Since we know that the graph passes through the point `(0,2),` we have `f(0)=c=2,` so we know the constant term and we get`f(x)=ax^2+bx+2.`
Now we plug in another point, say `(-2,8).` This gives
`8=f(-2)=a(-2)^2+b(-2)+2=4a-2b+2.` The last point gives
`20=f(2)=a(2)^2+b(2)+2=4a+2b+2,` so we end up with the system of equations
If we add them, the `b`'s cancel and we get `8a=24,` so `a=3.` Now we can plug this value of `a` into the top equation and get
`4(3)-2b=6,` so `b=3.`
The quadratic polynomial that goes through those points is `f(x)=3x^2+3x+2,` which can be easily checked by plugging in.