# How do you find the vertex and intercepts for y=2(x+2)^2-9?

May 8, 2017

see explanation.

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where ( h , k ) are the coordinates of the vertex and a is a constant.

$y = 2 {\left(x + 2\right)}^{2} - 9 \text{ is in this form}$

$\text{with " h=-2" and } k = - 9$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- 2 , - 9\right)$

$\textcolor{b l u e}{\text{to find intercepts}}$

• " let x = 0, in equation, for y-intercept"

• " let y = 0, in equation, for x-intercepts"

$x = 0 \to y = 2 {\left(2\right)}^{2} - 9 = - 1 \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to 2 {\left(x + 2\right)}^{2} - 9 = 0$

$\Rightarrow {\left(x + 2\right)}^{2} = \frac{9}{2}$

$\textcolor{b l u e}{\text{taking the square root of both sides}}$

$\sqrt{{\left(x + 2\right)}^{2}} = \textcolor{red}{\pm} \sqrt{\frac{9}{2}} \leftarrow \text{ note plus or minus}$

$\Rightarrow x + 2 = \pm \frac{3}{\sqrt{2}}$

$\Rightarrow x = - 2 \pm \frac{3 \sqrt{2}}{2}$

$x = - 2 + \frac{3 \sqrt{2}}{2} \approx 0.12 \leftarrow \textcolor{red}{\text{ x-intercept}}$

$x = - 2 - \frac{3 \sqrt{2}}{2} \approx - 4.12 \leftarrow \textcolor{red}{\text{ x-intercept}}$
graph{2(x+2)^2-9 [-28.87, 28.86, -14.43, 14.44]}