# How do you find the vertex and intercepts for  y = x^2 + 3x – 10?

Jun 22, 2018

$\text{vertex } = \left(- \frac{3}{2} , - \frac{49}{4}\right) , x = - 5 , x = 2$

#### Explanation:

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

•color(white)(x)y=a(x-h)^2+k

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

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

$y = {x}^{2} + 2 \left(\frac{3}{2}\right) x \textcolor{red}{+ \frac{9}{4}} \textcolor{red}{- \frac{9}{4}} - 10$

$\textcolor{w h i t e}{y} = {\left(x + \frac{3}{2}\right)}^{2} - \frac{49}{4}$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(- \frac{3}{2} , - \frac{49}{4}\right)$

$\text{to find the x-intercepts let y = 0}$

${x}^{2} + 3 x - 10 = 0$

$\text{the factors of - 10 which sum to + 3 are + 5 and - 2}$

$\left(x + 5\right) \left(x - 2\right) = 0$

$\text{equate each factor to zero and solve for } x$

$x = - 5 , x = 2 \leftarrow \textcolor{red}{\text{x-intercepts}}$