# What is the vertex form of x= 10y^2-31y+15 ?

##### 1 Answer
Feb 1, 2016

The vertex form for a parabola is a function of the form;

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

Where $\left(h , k\right)$ is the vertex of the parabola. To convert a quadratic to vertex form, we want to start by isolating the $y$ terms and completing the square.

$x = 10 \left({y}^{2} - \frac{31}{10} y\right) + 15$

Complete the square inside the parenthesis.

$x = 10 \left({y}^{2} - \frac{31}{10} y + \frac{961}{400} - \frac{961}{400}\right) + 15$

$x = 10 \left({y}^{2} - \frac{31}{10} y + \frac{961}{400}\right) - \frac{961}{40} + 15$

$x = 10 {\left({y}^{2} - \frac{31}{20}\right)}^{2} - 24 \frac{1}{400} + 15$

$x = 10 {\left({y}^{2} - \frac{31}{20}\right)}^{2} - 9 \frac{1}{400}$

This is the vertex form of the equation.