How to find the $f(3)=y$ with given points?

Question

If an quadratic equation $f(x)$ has max value $4$ when $x = 0$ and $f(-2)=0$, what is the value of $f(3) =$ ?

Answer 1

A quadratic function has the form $a\cdot (x-b)^2 + c$ Given the maximum value at $x=0$ and a negative opening, says $a<0$ and $b=0$.

From there, it should be easy to determine $a$ and $c$ from the two points given. After that, you plug $x=3$ in there.

Answer 2

Since the parabola is symmetric about the axis of symmetry, $2$ is another root of the polynomial (draw a picture). Hence the quadratic function can be written as $$ f(x)=a(x+2)(x-2) $$ where $a<0$ (why?). You can solve for $a$ since $f(0)=4$, and then find $f(3)$.

Answer 3

making the ansatz $$f(x)=ax^2+c$$ we get easily $$c=4$$ from $$f(-2)=0$$ we get $$a=-1$$ and our searched function is $$f(x)=-x^2+4$$