Question

suppose a parabola has an axis of symmetry at x=-2, a maximum height of 8, passes through the point (-6, 2). Wrire the equation of the parabola in vertex form.

Answer 1

Answer:

y = -3/8(x + 2)^2 + 8

Step-by-step explanation:

vertex form is

y = a(x - b)^2 + c     where a is a constant and (b,c) is the vertex.

The maximum is at (-2, 8)  because x 8 = height and x =-2 is equn. of symmetry

So here we have

y = a(x - (-2))^2 + 8

y = a(x + 2)^2 + 8

Now at the point (-6, 2):

2 = a(-6+2)^2 + 8

2 = 16a + 8

16a = -6

1 = -3/8.

So our equation is y = 3/8(x + 2)^2 + 8