# How do you find the vertex and the intercepts for y = x^2 + 10x - 7?

May 20, 2017

vertex (-5, -32)

#### Explanation:

y = x^2 + 10x - 7
x-coordinate of vertex:
x = -b/(2a) = -10/2 = -5
y-coordinate of vertex:
y(-5) = 25 - 50 - 7 = -32
To find x-intercepts, make y = 0 and solve the quadratic equation:
${x}^{2} + 10 x - 7 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 100 + 28 = 128$ --> $d = \pm 8 \sqrt{2}$
There are 2real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{10}{2} \pm \frac{8 \sqrt{2}}{2} = - 5 \pm 4 \sqrt{2}$
$x 1 = - 5 + 4 \sqrt{2}$
$x 2 = - 5 - 4 \sqrt{2}$