Question

Suppose 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.

Answer 1

The vertex form of a quadratic equation:
y = a(x - h)^2+ k

Axis of symmetry is said to be x=-7. This means that the x value of the vertex is at x=-7. The maximum height is 4 thus the y value of the vertex is 4.

in the form
y=a(x-h)^2+k
(h,k) represents the vertex

Thus, the equation becomes,
y=a(x-(-7))^2+4
y=a(x+7)^2+4

We obtain a by substituting the point (-6,0) for x and y in the equation.
0=a(-6+7)^2+4
0=a(1)^2+4
0=a+4
a = -4

Therefore, the equation in vertex form is:
y=-4(x+7)^2+4

Answer 2

axis of symmetry at x=-7, meaning that the x value of the vertex is at x=-7
means that it is a function
max height is 4
means that the y value of the vertex is 4
passes through the point (-6,0) or one point is x=-6, y=0

in the form
y=a(x-h)^2+k
(h,k) is vertex

we know that vertex is at (-7,4)
y=a(x-(-7))^2+4
y=a(x+7)^2+4
find a by plugging in the point (-6,0) aka subsitutin -6 for x and 0 for y
0=a(-6+7)^2+4
0=a(1)^2+4
0=a+4
-4=a

therefor the equation is
y=-4(x+7)^2+4