# How do you find the vertex and the intercepts for f(x) = x^2 + 7x -8?

Jul 19, 2016

Vertex $\left(- \frac{7}{2} , - \frac{81}{4}\right)$

#### Explanation:

$f \left(x\right) = {x}^{2} + 7 x - 8$
x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{7}{2}$
y-coordinate of vertex:
$y \left(- \frac{7}{2}\right) = \left(\frac{49}{4}\right) + 7 \left(- \frac{7}{2}\right) - 8 = \frac{49}{4} - \frac{49}{2} - 8 = - \frac{81}{4}$
Vertex (-7/2, - 81/4)
Make x = 0 --> y-intercept = - 8
To find x-intercepts, make x = 0 and solve f(x) = 0
Since a + b + c = 0, the, the 2 x-intercepts (real roots) are:
1 and $\frac{c}{a} = - 8$
graph{x^2 + 7x - 8 [-40, 40, -20, 20]}