Question

How do you find the vertex and the intercepts for `f(x)=x^2-8x-9`?

Answer 1

Vertex (4, - 25)

Explanation:

`f(x) = x^2 - 8x - 9`
x-coordinate of vertex:
`x = -b/(2a) = 8/2 = 4`
y-coordinate of vertex:
y(4) = 16 - 32 - 9 = -16 - 9 = - 25
Vertex (4, -25)
Make x = 0 --> y-intercept = - 9
To find x-intercepts, make y = 0 and solve the quadratic equation:
`y = x^2 - 8x - 9 = 0.`
Since a - b + c = 0, use shortcut. The 2 x-intercepts (real roots) are:
-1 and `(-c/a = 9)`