Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,17). Express your answer in the form "ax^2+bx+c".
Since the parabola has a vertical axis of symmetry, we know that the parabola opens up or down.
We can use vertex form y=a(x-h)^2+k
If the vertex is (h,k) then we can substitute to find:
We need to find a and we have only 1 equation, so we can use the other point for x and y:
The graph of the parabola in vertex form would be y=13(x-2)^2+4
Converting to Standard Form, we get: