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) $

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- Oct 21st 2009, 01:02 PMSHiFTQuadratic Function
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) $

- Oct 21st 2009, 01:40 PMpickslides
$\displaystyle f(x) = ax^2+bx+c$

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 $ - Oct 21st 2009, 02:05 PMSHiFT
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..

- Oct 21st 2009, 02:19 PMpickslides
- Oct 21st 2009, 02:53 PMSHiFT
so i need to set a-b and a+b in a equation to solve for them?

- Oct 21st 2009, 02:57 PMstapel
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. (Wink)