Question

Suppose a parabola has an axis of symmetry at x = -2, a minimum height at -6, and passes through the point (0, 10). Write the equation of the parabola in vertex form.

Answer 1

a(x-h)^2 + k is the vertex form

Answer 2

a(x+2)^2+k is the next step

Answer 3

vertex = (-2,-6) = (h,k)

plug in point (0,10) for (x,y)
solve for a

Answer 4

how is it a(x+2) and not a(x-2) ??

Answer 5

x-h , h = -2, x-(-2) = x+2

Answer 6

so 10 = a(0 + 2) ^2 - 6
10 = a(2)^2 - 6
10 = a * 4 - 6
10 = a* -2
a = -2 -10
a = 12
y = 12( x + 2)^2 - 6

Answer 7

@dumbcow

Answer 8

I mean a = -12 not 12

Answer 9

almost, go back to step:
4a -6 = 10

you cant add the 4 and -6
add 6 to both sides

4a = 16

Answer 10

oh ok so then a = 4
y = 4(x+2)^2 - 6? @dumbcow