# How do you find the vertex and x intercept of f(x)=x^2+11x+30?

Aug 3, 2016

Vertex $\left(- \frac{11}{2} , - \frac{1}{4}\right)$
x-intercepts: -5 and -6

#### Explanation:

$f \left(x\right) = {x}^{2} + 11 x + 30.$
x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{11}{2}$
y-coordinate of vertex:
$y \left(- \frac{11}{2}\right) = \frac{121}{4} - \frac{121}{2} + 30 = - \frac{121}{4} + \frac{120}{4} = - \frac{1}{4}$
Vertex $\left(- \frac{11}{2} , - \frac{1}{4}\right)$
To find the 2 x-intercepts, solve the quadratic equation f(x) = 0.
Find 2 real roots, knowing sum (-b = -11) and product (c = 30).
The 2 x-intercepts are: - 5 and -6 -> sum (- 11 = -b) --> product (30).