Relations and Functions
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These are the solutions to the NSSCO past questions on Relations and Functions.
The TI-84 Plus CE shall be used for applicable questions.
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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) $
Work out the equation of line B, giving your answer in the form y = mx + c.
Two lines are perpendicular to each other if the slope of one line is the negative reciprocal of the slope of the other line.
Let the other line = Line A
$ \underline{\text{For Line A}} \\[3ex] y = 2x - 3 \\[3ex] \text{Compare to: } y = mx + c \\[3ex] m = slope = 2 \\[5ex] \underline{\text{For Line B}} \\[3ex] Line\;B \perp Line\;A \\[3ex] \implies \\[3ex] slope = m = -\dfrac{1}{\text{slope of Line A}} \\[5ex] m = -\dfrac{1}{2} \\[5ex] Passes\;\;through\;\;(-2, 5) \\[3ex] x = -2 \\[3ex] y = 5 \\[3ex] y = mx + c \\[3ex] c = y - mx \\[3ex] c = 5 - \left(-\dfrac{1}{2} * -2\right) \\[5ex] c = 5 - 1 \\[3ex] c = 4 \\[3ex] \implies \\[3ex] y = mx + c \\[3ex] y = -\dfrac{1}{2}x + 4 $
