# How do you find the vertex and axis of symmetry for y = 1/2x^2 + 8x - 9?

##### 1 Answer
Aug 5, 2015

Convert the given equation into vertex form to get the vertex at $\left(- 8 , - 41\right)$

#### Explanation:

Vertex form of a quadratic:
$\textcolor{w h i t e}{\text{XXXX}}$$y = m {\left(x - a\right)}^{2} + b$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$with vertex at $\left(a , b\right)$

Given $y = \frac{1}{2} {x}^{2} + 8 x - 9$

Extract $m$
$\textcolor{w h i t e}{\text{XXXX}}$$y = \left(\frac{1}{2}\right) \left({x}^{2} + 16 x\right) - 9$

Complete the square
$\textcolor{w h i t e}{\text{XXXX}}$$y = \left(\frac{1}{2}\right) \left({x}^{2} + 16 x + 64\right) - 9 - \left(\frac{1}{2}\right) \left(64\right)$

Simplify into vertex form:
$\textcolor{w h i t e}{\text{XXXX}}$$y = \left(\frac{1}{2}\right) {\left(x + 8\right)}^{2} - 41$
$\textcolor{w h i t e}{\text{XXXX}}$or
$\textcolor{w h i t e}{\text{XXXX}}$$y = \left(\frac{1}{2}\right) \left(x - \left(- 8\right)\right) + \left(- 41\right)$