# How do you find the vertex and the intercepts for f(x)=x^2+x-2?

Mar 14, 2016

$v e r t e x \left(- \frac{1}{2} , - \frac{9}{4}\right)$

#### Explanation:

x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{1}{2}$
y-coordinate of vertex:
$y \left(- \frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} - 2 = - \frac{9}{4}$
$V e r t e x \left(- \frac{1}{2} , - \frac{9}{4}\right)$.
To find y-intercept, make x = 0 --> y = -2.
To find x-intercepts, make y = 0 and solve the quadratic equation
$y = {x}^{2} + x - 2 = 0$
Since a + b + c = 0, use shortcut.
The 2 real roots are 1 and $\frac{c}{a} = - 2.$