# How do you find the vertex of y= x^2 + 8x -9?

May 20, 2018

$\text{vertex } = \left(- 4 , - 25\right)$

#### Explanation:

$\text{given a parabola in "color(blue)"standard form}$

•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0

$\text{then the x-coordinate of the vertex is}$

•color(white)(x)x_(color(red)"vertex")=-b/(2a)

$y = {x}^{2} + 8 x - 9 \text{ is in standard form}$

$\text{with "a=1,b=8" and } c = - 9$

$\Rightarrow {x}_{\text{vertex}} = - \frac{8}{2} = - 4$

$\text{substitute this value into the equation for y}$

${y}_{\text{vertex}} = {\left(- 4\right)}^{2} + 8 \left(- 4\right) - 9 = - 25$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- 4 , - 25\right)$

May 20, 2018

The vertex is $\left(- 4 , - 25\right)$

#### Explanation:

You can change the standard form which is given into vertex form which is $y = a {\left(x + p\right)}^{2} + q$ by the process of completing the square. The vertex is $\left(- p , q\right)$

$y = {x}^{2} + 8 x - 9$

To complete the square, add and subtract $\textcolor{b l u e}{{\left(\frac{b}{2}\right)}^{2} = 16}$
$y = {x}^{2} + 8 x \text{ } \textcolor{b l u e}{+ 16 - 16} - 9$

$y = \left({x}^{2} + 8 x \textcolor{b l u e}{+ 16}\right) + \left(\textcolor{b l u e}{- 16} - 9\right)$

$y = {\left(x + 4\right)}^{2} - 25$

The vertex is $\left(- 4 , - 25\right)$