f is a quadratic function such that $f(0) = 0, f(1) = 2,$ and $f(-1) = 2.$ Find an equation for $f(x)$

2. $f(x) = ax^2+bx+c$

or

$f(x) = a(x-b)^2+c$

Choose one form and use your co-ordinates to substitute and then solve.

Here's a start...

$f(x) = ax^2+bx+c$

$f(0) = 0 \Rightarrow 0 = a\times 0^2+b\times 0 +c \Rightarrow c=0$

So we now have $f(x) = ax^2+bx+0$

$f(1) = 2 \Rightarrow 2 = a\times 1^2+b\times 1 \Rightarrow 2 = a+b$

$f(-1) = 2$

3. Okay, well I understand what to do now, but I have some questions. Why for f(1) do you not include c, and I'm assuming for f(-1) you don't include it either. Then after you have solved all of them, do you have to plug them back into the quadratic or can you just leave it as c=0, 2= a+b, etc..

4. Originally Posted by SHiFT
Okay, well I understand what to do now, but I have some questions. Why for f(1) do you not include c,
because we have already found c to be zero

Originally Posted by SHiFT
and I'm assuming for f(-1) you don't include it either.
True for the same reason above.

Originally Posted by SHiFT
Then after you have solved all of them, do you have to plug them back into the quadratic or can you just leave it as c=0, 2= a+b, etc..
Yep after you have a solution for each a,b and c.

You need to solve for a and b with simultaneous equations first.

5. so i need to set a-b and a+b in a equation to solve for them?

6. Originally Posted by SHiFT
so i need to set a-b and a+b in a equation to solve for them?
I'm not sure what you mean by this...?

Instead, try using the set-up that the other helper provided to you: Plug the given x-values into the general formula for a quadratic, simplify, and set equal to the given f-values. Solve the resulting system of linear equations by whatever method you prefer of the ones you learned in an earlier algebra course.