Question

What is the quadratic function that is created with roots -10 and -6 and a vertex at (-8, -8)
Format using this sample answer:
f(x)=(3)(x^2-21x-19)

Answer 1

well the way i would do it would be:

if it has roots -10 and -6 we know that when you factor it it looks like:

(x+6)(x+10)

which equals x^2 +16x+60

and if the vertex is at -8,-8 then -b/2a = -8

right now -b/2a is -16/2 which ='s -8 so thats good!

then lets check the y coordinate of the vertex

our equation is y = x^2+16x + 60

and if we plug in a "-8" we get:

y = (64) - 128 + 60 = -4

darn well thats half

so lets times the whole thing by 2

2(x+6)(x+10)

Answer 2

check that through the same process and sure enough it works!!

Answer 3

i feel like there may be a simpler way... but I am not sure what it would be