# How do you find the vertex and intercepts for y = 2(x + 4)^2 - 21?

May 9, 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.

$y = 2 {\left(x + 4\right)}^{2} - 21 \text{ is in this form}$

$\text{with " h=-4" and } k = - 21$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- 4 , - 21\right)$

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

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

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

$x = 0 \to y = 2 {\left(4\right)}^{2} - 21 = 32 - 21 = 11 \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to 2 {\left(x + 4\right)}^{2} - 21 = 0$

$\Rightarrow 2 {\left(x + 4\right)}^{2} = 21$

$\Rightarrow {\left(x + 4\right)}^{2} = \frac{21}{2}$

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

$\sqrt{{\left(x + 4\right)}^{2}} = \textcolor{red}{\pm} \sqrt{\frac{21}{2}} \leftarrow \text{ note plus or minus}$

$\Rightarrow x + 4 = \pm \sqrt{\frac{21}{2}}$

$\text{subtract 4 from both sides}$

$x \cancel{+ 4} \cancel{- 4} = + \sqrt{\frac{21}{2}} - 4$

$\Rightarrow x = - 4 \pm \sqrt{\frac{21}{2}} \leftarrow \textcolor{red}{\text{ x-intercepts}}$

$\text{these are the exact values, which may be approximated to}$

$x = - 7.24 \text{ and " x=-0.76" to 2 dec. places}$
graph{2(x+4)^2-21 [-46.3, 46.16, -23.07, 23.18]}