Question

What is the quadratic function that is created with roots -3 and 1 and a vertex at (-1, -8)?

Answer 1

Plug the vertex and one of the roots into the standard form, then solve for the missing variable.

Answer 2

I have done it several times and I am not getting the right equation.

Answer 3

Hm, what do you get when you substitute the vertex in?

Answer 4

I tried it a few different ways and once I got the final equation as f(x)=4x^2-8x+12 and another time I got f(x)=2x^2+2x-6 neither are right when I check the graph.

Answer 5

@ig0p0g0

Answer 6

Try y=a(x-h)^2 +k, where h,k is the vertex. What do you get when you plug in the vertex?

Answer 7

y=a(x-(-1))^2 + -8?

Answer 8

Yep...probably the easiest next step is to plug in one of the roots. What do you get?

Answer 9

y=a(-1 - (-1))^2 +8?

Answer 10

What's the value of y if that's a root?

Answer 11

the other root?

Answer 12

Think about...what does it mean for something to be a root?

Answer 13

where it hits the line on the x axis

Answer 14

@ig0p0g0

Answer 15

Yes! What's the y value there?

Answer 16

-8? @ig0p0g0

Answer 17

The y value of a point on the x axis....

Answer 18

....any point on the x axis...

Answer 19

ok any one at all

Answer 20

(4,0) is on the x-axis. (4,3) is not. There's something you know about any point on the x-axis.

Answer 21

no y valuee

Answer 22

well, zero for y.

Answer 23

So go back to this...

y=a(-1 - (-1))^2 +8?

and plug in zero for y.

Answer 24

0=a(-1 - (-1))^2 +8

Answer 25

OK, what's a?

Answer 26

Ooo...something's wrong there...

Answer 27

Oh, the root is 1, not -1

Answer 28

So you want

0=a(1 - (-1))^2 +8

Answer 29

Gotta go but you're almost done.