First, f(x) can be written as a y, so y=x^{2}-6x-16

Next, you can have two options here: 1) change this equation into the vertex form of a parabola or 2) find the coordinate for the vertex.

1) The first option is more detailed - The vertex form of a parabola is: y=(x-h)^{2}+k where the (h,k) is the vertex of the parabola.

We will have to use “completing the square” method to transform the equation from the general form to the vertex form.

y+16+9 = x^{2}-6x+9

y+25 = (x-3)^{2}

y=(x-3)^{2}-25, so the vertex is (3, -25)

2) The second option uses the coordinate (-b/2a, f(-b/2a)), where ax^{2}+bx+c is your equation in standard form.

So, a=1 and b=-6 therefore, (3, f(3)) yields a vertex at (3, -25)