Complete the general form equation of the parabola that passes through (1,20) with vertex at (−2,2) .

Answer 1

Vertex form = y-k=4p(x-h)² where (h,k) = vertex. Replacing h & k with the given location of the vertex yields y-2=4P(x+2)². Inserting the given point of (1,20) yields 20-2=4p(1+2)². Solving, 4P=2 thus the vertex form of this parabola is y-2=2(x+2)².

Answer 2

it's an upward opening or a rightward opening parabola

y-2 = a(x+2)^2 plug in the point (1,20) to solve for a

20-2 = a(1+2)^2 = 9a

a = 18/9 = 2

y-2 =2(x+2)^2

but for a rightward opening parabola

x+2 =a(y-2)^2

1+2 =a(20-2)^2

3 = 324a

a = 3/324 = 1/108

x+2 = (1/108)(y-2)^2