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:
y=a(x-2)^2+4
We need to find a and we have only 1 equation, so we can use the other point for x and y:
17=a(1-2)^2+4
17=a+4
a=13
The graph of the parabola in vertex form would be y=13(x-2)^2+4
Converting to Standard Form, we get:
y=13(x-2)^2+4
y=13(x^2-4x+4)+4
y=13x^2-52x+52+4
y=13x^2-52x+56
+36916 Vertex (2,4) form of parabola
y = a (x -2)^2 + 4 sub in the point (1,17) to calculate 'a'
17 = a (1-2)^2 + 4 so a = 13
y = 13 ( x-2)^2 + 4 expand
y = 13x^2 -52x + 56
