Relations and Functions

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These are the solutions to the GCSE past questions on Relations and Functions.
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Formulas for Linear Functions

$ (1.)\;\; \text{Slope,}\;\; m = \dfrac{\Delta y}{\Delta x} = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{rise}{run} \\[5ex] (2.)\;\; \text{Slope-Intercept Form:}\;\; y = mx + b \\[5ex] (3.)\;\; \text{Point-Slope Form:}\;\; y - y_1 = m(x - x_1) \\[5ex] (4.)\;\; \text{Two-Points Form:}\;\; \dfrac{y - y_1}{x - x_1} = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] $

Formulas for Quadratic Functions

$ (1.)\;\; \text{Discriminant}\;\; = b^2 - 4ac \\[5ex] (2.)\;\; \text{Standard Form:}\;\; f(x) = ax^2 + bx + c \\[5ex] (3.)\;\; \text{Vertex from Standard Form:}\;\; \left[-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right] \\[7ex] (4.)\;\; \text{Vertex Form:}\;\; f(x) = a(x - h)^2 + k \\[5ex] (5.)\;\; \text{Vertex from Vertex Form:}\;\; Vertex = (h, k) \\[5ex] (6.)\;\; \text{Extended Vertex Form:}\;\; f(x) = a\left(x + \dfrac{b}{2a}\right)^2 + \dfrac{4ac - b^2}{4a} \\[7ex] (7.)\;\; \text{Vertex from Extended Vertex Form:}\;\; Vertex = \left(-\dfrac{b}{2a}, \dfrac{4ac - b^2}{4a}\right) \\[7ex] $

Factoring Formulas

$ \underline{Difference\;\;of\;\;Two\;\;Squares} \\[3ex] (1.)\;\;x^2 - y^2 = (x + y)(x - y) \\[5ex] \underline{Difference\;\;of\;\;Two\;\;Cubes} \\[3ex] (2.)\;\; x^3 - y^3 = (x - y)(x^2 + xy + y^2) \\[5ex] \underline{Sum\;\;of\;\;Two\;\;Cubes} \\[3ex] (3.)\;\; x^3 + y^3 = (x + y)(x^2 - xy + y^2) $

(1.) Write a possible equation in factored form for the given graph.

Number 1

n = degree
a = leading coefficient
x = independent variable
y = f(x) = dependent variable

1st: Turning Points:
If: Turning points: n − 1;
then Degree: at least n
Turning points = 3
n − 1 = 3
n = 3 + 1
n = 4

2nd: Degree: ≥ 4
Assume: degree = 4

3rd: **Zeros** and **Multiplicities of Zeros**
Zeros	Action	Multiplicity
$x = -1$	touch	even : touch multiplicity = 2
$x = 1$	touch	even : touch multiplicity = 2

4th: Leading coefficient and Factored Form

$ f(x) = a(x + 1)^2(x - 1)^2 \\[3ex] \text{passes through }y-intercept\;(0, -2) \\[3ex] x = 0 \\[3ex] y = f(x) = f(0) = -2 \\[3ex] \implies \\[3ex] f(0) = a(0 + 1)^2(0 - 1)^2 \\[3ex] -2 = a(1)^2(-1)^2 \\[3ex] -2 = a(1)(1) \\[3ex] -2 = a \\[3ex] a = -2 \\[3ex] \implies \\[3ex] f(x) = -2(x + 1)^2(x - 1)^2...\text{factored form} $

(2.) Write a possible equation in factored form for the given graph.

Number 2

n = degree
a = leading coefficient
x = independent variable
y = f(x) = dependent variable

1st: Turning Points:
If: Turning points: n − 1;
then Degree: at least n
Turning points = 4
n − 1 = 4
n = 4 + 1
n = 5

2nd: Degree: ≥ 5
Assume: degree = 5

3rd: **Zeros** and **Multiplicities of Zeros**
Zeros	Action	Multiplicity
$x = -2$	touch	even : touch multiplicity = 2
$x = 0$	cross	odd : cross multiplicity = 1
$x = 2$	touch	even : touch multiplicity = 2

4th: Leading coefficient and Factored Form

$ f(x) = ax(x + 2)^2(x - 2)^2 \\[3ex] \text{passes through the point } (2.5, -25) \\[3ex] x = 2.5 \\[3ex] y = f(x) = f(2.5) = -25 \\[3ex] \implies \\[3ex] f(2.5) = a(2.5)(2.5 + 2)^2(2.5 - 2)^2 \\[3ex] -25 = a(2.5)(4.5)^2(0.5^2) \\[3ex] -25 = 12.65625a \\[3ex] a = -\dfrac{25}{12.65625} \\[5ex] a = -1.975308642 \\[3ex] a \approx - 2...\text{to the nearest integer} \\[3ex] \implies \\[3ex] f(x) = -2x(x + 2)^2(x - 2)^2 ...\text{factored form} $

(3.) Calculate the distance between each pair of points.
Write the exact values.

$ \text{(1.) } (-5, 5) \text{ and } (1, -2) \\[3ex] \text{(2.) } (-5, 1) \text{ and } (4, 0) \\[3ex] \text{(3.) } (-2, 4) \text{ and } (0, 3) \\[3ex] \text{(4.) } (-4, 2) \text{ and } (-4, -4) \\[3ex] \text{(5.) } (-5, -4) \text{ and } (-1, -4) \\[3ex] \text{(6.) } (-2, 0) \text{ and } (3, -5) \\[3ex] $

$distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Question	Point 1	Point 2	$x_2 - x_1$	$(x_2 - x_1)^2$	$y_2 - y_1$	$(y_2 - y_1)^2$	$(x_2 - x_1)^2 \\ + \\ (y_2 - y_1)^2$	$\sqrt{(x_2 - x_1)^2 \\ + \\ (y_2 - y_1)^2}$
(1.)	$ (-5, 5) \\[3ex] x_1 = -5 \\[3ex] y_1 = 5 $	$ (1, -2) \\[3ex] x_2 = 1 \\[3ex] y_2 = -2 $	$ 1 - (-5) \\[3ex] 1 + 5 \\[3ex] 6 $	$6^2 = 36$	$ -2 - 5 \\[3ex] -7 $	$(-7)^2 = 49$	$ 36 + 49 \\[3ex] 85 $	$ \sqrt{85} $
(2.)	$ (-5, 1) \\[3ex] x_1 = -5 \\[3ex] y_1 = 1 $	$ (4, 0) \\[3ex] x_2 = 4 \\[3ex] y_2 = 0 $	$ 4 - (-5) \\[3ex] 4 + 5 \\[3ex] 9 $	$9^2 = 81$	$ 0 - 1 \\[3ex] -1 $	$(-1)^2 = 1$	$ 81 + 1 \\[3ex] 82 $	$ \sqrt{82} $
(3.)	$ (-2, 4) \\[3ex] x_1 = -2 \\[3ex] y_1 = 4 $	$ (0, 3) \\[3ex] x_2 = 0 \\[3ex] y_2 = 3 $	$ 0 - (-2) \\[3ex] 0 + 2 \\[3ex] 2 $	$2^2 = 4$	$ 3 - 4 \\[3ex] -1 $	$(-1)^2 = 1$	$ 4 + 1 \\[3ex] 5 $	$ \sqrt{5} $
(4.)	$ (-4, 2) \\[3ex] x_1 = -4 \\[3ex] y_1 = 2 $	$ (-4, -4) \\[3ex] x_2 = -4 \\[3ex] y_2 = -4 $	$ -4 - (-4) \\[3ex] -4 + 4 \\[3ex] 0 $	$0^2 = 0$	$ -4 - 2 \\[3ex] -6 $	$(-6)^2 = 36$	$ 0 + 36 \\[3ex] 36 $	$ \sqrt{36} = 6 $
(5.)	$ (-5, -4) \\[3ex] x_1 = -5 \\[3ex] y_1 = -4 $	$ (-1, -4) \\[3ex] x_2 = -1 \\[3ex] y_2 = -4 $	$ -1 - (-5) \\[3ex] -1 + 5 \\[3ex] 4 $	$4^2 = 16$	$ -4 - (-4) \\[3ex] -4 + 4 \\[3ex] 0 $	$0^2 = 0$	$ 16 + 0 \\[3ex] 16 $	$ \sqrt{16} \\[3ex] 4 $
(6.)	$ (-2, 0) \\[3ex] x_1 = -2 \\[3ex] y_1 = 0 $	$ (3, -5) \\[3ex] x_2 = 3 \\[3ex] y_2 = -5 $	$ 3 - (-2) \\[3ex] 3 + 2 \\[3ex] 5 $	$5^2 = 25$	$ -5 - 0 \\[3ex] -5 $	$(-5)^2 = 25$	$ 25 + 25 \\[3ex] 50 $	$ \sqrt{50} \\[3ex] \sqrt{25 * 2} \\[3ex] \sqrt{25} * \sqrt{2} \\[3ex] 5 * \sqrt{2} \\[3ex] 5\sqrt{2} $

(4.) Write the domain and range for these functions.

Number 4

Let:
vertical asymptote = VA
horizontal asymptote = HA
A graph may touch or cross the horizontal asymptote, but will never touch or cross the vertical asymptote.
The vertical asymtpote(s) help us to determine the domain, as they are excluded from the domain.
The horizontal asymptote(s) indicate the boundaries for the range, which are usually but not necessarily excluded from the range.
If the graph does not touch the horizontal asymptote, that value is excluded from the range.
If the graph touches or crosses the horizontal asymptote, that value is included in the range.

Let:
domain = D
range = R
The domain is the set of all input values, x for which the function produces an output, y.
The range is the set of all output values, y that result from those input values, x.

$ (a.) \text{ Rational Function} \\[3ex] VA:\;x = 2 \\[3ex] D = (-\infty, 2) \cup (2, \infty) \\[3ex] HA:\; y = -1 \\[3ex] R = (-\infty, -1) \cup (-1, \infty) \\[5ex] (b.) \text{ Rational Function} \\[3ex] VA:\;x = -2, 2 \\[3ex] D = (-\infty, -2) \cup (-2, 2) \cup (2, \infty) \\[3ex] HA:\; y = -2 \\[3ex] R = (-\infty, -2) \cup (-2, \infty) \\[5ex] (c.) \text{ Rational Function} \\[3ex] VA:\;x = -2, 1 \\[3ex] D = (-\infty, -2) \cup (-2, 1) \cup (1, \infty) \\[3ex] HA:\; y = 0 \\[3ex] \text{However, the graph crosses the HA, hence that value is included in the range} \\[3ex] R = (-\infty, \infty) \\[3ex] $ (d.) Piecewise Function
For the domain: the graph is open at $x = -3$ but closed at $x = 2$
D = (−∞, −3) ⋃ [−2, ∞)
For the range: the graph is open at $y = 1$ in the 1st piece but closed at $y = 1$ in the 2nd piece.
So, 1 is included in the range.
$R = (-\infty, \infty)$

(e.) Piecewise Function
For the domain: the graph is open and closed at $x = -2$ and $x = 2$
So, −2 and 2 are included in the domain.
$D = (-\infty, \infty)$
For the range:
From Left to Right
1st piece: the graph is open at $y = -2$
2nd piece: the graph is closed at $y = 1$ and open at $y = 1$
But the maximum is $y = 3$ (included).
3rd piece: the graph is closed at $y = -2$
In summary, we see that the graph does not include any value in-between from −2 to 1, and anything above 3.
R = (−∞, −2] ⋃ [1, 3]

(f.) Piecewise Function
For the domain:
1st piece: the graph is open at $x = -1$
2nd piece: the graph is closed at $x = -1$ and open at $x = 3$
−1 is included in the domain but 3 is not.
D = (−∞, −1] ⋃ [−1, 3)
So, we combine it to have:
$D = (-\infty, 3)$
For the range:
1st piece: the graph is open at $y = -2$
But the minimum is $y = -3$ (included). This implies that $y = -2$ is also included.
2nd piece: the graph is closed at $y = 1$ and open at $y = 1$
1 is included in the range.
In summary, the 1st piece also includes the 2nd piece for the range.
$R = [-3, \infty)$

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