f is a quadratic function such that $\displaystyle f(0) = 0, f(1) = 2, $ and $\displaystyle f(-1) = 2. $ Find an equation for $\displaystyle f(x) $
$\displaystyle f(x) = ax^2+bx+c$
or
$\displaystyle f(x) = a(x-b)^2+c$
Choose one form and use your co-ordinates to substitute and then solve.
Here's a start...
$\displaystyle f(x) = ax^2+bx+c$
$\displaystyle f(0) = 0 \Rightarrow 0 = a\times 0^2+b\times 0 +c \Rightarrow c=0 $
So we now have $\displaystyle f(x) = ax^2+bx+0$
$\displaystyle f(1) = 2 \Rightarrow 2 = a\times 1^2+b\times 1 \Rightarrow 2 = a+b $
Now your turn for
$\displaystyle f(-1) = 2 $
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..
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.