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".

Guest Jan 4, 2022

#1**0 **

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

Guest Jan 4, 2022

#2**0 **

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

ElectricPavlov Jan 4, 2022