# How do you find the axis of symmetry and vertex point of the function: f(x)=2x^2+x-3?

Oct 22, 2015

See explanation.

#### Explanation:

Firstly, you change this equation in the form of $a {\left(x - h\right)}^{2} + k$ which is also called as complete square form
So,

taking common the coefficient of ${x}^{2}$

$2 \left({x}^{2} + \frac{x}{2} - \frac{3}{2}\right)$

$2 \left[{x}^{2} + 2 \cdot \frac{x}{4} + {\left(\frac{1}{4}\right)}^{2} - \left(\frac{3}{2}\right) - {\left(\frac{1}{4}\right)}^{2}\right]$

$2 \left[{\left({x}^{2} + \frac{1}{4}\right)}^{2} - \left(\frac{1}{16} + \frac{3}{2}\right)\right]$

$2 \left[{\left({x}^{2} + \frac{1}{4}\right)}^{2} - \frac{25}{16}\right]$

Opening the bracket and multiplying by 2.

$2 {\left({x}^{2} + \frac{1}{4}\right)}^{2} - 2 \cdot \frac{25}{16}$

$2 {\left({x}^{2} + \frac{1}{4}\right)}^{2} - \frac{25}{8}$ is in the form of $a {\left(x - h\right)}^{2} + k$

where $a = 2$ , $h = - \frac{1}{4}$ and $k = - \frac{25}{8}$

The vertex is given by $\left(h , k\right)$

So the vertex is $\left(- \frac{1}{4} , - \frac{25}{8}\right)$

and the line of symmetry is given by $\left(x - h\right) = 0$
So

$x - \left(- \frac{1}{4}\right) = 0$

$x + \frac{1}{4} = 0 \implies x = - \frac{1}{4}$