Question

What is the quadratic function that is created with roots -10 and -6 and a vertex at (-8, -8)?

Answer 1

Answer:

f(x) = x^2 + 16x + 56.

Step-by-step explanation:

The vertex form is:

y = a(x + 8)^2 - 8 where  a is a constant.

When x = -6, y = 0 so:

0 = a (-6+8)^2 - 8

4^2a  = 8

a = 1

So the function is y = (x + 8)^2 - 8

in standard form this is

y = x^2 + 16x + 64 - 8

y = x^2 + 16x + 56.