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

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.