# How do you find the vertex and the intercepts for y = x^2 + 8x + 15?

Jun 29, 2016

Vertex (-4, -1)

#### Explanation:

x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{8}{2} = - 4$
y-coordinate of vertex:
y(-4) = 16 - 32 + 15 = - 1
Vertex (-4, -1)
To find y-intercept, make x = 0 --> y = 15.
To find x-intercepts, make y= 0, and solve the quadratic equation
$y = {x}^{2} + 8 x + 15 = 0$
Find 2 numbers knowing sum (-b = -8) and product (c = 15).
The 2 real roots (x-intercepts) are: - 3 and - 5.
graph{x^2 + 8x + 15 [-10, 10, -5, 5]}