# How do you find the vertex, axis of symmetry and intercepts of f(x)=2x^2+x=3?

Dec 21, 2017

$\text{see explanation}$

#### Explanation:

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

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

$\text{then the x-coordinate of the vertex which is also the}$
$\text{axis of symmetry is}$

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

$f \left(x\right) = 2 {x}^{2} + x - 3 \text{ is in standard form}$

$\text{with } a = 2 , b = 1 , c = - 3$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{1}{4}$

$\Rightarrow \text{axis of symmetry is } x = - \frac{1}{4}$

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

${y}_{\textcolor{red}{\text{vertex}}} = 2 {\left(- \frac{1}{4}\right)}^{2} + \left(- \frac{1}{4}\right) - 3 = - \frac{25}{8}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- \frac{1}{4} , - \frac{25}{8}\right)$

$\text{for "color(blue)"intercepts}$

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

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

$x = 0 \to y = - 3 \leftarrow \textcolor{red}{\text{y-intercept}}$

$y = 0 \to 2 {x}^{2} + x - 3 = 0$

$\Rightarrow \left(x - 1\right) \left(2 x + 3\right) = 0$

$\Rightarrow x = 1 \text{ and "x=-3/2larrcolor(red)"x-intercepts}$
graph{(y-2x^2-x+3)(y-1000x-250)=0 [-10, 10, -5, 5]}