# How do you find the vertex and intercepts for y=x^2-6x+15?

Mar 29, 2018

Vertex (3, 6)
No x-intercepts.

#### Explanation:

$y = {x}^{2} - 6 x + 15$
x-coordinate of vertex:
$x = - \frac{b}{2 a} = \frac{6}{2} = 3$
y-coordinate of vertex:
$y \left(3\right) = 9 - 18 + 15 = 6$
Vertex (3, 6)
To find the x-intercepts, make y = 0, and solve the quadratic equation by the improved quadratic formula.
${x}^{2} - 6 x + 15 = 0$
$D = {b}^{2} - 4 a c = 36 - 60 = - 24 < 0$.
There are no x-intercepts (no real roots). The upward parabola
doesn't intersect the x-axis. It is completely above the x-axis.