Question

What is the quadratic function that is created with roots -10 and -4 and a vertex at (-7, -9)?

Answer 1

In vertex form it's
`y=(x+7)^2-9`

In standard form it's
`y=x^2+14x+40`

Explanation:

We can use the vertex form to solve for a in

`y=a(x-h)^2+k`

"a" is the number out front that dictates the steepness, or lack thereof, in a parabola. That means that we need h and k (which we have in the vertex) and we need x and y (which we have in the form of one of the zeros). Filling in using a vertex of (-7, -9) and a coordinate point (-7, 0):

`0=a(-4-(-7))^2-9` simplifies a bit to

`0=a(-4+7)^2-9` simplifies a bit more to

`0=a(3)^2-9` and

`0=a(9)-9` so

`0=9a-9`,

`9=9a` and finally,

a = 1

Phew! So there is the a value. Now we can simply fill in the formula completely, using the vertex as our guide:

`y=(x+7)^2-9`

In standard ofrm that is

`y=x^2+14x+40`

We can check ourselves for accuracy by factoring that standard polynomial using whatever method of factoring you like for quadratics and get that the roots are in fact x = -10, -4

The only reason we needed the zeros is to use one of them as the x and y to solve for a.