# How do you find the vertex and the intercepts for y = x^2 + 6x − 16?

Sep 16, 2017

$\text{see explanation}$

#### Explanation:

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

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

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

• " ensure coefficient of "x^2" term is 1"

• " add/subtract "(1/2"coefficient of x-term")^2" to "x^2+6x

$\Rightarrow y = {x}^{2} + 2 \left(3\right) x \textcolor{red}{+ 9} \textcolor{red}{- 9} - 16$

$\textcolor{w h i t e}{\Rightarrow y} = {\left(x + 3\right)}^{3} - 25$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- 3 , - 25\right)$

$\textcolor{b l u e}{\text{Intercepts}}$

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

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

$x = 0 \to y = - 16 \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to {\left(x + 3\right)}^{2} - 25 = 0$

$\Rightarrow {\left(x + 3\right)}^{2} = 25$

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

$\Rightarrow x + 3 = \pm 5 \leftarrow \text{ note plus or minus}$

$\Rightarrow x = \pm 5 - 3$

$\Rightarrow x = 2 \text{ or "x=-8larrcolor(red)" x-intercepts}$