Find the quadratic equation, given three points on the parabola

Question

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 $c$?

Answer 1

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: $$f(x)=-2(x-1)(x-5)-1$$ $$f(x)=-2(x-3+2)(x-3-2)-1$$ $$f(x)=-2((x-3)^2-4)-1$$ $$f(x)=-2(x-3)^2+7$$