Question

What is the vertex form of `y=(3x-5)(6x-2)`?

Answer 1

The vertex form of `y=(3x-5)(6x-2) = 30(x-0.6)^2-0.8`

Explanation:

First we must know what is meant by the vertex form of a quadratic function, which is
`y=a(x-h)^2+k` (https://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html)
We, therefore, want `(3x-5)(6x-2)` on the above form.
We have `(3x-5)(6x-2)=30x^2-36x+10`

Therefore `a=30`

`30(x-h)^2+k=30(x^2-2hx+h^2)+k= 30x^2-36x+10=30(x^2-1,2x)+10`
Therefore `2h=1,2`

The quadratic part, therefore, is
`30(x-0.6)^2=30(x^2-1.2x+0.36)=30x^2-36x+10.8`
This gives
`30x^2-36x+10 = (30x^2-36x+10.8)-0.8`
Therefore,
`(3x-5)(6x-2)=30(x-0.6)^2-0.8`

Answer 2

`y=18(x-1)^2-8`

Explanation:

`"the equation of a parabola in "color(blue)"vertex form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))`

`"where "(h,k)" are the coordinates of the vertex and a"`
`"is a multiplier"`

`"to obtain this form use "color(blue)"completing the square"`

`"expand the factors"`

`rArry=18x^2-36x+10`

`• " the coefficient of the "x^2" term must be 1"`

`"factor out 18"`

`y=18(x^2-2x+5/9)`

`• " add/subtract "(1/2"coefficient of the x-term")^2" to"`
`x^2-2x`

`y=18(x^2+2(-1)x color(red)(+1)color(red)(-1)+5/9)`

`color(white)(y)=18(x-1)^2+18(-1+5/9)`

`color(white)(y)=18(x-1)^2-8larrcolor(red)"in vertex form"`