# How do you find the vertex and the intercepts for y=x^2 - 8x + 3?

Jul 14, 2016

Vertex (4, -13)

#### Explanation:

$y = {x}^{2} - 8 x + 3$
x-coordinate of vertex:
$x = - \frac{b}{2 a} = \frac{8}{2} = 4$
y-coordinate of vertex:
y(4) = 16 - 32 + 3 = -13
Vertex (4, -13)
To find y-intercepts, make x = 0 --> y = 3
To find x-intercepts, solve the quadratic equation y = 0
$D = {d}^{2} = {b}^{2} - 4 a c = 64 - 12 = 52$ --> $d = \pm 2 \sqrt{13}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{8}{2} \pm \frac{2 \sqrt{13}}{2} = 4 \pm \sqrt{13}$