Question 1102367: What is the quadratic function that is created with roots -10 and -4 and a vertex at (-7, -9)? Please help!!! Found 2 solutions by richwmiller, ikleyn:Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! y = a(x – h)^2 + k
y = a(x +7)^2- 9
-4 = a(-10 +7)^2 - 9
-4 = a(-3)^2 - 9
5= a(-3)^2
5= 9a
5/9=a
y = 5/9(x + 7)^2 - 9
oops!
0 = a(-10 +7)^2 - 9
a=1
so
y = (x + 7)^2 - 9
y = a(x – h)^2 + k <<<---=== it is the (general) vertex form of a quadratic function
y = a(x +7)^2- 9 <<<---=== it is the vertex form of a GIVEN quadratic function written specifically for the GIVEN vertex.
The coefficient "a" still remains undetermined.
And our goal is to determine it.
0 = a(-10 +7)^2 - 9 <<<---=== To determine the coefficient "a", use the condition that x= -10 is the root.
At this point my solution becomes different from @richwmiller writing, who made a mistake in this point.
0 = a(-3)^2 - 9
9 = a(-3)^2
9 = 9a ====>
a = = 1.
Thus (and finally), your function is
y = (x + 7)^2 - 9
Check. a) (-7,-9) is the vertex; it is CLEAR.
b) the function is ZERO at x= -10: y = (-10+7)^2 - 9 = 9 - 9 = 0.
and at x = -4: y = (-4-7)^2 - 9 = 9 - 9 = 0.
Answer. Your function is y = (x+7)^2 - 9.
