Find the quadratic equation, given three points on the parabola 
Given that $f(x)=a(x-b)^2+c$, and that $f(1)=f(5)=-1$ and $f(0)=-11$, find the values of $a,b$,and $c$. 

I found the axis of symmetry is $3$, because $f$ has the same value at $1$ and $5$, and $(1+5)/2=3$. Therefore $b=3$. How to find $a$ and $c$?

You have found $b$ correctly as 3. I.e. $f(x)=a(x-3)^2+c=0$.
From this point you could solve simultaneously to get $a$ and $c$. I.e. solving:
$$f(1)=a(1-3)^2+c=-1$$
$$f(0)=a(0-3)^2+c=-11$$
But there is an even easier way thinking about it graphically:
As you have two points with the same y-value we could also express the quadratic as: $f(x)=a(x-1)(x-5)-1$. Now we only need to solve one equation in one variable:
$$f(0)=a(0-1)(0-5)-1=-11$$
$$5a-1=-11$$
$$5a=-10$$
$$a=-2$$
So the equation is:
$$f(x)=-2(x-1)(x-5)-1$$
$$f(x)=-2(x-3+2)(x-3-2)-1$$
$$f(x)=-2((x-3)^2-4)-1$$
$$f(x)=-2(x-3)^2+7$$
