# What is the vertex form of x = (5y +3 )^2 + 2x+1?

Oct 23, 2017

The vertex form for a parabola of this type is:

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

where $\left(h , k\right)$ is the vertex and "a" is the leading coefficient.

#### Explanation:

Given: $x = {\left(5 y + 3\right)}^{2} + 2 x + 1$

Here is a graph of the original equation:

graph{x = (5y +3 )^2 + 2x+1 [-10, 10, -5, 5]}

Factor out a 5 from within the square:

$x = {\left(5 \left(y + \frac{3}{5}\right)\right)}^{2} + 2 x + 1$

Use the rules of squaring:

$x = {\left(5\right)}^{2} {\left(y + \frac{3}{5}\right)}^{2} + 2 x + 1$

$x = 25 {\left(y + \frac{3}{5}\right)}^{2} + 2 x + 1$

The sign within the square must be negative:

$x = 25 {\left(y - \left(- \frac{3}{5}\right)\right)}^{2} + 2 x + 1$

Subtract $2 x$ from both sides:

$- x = 25 {\left(y - \left(- \frac{3}{5}\right)\right)}^{2} + 1$

Multiply both sides by -1:

$x = - 25 {\left(y - \left(- \frac{3}{5}\right)\right)}^{2} - 1$

The is the same form as equation [1] where $a = - 25 , k = - \frac{3}{5} \mathmr{and} h = - 1$

Here is a graph of the vertex form:

graph{x = -25(y - (-3/5))^2-1 [-10, 10, -5, 5]}

Please observe that the graphs are identical.