the maximum value of a function $f(x)=ax^2+bx+c$ is $10$. Given that $f(3)=f(-1)=2$ find $f(2)$

Question

could you please help solving this?

The maximum value of $f(x)=ax^2+bx+c$ is $10$. Given that $f(3)=f(-1)=2$ find $f(2)$

I've realized that $f(1)$ is the middle between two of those which will be the vertex, however, how do I proceed from that?

Question was asked a month ago, however, not answered properly

Answer 1

As you said the max. is attend at $x=1$ so $f(1)=10$ and thus $f(x)=a(x-1)^2+10$. Since $f(3)=2$ we have $2 = 4a+10$ so $a=-2$, Finally we have $f(2)=-2+10 =8$.

Answer 2

John Watson has an easier method (I think he made a small mistake, he took $f(3)=-2$), but you can set a system of three equations: $f(1) = 10$, $f(3)=2$, and $f(-1)=2$, therefore: $$9a+3b+c=2$$ $$a+b+c=10$$ $$a-b+c=2$$ which has solution $(-2,4,8)$. Therefore $f(2)=4a+2b+c=8$.