# How do you find the vertex and intercepts for y= -2x^2 + 12x - 13?

Jul 8, 2017

Vertex (3, 5)

#### Explanation:

$y = - 2 {x}^{2} + 12 x - 13$
x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{12}{-} 4 = 3$
y-coordinate of vertex:
y(3) = - 2(9) + 12(3) - 13 = - 18 + 36 - 13 = 5
Vertex (3, 5)
To find y-intercept, make x = 0 --> y-intercept = - 13.
To find the 2 intercept, make y = 0 and solve:
$y = - 2 {x}^{2} + 12 x - 13 = 0$
$D = {b}^{2} - 4 a c = 144 - 104 = 40$ --> $d = \pm 2 \sqrt{10}$
There are 2 x-intercepts (real roots):
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{12}{-} 4 \pm \frac{2 \sqrt{10}}{4} = 3 \pm \frac{\sqrt{10}}{2}$
graph{-2 x^2 + 12x - 13 [-20, 20, -10, 10]}