Question

How do you write the vertex form of the equation where the vertex is `(-3,-9)` and the directrix is `x=-49/16`?

Answer 1

Please observe that the directrix is a vertical line, `x = -49/16`; this means that the parabola is horizontally oriented and it opens either to the left or to the right.

Explanation:

The vertex form of the equation for a parabola of this type is:

`x = 1/(4f)(y-k)^2+h" [1]"`

where `(h,k)` is the vertex and the equation of the directrix is `x = h-f`.

Substitute `(h,k) = (-3,-9)` into equation [1]:

`x = 1/(4f)(y--9)^2 -3" [1.1]"`

We are given:

`x = -49/16" [2]"`

Substitute `x = h-f` into the equation [2]:

`h-f = -49/16"`

Substitute `h= -3`:

`-3-f = -49/16"`

Solve for `f`:

`3+f = 49/16"`

`f = 49/16-3`

`f = 1/16" [2.1]"`

Substitute equation [2.1] into equation [1.1]:

`x = 1/(4(1/16))(y--9)^2 -3`

`x = 16/4(y--9)^2 -3`

`x = 4(y--9)^2 -3 larr` this the vertex form.