Question

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

Answer 1

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 `(h, k)` is the vertex of the parabola and `a` is a multiplier determining which way up the parabola is and its steepness.

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 `2^2 * 3 = 12`. We will divide by `24` at the end:

`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 `24` we find:

`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 `x^2` with vertex `(h, k) = (-5/6, 119/24)`

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]}