Quadratic Function Equation Help..........

The graph of a quadratic function has a vertex at (3, 3) and passes the point (0, -6) . Find the quadratic function.



If you could please explain how you got the function that would be great. I have worked the problem and gotten the wrong answer several times. Thank You!

Use vertex form:
-(x-3)^2 + 3

-9 + 3 = -6

I am not sure if they want you to expand it or not.

But anyways, my first guess was:
(x-3)^2 + 3

But when i tried plugging in the numbers, i go 9 + 3 = 12 ≠ -6
So i tagged a negative in front and got -9 + 3 = -6

So we know we have an upside down parabola shifted right 3 and up 3

f(x)= ax^2 + bx +c

To find the vertex, we have to get the derivative of the function and then equate it to zero. At this point we find (x) of the vertex. That is in the normal situations. Let us do it here:

f ' (x)= 2xa + b =0 (we have the x=3 for the vertex) so: 2(3)a+b=0 and hence: 6a= -b equation1

Now, for x=3, as given y=3. so we replace in the equation: 3= 9a +3b +c equation2

From the given point we also get another equation for y=-6 when x=0

It gives that c= -6 so, we now replace in the previous equations:

9a + 3(-6a) -6=3 which gives a= -1 and hence b= 6

The complete quadratic equation becomes: f(x)= -x^2 + 6x -6