# How do you write  y=x^2+8x+14 in vertex form and identify the vertex, y intercept and x intercept?

May 4, 2018

$\text{see explanation}$

#### Explanation:

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$\text{to obtain this form "color(blue)"complete the square}$

$y = {x}^{2} + 2 \left(4\right) x \textcolor{red}{+ 16} \textcolor{red}{- 16} + 14$

$\Rightarrow y = {\left(x + 4\right)}^{2} - 2 \leftarrow \textcolor{b l u e}{\text{in vertex form}}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- 4 , - 2\right)$

$\text{to obtain the intercepts}$

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

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

$x = 0 \Rightarrow y = 0 + 0 + 14 = 14 \leftarrow \textcolor{red}{\text{y-intercept}}$

$y = 0 \Rightarrow {\left(x + 4\right)}^{2} - 2 = 0 \leftarrow \text{add 2 to both sides}$

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

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

$\sqrt{{\left(x + 4\right)}^{2}} = \pm \sqrt{2} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x + 4 = \pm \sqrt{2} \leftarrow \text{subtract 4 from both sides}$

$\Rightarrow x = - 4 \pm \sqrt{2} \leftarrow \textcolor{red}{\text{exact values}}$
graph{(y-x^2-8x-14)((x+4)^2+(y+2)^2-0.04)=0 [-10, 10, -5, 5]}