Relations and Functions

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These are the solutions to the CSEC Additional Mathematics 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) \\[4ex] $

Formula Sheet: List of Formulae

(1.) For the quadratic function $g(x) = -5x^2 - 4x + 2$, determine the exact value of the maximum point and the range using the method of completing the square, or otherwise.

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

$ \underline{\text{1st Approach: Completing the Square Method}} \\[3ex] g(x) = -5x^2 - 4x + 2 \\[4ex] = -5\left(x^2 + \dfrac{4x}{5} - \dfrac{2}{5}\right) \\[5ex] \text{Coefficient of }x = \dfrac{4}{5} \\[5ex] \text{Half of it} = \dfrac{1}{2} * \dfrac{4}{5} = \dfrac{2}{5} \\[5ex] \text{Square it} = \left(\dfrac{2}{5}\right)^2 \\[6ex] \implies \\[3ex] = -5\left[x^2 + \dfrac{4x}{5} + \left(\dfrac{2}{5}\right)^2 - \dfrac{2}{5} - \left(\dfrac{2}{5}\right)^2\right] \\[6ex] = -5\left[\left(x + \dfrac{2}{5}\right)^2 - \dfrac{2}{5} - \dfrac{4}{25}\right] \\[6ex] = -5\left[\left(x + \dfrac{2}{5}\right)^2 - \dfrac{10}{25} - \dfrac{4}{25}\right] \\[6ex] = -5\left[\left(x + \dfrac{2}{5}\right)^2 - \dfrac{14}{25}\right] \\[6ex] = -5\left(x + \dfrac{2}{5}\right)^2 + \dfrac{14}{5} \\[6ex] = a(x - h)^2 + k \\[4ex] \implies \\[3ex] (i.) \\[3ex] \text{Vertex = Maximum Point} = (h, k) = \left(-\dfrac{2}{5}, \dfrac{14}{5}\right) \\[7ex] \underline{\text{2nd Approach: Vertex from Standard Form Method}} \\[3ex] g(x) = -5x^2 - 4x + 2...\text{Standard Form} \\[4ex] \text{Compare to: } g(x) = ax^2 + bx + c \\[4ex] a = -5 \\[3ex] b = -4 \\[3ex] c = 2 \\[3ex] -\dfrac{b}{2a} = \dfrac{-(-4)}{2(-5)} = \dfrac{2}{-5} = -\dfrac{2}{5} \\[5ex] g\left(-\dfrac{b}{2a}\right) = g\left(-\dfrac{2}{5}\right) \\[3ex] = -5\left(-\dfrac{2}{5}\right)^2 - 4\left(-\dfrac{2}{5}\right) + 2 \\[6ex] = -5\left(\dfrac{4}{25}\right) + \dfrac{8}{5} + \dfrac{10}{5} \\[5ex] = \dfrac{-4}{5} + \dfrac{8}{5} + \dfrac{10}{5} \\[5ex] = \dfrac{-4 + 8 + 10}{5} \\[5ex] = \dfrac{14}{5} \\[5ex] \text{Vertex = Maximum Point} = \text{Vertex from Standard Form:} \\[3ex] \implies \\[3ex] \left[-\dfrac{b}{2a}, g\left(-\dfrac{b}{2a}\right)\right] = \left(-\dfrac{2}{5}, \dfrac{14}{5}\right) \\[7ex] (ii.) \\[3ex] Range = \left(-\infty, \dfrac{14}{5}\right] $

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