Question

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

Answer 1

Vertex (3, 6)
No x-intercepts.

Explanation:

`y = x^2 - 6x + 15`
x-coordinate of vertex:
`x = -b/(2a) = 6/2 = 3`
y-coordinate of vertex:
`y(3) = 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 - 6x + 15 = 0`
`D = b^2 - 4ac = 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.