# What is the vertex form of 2y = 3x^2+5x+12?

Jun 9, 2018

Vertex form is:

$y = \frac{3}{2} {\left(x + \frac{5}{6}\right)}^{2} + \frac{119}{24}$

or more strictly:

$y = \frac{3}{2} {\left(x - \left(- \frac{5}{6}\right)\right)}^{2} + \frac{119}{24}$

#### Explanation:

Vertex form looks like this:

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

where $\left(h , k\right)$ is the vertex of the parabola and $a$ is a multiplier determining which way up the parabola is and its steepness.

Given:

$2 y = 3 {x}^{2} + 5 x + 12$

we can get this into vertex form by completing the square.

To avoid some fractions during the calculations, first multiply by ${2}^{2} \cdot 3 = 12$. We will divide by $24$ at the end:

$24 y = 12 \left(2 y\right)$

$\textcolor{w h i t e}{24 y} = 12 \left(3 {x}^{2} + 5 x + 12\right)$

$\textcolor{w h i t e}{24 y} = 36 {x}^{2} + 60 x + 144$

$\textcolor{w h i t e}{24 y} = {\left(6 x\right)}^{2} + 2 \left(6 x\right) \left(5\right) + {\left(5\right)}^{2} + 119$

$\textcolor{w h i t e}{24 y} = {\left(6 x + 5\right)}^{2} + 119$

$\textcolor{w h i t e}{24 y} = 36 {\left(x + \frac{5}{6}\right)}^{2} + 119$

Then dividing both ends by $24$ we find:

$y = \frac{3}{2} {\left(x + \frac{5}{6}\right)}^{2} + \frac{119}{24}$

If we are strict about the signs of the coefficients, then for vertex form we could instead write:

$y = \frac{3}{2} {\left(x - \left(- \frac{5}{6}\right)\right)}^{2} + \frac{119}{24}$

Comparing this with:

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

we find that the parabola is upright, 3/2 as steep as ${x}^{2}$ with vertex $\left(h , k\right) = \left(- \frac{5}{6} , \frac{119}{24}\right)$

graph{(y-1/2(3x^2+5x+12))((x+5/6)^2+(y-119/24)^2-0.001) = 0 [-3.24, 1.76, 4.39, 6.89]}