Question

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

Answer 1

Answer:

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

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

Step-by-step 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.