Question

What is the vertex form of `7y=3x^2 − 2x + 12`?

Answer 1

We will have to complete the square for this quadratic which will put the equation in vertex form.

First lets solve for the y variable by dividing both sides by 7

`(cancel7y)/cancel7=3/7x^2-2/7x+12/7`

Set the equation equal to zero.

`0=3/7x^2-2/7x+12/7`

Subtract `12/7` from both sides

`0color(red)(-12/7)=3/7x^2-2/7x+12/7color(red)(-12/7)`

Simplify

`color(red)(-12/7)=3/7x^2-2/7x`

Factor out `3/7`

`-12/7=3/7(x^2-2/cancel7(cancel7/3)x)`

Simplify

`-12/7=3/7(x^2-2/3x)`

Take the coefficient of x and divide it by 2 and then square it

`((-2/3)/2)^2=(-2/3*1/2)^2=(-2/6)^2=(-1/3)^2=1/9`

Add `1/9` to the right side and add `3/7(1/9)` to the left side because we factored out `3/7` in the beginning. This process will keep the equation balanced.

`color(red)(3/7(1/9))-12/7=3/7(x^2-2/3x+color(red)(1/9))`

Simply

`color(red)(cancel3/7(1/(cancel9 3)))-12/7=3/7(x^2-2/3x+color(red)(1/9))`

`1/21-12/7=3/7(x^2-2/3x+color(red)(1/9))`

Find Common Denominator

`1/21-12/7*color(red)(3/3)=3/7(x^2-2/3x+color(red)(1/9))`

`1/21-36/21=3/7(x^2-2/3x+color(red)(1/9))`

The right side is a perfect square trinomial

`1/21-36/21=3/7(x-1/3)^2`

`-35/21=3/7(x-1/3)^2`

`-(cancel35 5)/(cancel 21 3)=3/7(x-1/3)^2`

`-5/3=3/7(x-1/3)^2`

Add `5/3` from both sides

`color(red)(5/3)-5/3=3/7(x-1/3)^2color(red)(+5/3)`

`0=3/7(x-1/3)^2color(red)(+5/3)`

Vertex form `=> y=(x-h)^2+k`

Vertex `=> (h,k) => (1/3,5/3)`

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