 \\[3ex] g(0) = +73 $

Parent Function
$ f(x) = x^2 \\[3ex] (2, 4) $

1st Transformation: Reflection across the x-axis
Also known as Vertical Reflection
This will flip the graph vertically.
$(2, 4) \rightarrow (2, -4)$

2nd Transformation: Horizontal Stretch will stretch the graph horizontally, making it wider.
This implies that the correct answer is Option D.

$ \left(\dfrac{3}{2}p + 1\right)\left(\dfrac{1}{2}p + 3\right) \\[5ex] \dfrac{3}{4}p^2 + \dfrac{9}{2}p + \dfrac{1}{2}p + 3 \\[5ex] \dfrac{3}{4}p^2 + \dfrac{10}{2}p + 3 \\[5ex] \dfrac{3}{4}p^2 + 5p + 3 $

Because of the dashed line, the inequality will not contain an equality sign.

$ \text{When } x = 0, y = 7 \implies 7x \\[3ex] \text{When } y = 0, x = 4 \implies 4y \\[3ex] xy = 7(4) = 28 \\[3ex] \text{shaded region is below the line} \implies \lt \\[3ex] \implies \\[3ex] 7x + 4y \lt 28 $

The situation that shows causation is
B. The amount of time a cell phone is used affects the charge of its battery.
In other words, cell phone use (while not being charged) causes the battery to be used, and hence causes the charge level of the battery to decrease.
Using a cell phone directly reduces its battery charge.

$ 16w^2 + 24w + 9 \\[3ex] 16w^2 \cdot 9 = 144w^2 \\[3ex] \text{Factors are: } 12w \text{ and } 12w \\[3ex] \implies \\[3ex] 16w^2 + 12w + 12w + 9 \\[3ex] 4w(4w + 3) + 3(4w + 3) \\[3ex] (4w + 3)(4w + 3) \\[3ex] (4w + 3)^2 $

The domain is the set of all the input values, x for which the function has an output.
The closed circle represents a closed interval. It indicates that the endpoint is included.
The arrow represents infinity (positive or negative infinity as applicable).
Let us analyze each option and discard the incorrect ones.

Option A.
The domain is the set of all real numbers less than or equal to 6
This is the correct option.

Option B.
The domain is the set of all real numbers greater than or equal to –3.
Discard.

Option C.
The domain is the set of all real numbers greater than or equal to 6
Discard.

Option D.
The domain is the set of all real numbers less than or equal to 3
Discard.

The maximum value of the graph of the function is +225 feet.

$ \left(15a^0b^2c^{34}\right)\left(3a^{16}b^{-29}c^0\right) \\[4ex] \text{DISSOCIATE and SOLVE} \\[3ex] 15 \cdot a^0 \cdot b^2 \cdot c^{34} \cdot 3 \cdot a^{16} \cdot b^{-29} \cdot c^0 \\[4ex] 15 \cdot 1 \cdot b^2 \cdot c^{34} \cdot 3 \cdot a^{16} \cdot b^{-29} \cdot 1...\text{Law 4 Exp} \\[4ex] 15 \cdot 3 \cdot a^{16} \cdot b^2 \cdot b^{-29} \cdot c^{34} \\[4ex] 45 \cdot a^{16} \cdot b^{2 + (-29)} \cdot c^{34}...\text{Law 1 Exp} \\[4ex] 45 \cdot a^{16} \cdot b^{-27} \cdot c^{34} \\[4ex] 45 \cdot a^{16} \cdot \dfrac{1}{b^{27}} \cdot c^{34}...\text{Law 6 Exp} \\[5ex] \dfrac{45a^{16}c^{34}}{b^{27}} $

The rate of change of the contractor's total earnings in dollars with respect to the number of hours worked is the slope of the function.

$ \text{Point 1: } (0, 20) \\[3ex] x_1 = 0 \\[3ex] y_1 = 20 \\[5ex] \text{Point 2: } (40, 370) \\[3ex] x_2 = 40 \\[3ex] y_2 = 370 \\[5ex] \text{slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{370 - 20}{40 - 0} \\[5ex] m = \dfrac{350}{40} \\[5ex] m = 8.75 \\[3ex] $ The rate of change of the contractor's total earnings in dollars with respect to the number of hours worked is span>$8.75 per hour worked.

$ 2x^2 + 15 = -11x \\[3ex] 2x^2 + 11x + 15 = 0 \\[3ex] 2x^2 \cdot 15 = 30x^2 \\[3ex] \text{Factors are: } 6x \text{ and } 5x \\[3ex] \implies \\[3ex] 2x^2 + 6x + 5x + 15 = 0 \\[3ex] 2x(x + 3) + 5(x + 3) = 0 \\[3ex] (x + 3)(2x + 5) = 0 \\[3ex] x + 3 = 0 \text{ or } 2x + 5 = 0 ...\text{Zero Product Property} \\[3ex] x = -3 \text{ or } 2x = -5 \\[3ex] x = -3 \text{ or } x = -\dfrac{5}{2} \\[5ex] x = -3 \text{ or } x = -2.5 $

The table represents some points on the graph of an exponential function.
Let us use the points that will give us integers and simplified fractions similar to the answers in the options.

$ \underline{\text{Exponential Function}} \\[3ex] f(x) = ab^x \\[3ex] a = \text{initial value} \\[3ex] b = \text{base} \\[5ex] \text{Point 1: } (2, 36) \\[3ex] x = 2 \\[3ex] f(x) = 36 \\[3ex] 36 = a * b^{2} \\[3ex] ab^2 = 36...eqn.(1) \\[5ex] \text{Point 2: } (3, 54) \\[3ex] x = 3 \\[3ex] f(x) = 54 \\[3ex] 54 = a * b^{3} \\[4ex] ab^3 = 54...eqn.(2) \\[5ex] eqn.(2) \div eqn.(1) \implies \\[3ex] \dfrac{ab^3}{ab^2} = \dfrac{54}{36} \\[5ex] b^{3 - 2} = \dfrac{3}{2}...\text{Law 2 Exp} \\[5ex] b = \dfrac{3}{2} \\[5ex] \text{Substitute for } b \text{ in } eqn.(1) \\[3ex] a * \left(\dfrac{3}{2}\right)^2 = 36 \\[5ex] a * \dfrac{9}{4} = 36 \\[5ex] a = 36 \div \dfrac{9}{4} \\[5ex] a = 36 * \dfrac{4}{9} \\[5ex] a = 16 \\[5ex] \implies \\[3ex] f(x) = 16\left(\dfrac{3}{2}\right)^x $

C. The equation of the line is y = 2.5, and the line is parallel to the x-axis.

$ \underline{\text{Difference of Two Squares}} \\[3ex] 9n^2 - 25 \\[3ex] 3^2n^2 - 5^2 \\[3ex] (3n)^2 - 5^2 \\[3ex] (3n + 5)(3n - 5) $

The x-intercept is the point where the graph intersects the x-axis.
The y-intercept is the point where the graph intersects the y-axis.
Let us analyze the options and discard the incorrect ones.

Option A.
x-intercept: (5, 0)
y-intercept: (0, –1)
Discard.

Option B.
x-intercept: (1, 0)
y-intercept: (0, 5)
Discard.

Option A.
x-intercept: (–1, 0)
y-intercept: (0, –5)
Discard.

Option A.
x-intercept: (–1, 0)
y-intercept: (0, 5)
This is the correct answer.

$ y \propto x \\[3ex] y = kx ...k\text{ is the constant of variation} \\[3ex] 14 = k * 3.5 \\[3ex] 3.5k = 14 \\[3ex] k = \dfrac{14}{3.5} \\[5ex] k = 4 \\[5ex] \implies \\[3ex] y = 4x \\[3ex] x = 28 \\[3ex] y = 4(28) \\[3ex] y = 112 \\[3ex] y = +112 $

The arrow represents infinity (positive or negative infinity as applicable).

The domain is the set of all the input values, x for which the function has an output.
On the graph, the domain includes both positive infinity and negative infinity
This implies that the domain is all real numbers.

The range of a function is the set of all output values that result from applying the function to its input values.
On the graph, it begins from 5 up to positive infinity
This implies that the range is all real numbers greater than or equal to 5.

B. Domain: All real numbers
  Range: y ≥ 5

$ \text{Point 1: } (-3, -130) \\[3ex] x_1 = -3 \\[3ex] y_1 = -130 \\[5ex] \text{Point 2: } (11, 234) \\[3ex] x_2 = 11 \\[3ex] y_2 = 234 \\[5ex] \text{Slope, } m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{234 - (-130)}{11 - (-3)} \\[5ex] m = \dfrac{364}{14} \\[5ex] m = 26 \\[5ex] \text{Point–Slope Form} \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - (-130) = 26[x - (-3)] \\[3ex] y + 130 = 26(x + 3) \\[3ex] y + 130 = 26x + 78 \\[3ex] y = 26x + 78 - 130 \\[3ex] y = 26x - 52 \\[3ex] y = 26(x - 2) $

$ x^2 + 10x + 24 \\[3ex] \text{Factors are: } 4 \text{ and } 6 \\[3ex] \implies \\[3ex] (x + 4)(x + 6) $

Because the inequality contains an equal sign, the line is a solid line.

$ y \le \dfrac{3}{4}x - 4 \\[5ex] \text{When } x = 0 \\[3ex] y \le \dfrac{3}{4} * 0 - 4 \\[5ex] y \le 0 - 4 \\[3ex] y \le -4 \\[5ex] \text{When } x = 4 \\[3ex] y \le \dfrac{3}{4} * 4 - 4 \\[5ex] y \le 3 - 4 \\[3ex] y \le -1 \\[3ex] $ Less than or equal to implies that the shaded region is below the solid line.
Option A. is the correct answer.

There are at least two approaches to solve the question.
We can find the answer by analyzing the options.
We can also write a 3 by 3 linear system using the information given to us.
I recommend that we analyze the options to get the answer, rather than solving a 3 by 3 linear system to get the function.
Analyzing the options: let us test the first option.

$ \text{Point 1: } (2.25, 145) \\[3ex] x = 2.25 \\[3ex] y = 145 \\[5ex] \text{Option A.} \\[3ex] y = 0.001(2.25)^2 - 0.426(2.25) + 35.672 \\[3ex] y = 34.7185625 \ne 145 \\[3ex] \text{Discard} \\[5ex] \text{Option B.} \\[3ex] y = -60.4(2.25)^2 + 348.1(2.25) - 334.2 \\[3ex] y = 143.25 \\[3ex] \text{Do not discard because 143.25 is close to 145} \\[5ex] \text{Option C.} \\[3ex] y = 0.001(2.25)^2 + 35.672 \\[3ex] y = 35.6770625 \ne 145 \\[3ex] \text{Discard} \\[5ex] \text{Option D.} \\[3ex] y = -60.4(2.25)^2 - 334.2 \\[3ex] y = -639.975 \ne 145 \\[3ex] \text{Discard} \\[5ex] \text{Just to confirm, test another value for Option B.} \\[3ex] \text{Point 2: } (3.75, 126) \\[3ex] x = 3.75 \\[3ex] y = 126 \\[5ex] \text{Option B.} \\[3ex] y = -60.4(3.75)^2 + 348.1(3.75) - 334.2 \\[3ex] y = 121.8 \\[3ex] \text{121.8 is close to 126} \\[3ex] \text{This is the correct answer.} $

$ 35m^2 - 63 \\[3ex] \text{GCF} = 7 \\[3ex] \implies \\[3ex] 7(5m^2 - 9) $

The open circle represents an open interval.
The arrow represents infinity.

The domain is the set of all the input values, x for which the function has an output.
On the graph, it is from –2 (not included) to positive infinity

Domain:
–2 < x < ∞
Also written as: x > –2

$ 6y + x = -59...eqn.(1) \\[3ex] x = -2y + 9...eqn.(2) \\[5ex] \text{Substitution Method} \\[3ex] \text{From } eqn.(2) \\[3ex] \text{Substitute for } x \text{ in } eqn. (1) \\[3ex] 6y + (-2y + 9) = -59 \\[3ex] 6y - 2y + 9 = -59 \\[3ex] 4y = -59 - 9 \\[3ex] 4y = -68 \\[3ex] y = -\dfrac{68}{4} \\[5ex] y = -17 $

The correct answer is Option C.
The x-intercept is 60, which represents the number of monthly payments needed to repay the loan.

Parent Function: $f(x)$
$f(x) = x^2$

Child Function: $g(x)$
Tranformation:
Horizontal Shift of 4.5 units to the left
Transformed function: $g(x) = (x + 4.5)^2$

Exponential growth produces a rising curve, whereas exponential decay produces a falling curve.

$ \underline{\text{Exponential Function}} \\[3ex] y = 5\left(\dfrac{1}{3}\right)^x \\[5ex] \dfrac{1}{3} \lt 1 ...\text{This means that the function is an exponential decay function} \\[5ex] \text{When } x = 0 \\[3ex] y = 5\left(\dfrac{1}{3}\right)^0 \\[5ex] y = 5 \cdot 1 ...\text{Law 3 Exp} \\[3ex] y = 5 ...\text{initial value} \\[3ex] $ This implies that the correct answer is Option D.

$ -4(2m - 7) = 3(52 - 4m) \\[3ex] -8m + 28 = 156 - 12m \\[3ex] -8m + 12m = 156 - 28 \\[3ex] 4m = 128 \\[3ex] m = \dfrac{128}{4} \\[5ex] m = 32 $