# What is the axis of symmetry and vertex for the graph y = x^2 – 4x – 12?

Jul 24, 2016

Axis of symmetry $\to x = + 2$

$\text{Vertex} \to \left(x , y\right) = \left(2 , - 16\right)$

#### Explanation:

color(blue)("Using a bit of a cheat to find "x_("vertex"))

Given$\text{ } y = {x}^{2} \textcolor{m a \ge n t a}{- 4} x - 12$.....................Equation (1)

$\underline{\text{Axis of symmetry is the x value of the vertex}}$

$\textcolor{g r e e n}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(\textcolor{m a \ge n t a}{- 4}\right) = + 2}$

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$\textcolor{b r o w n}{\text{A note about what I have just done:}}$
Consider the standard form $y = a {x}^{2} + b x + c$

Write as $y = a \left({x}^{2} + \frac{b}{a} x\right) + c$

Then ${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a}$

In the case of this question $a = 1$
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color(blue)("Determine "y_("vertex"))

Substitute $x = 2$ into Equation (1)

$\textcolor{b r o w n}{y = {x}^{2} - 4 x - 12 \text{ "->" "y_("vertex}} = {\left(\textcolor{b l u e}{2}\right)}^{2} - 4 \left(\textcolor{b l u e}{2}\right) - 12$

$\textcolor{g r e e n}{{y}_{\text{vertex}} = 4 - 8 - 12 = - 16}$

$\text{VERTEX} \to \left(x , y\right) = \left(2 , - 16\right)$