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

Jan 5, 2017

#### Explanation:

The vertex form of a parabola of this type is:

$x = a {\left(y - k\right)}^{2} + h \text{ [1]}$

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

Given: $x = 3 {y}^{2} + 5 y + 9 \text{ [2]}$

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

$x = a {y}^{2} - 2 a k y + a {k}^{2} + h \text{ [1a]}$
$x = 3 {y}^{2} + 5 y + 9 \text{ [2]}$

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

$x = a {y}^{2} - 2 a k y + a {k}^{2} + h \text{ [1a]}$
$x = 3 {y}^{2} + 5 y + 3 {k}^{2} - 3 {k}^{2} + 9 \text{ [2a]}$

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:

$- 2 a k y = 5 y$

Substitute 3 for a:

$- 2 \left(3\right) k y = 5 y$

$k = - \frac{5}{6}$

This tells us that the first three terms in equation [2a] can be replaced by $3 {\left(y - - \frac{5}{6}\right)}^{2}$:

$x = 3 {\left(y - - \frac{5}{6}\right)}^{2} - 3 {k}^{2} + 9 \text{ [2b]}$

Substitute $\left(- \frac{5}{6}\right)$ for k:

$x = 3 {\left(y - - \frac{5}{6}\right)}^{2} - 3 {\left(- \frac{5}{6}\right)}^{2} + 9 \text{ [2c]}$

Simplify:

$x = 3 {\left(y - - \frac{5}{6}\right)}^{2} + \frac{83}{12} \text{ [2d]}$

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 $\left(\frac{83}{12} , - \frac{5}{6}\right)$ from equation [2d].