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

Question

show work

Answer 1

The vertex form of a parabola is given by:

f(x) = a(x-h)^2 + k

where (h,k) is the vertex of the parabola.

In this case, we are given that the axis of symmetry is x = -2, which means the vertex has an x-coordinate of -2. Therefore, h = -2.

We are also given that the maximum height is 8, which means the vertex has a y-coordinate of 8. Therefore, k = 8.

Substituting these values into the vertex form equation, we have:

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

Now we need to find the value of 'a'.

We are given that the parabola passes through the point (-6,2). Substituting these values into the equation, we get:

2 = a((-6) + 2)^2 + 8
2 = a(4)^2 + 8
2 = a * 16 + 8
2 = 16a + 8
16a = -6
a = -6/16
a = -3/8

So the equation of the parabola in vertex form is:

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