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

Question

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

The answer is $f(2)=8$

I thought that by maximum value it meant that c=10, but the equation I got gave as a result $f(2)=10$

Any hint on how to solve it?

Answer 1

you will get the following equations $$9a+3b+c=2$$ $$a-b+c=2$$ and $$c-\frac{b^2}{4a}=10$$ can you solve this?

Answer 2

You can answer this with geometry. The parabola opens down and since $x=-1$ and $x=3$ have the same value $y=2$ the vertex is midway between at $x=1$ with height $y=10$ which we are given as maximum. Since the decrease from $x=1$ to $x=3$ is $10-2=8$ then the decrease to $x=2$ is a quarter of that and so we get $y=10-(8/4)=8$ which is the value at $x=0$ also.