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

Vertex form: `y=a\left(x-h\right)^{2}+k`
<-------SHOW YOUR WORK

Answer 1

First, we know that the axis of symmetry is at `x=-2` which means that the vertex of the parabola is at `(-2,k)`. Since the maximum height is at `y=8`, this means that the vertex is at `(-2,8)`.

Using the vertex form of a parabola, we have `y=a(x-h)^2 + k`. Plugging in the vertex coordinates, we get `y=a(x-(-2))^2 + 8`. Simplifying this, we have `y=a(x+2)^2 + 8`.

Now we need to find the value of `a`. We can do this by using the point given that the parabola passes through `(-6,2)`. Plugging in these coordinates into our equation, we get `2=a((-6)+2)^2 + 8`. Simplifying this further, we have `2=a(-4)^2 + 8`.

Solving for `a`, we subtract 8 from both sides to get `2-8 = a(-4)^2`. Simplifying this, we have `-6 = 16a`. Dividing both sides by 16, we find that `a = -3/8`.

Plugging this value of `a` back into the equation `y=a(x+2)^2 + 8`, we have `y=(-3/8)(x+2)^2 + 8`.

Therefore, the equation of the parabola is `y=(-3/8)(x+2)^2 + 8` in vertex form.