For (f(x)=ax^2+bx+c), use (x=-frac{b}{2a})
Substitute that (x)-value into the function to get (y=f!left(-frac{b}{2a}right))
The vertex is (left(-frac{b}{2a},,f!left(-frac{b}{2a}right)right))
If the quadratic is in vertex form (f(x)=a(x-h)^2+k), the vertex is ((h,k))
If the quadratic is in factored form (f(x)=a(x-r_1)(x-r_2)), use (x=frac{r_1+r_2}{2}) and then find (y=f(x))