Question

How do you write the equation in vertex form `x=3y^2+5y-9`?

Answer 1

Please see the explanation.

Explanation:

The vertex form of a parabola of this type is:

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

where, x and y correspond to any point, `(x, y)`, on parabola, h and k correspond to the point that is the vertex, `(h, k)`, and a is the leading coefficient of the `y^2` term in standard form.

Given: `x = 3y^2 + 5y + 9" [2]"`

Let's expand the square in equation [1], multiply the terms by a, and write it directly above equation [2]:

`x = ay^2 - 2aky + ak^2 + h" [1a]"`
`x = 3y^2 + 5y + 9" [2]"`

Please observe that `a = 3` so lets add zero to equation [2] in the form `3k^2 - 3k^2`:

`x = ay^2 - 2aky + ak^2 + h" [1a]"`
`x = 3y^2 + 5y + 3k^2 - 3k^2 + 9" [2a]"`

If we set second term in equation [1a] equal to the second term in equation [2a] and we substitute 3 for a, then we can solve for the value of k:

`-2aky = 5y`

Substitute 3 for a:

`-2(3)ky = 5y`

`k = -5/6`

This tells us that the first three terms in equation [2a] can be replaced by `3(y - -5/6)^2`:

`x = 3(y - -5/6)^2 - 3k^2 + 9" [2b]"`

Substitute `(-5/6)` for k:

`x = 3(y - -5/6)^2 - 3(-5/6)^2 + 9" [2c]"`

Simplify:

`x = 3(y - -5/6)^2 + 83/12" [2d]"`

Equation [2d] is the vertex form for a parabola that opens to the right or left.

You can read that the vertex is at the point `(83/12, -5/6)` from equation [2d].