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

##### 1 Answer

Vertex form is:

#y = 3/2(x+5/6)^2+119/24#

or more strictly:

#y = 3/2(x-(-5/6))^2+119/24#

#### Explanation:

Vertex form looks like this:

#y = a(x-h)^2+k#

where

Given:

#2y = 3x^2+5x+12#

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

To avoid some fractions during the calculations, first multiply by

#24y = 12(2y)#

#color(white)(24y) = 12(3x^2+5x+12)#

#color(white)(24y) = 36x^2+60x+144#

#color(white)(24y) = (6x)^2+2(6x)(5)+(5)^2+119#

#color(white)(24y) = (6x+5)^2+119#

#color(white)(24y) = 36(x+5/6)^2+119#

Then dividing both ends by

#y = 3/2(x+5/6)^2+119/24#

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

#y = 3/2(x-(-5/6))^2+119/24#

Comparing this with:

#y = a(x-h)^2+k#

we find that the parabola is upright, 3/2 as steep as

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]}