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

## a parabola has an axis of symmetry at x=-7, a maximum height of 4 and also passes through the point (-6,0). Write the equation of the parabola in vertex form.

# 1 Answer

This is the deal:

Since the axis of symmetry is a vertical line, x = -7 and the parabola has a maximum height of 4 units above the x axis; we can safely conclude that it opens down and its vertex is at (-7, 4). Besides, it passes through the point (-6, 0), which is below the vertex; definitively, it opens down. Furthermore, if it would open up, it would have a minimum height, not a maximum; so, no arguments, it opens down.

The vertex form of the equation is y = a(x - h)^{2} + k

Since the vertex is defined as (h, k), h = -7 and k = 4:

y = a(x - (-7))^{2} + 4 or: y = a(x + 7)^{2} + 4

We just need to find the value of "a" (which must be negative, remember, it opens down). Here is where we use the point through which the parabola passes through: (-6, 0).

We substitute the values x = -6, y = 0 in the equation and solve for a:

0 = a(-6 + 7)^{2} + 4 simplify and subtract 4 on both sides of the equation:

-4 = a(1)^{2 } or a = -4 Now we are ready to write our completed equation:

y = -4(x + 7)^{2} + 4 A lot of fun, right?