Question

What is the vertex form of `y= 6x^2-9x+3`?

Answer 1

`y = 6(x-3/4)^2 - 3/8`

Explanation:

To complete the square of the equation, first take out the 6:

`y = 6(x^2 - 3/2x + 1/2)`

Then do the bit in the brackets:

`y = 6[(x-3/4)^2 - 9/16 + 1/2]`

`y = 6[(x-3/4)^2 - 1/16]`

`y = 6(x-3/4)^2 - 3/8`, as required.

Answer 2

`y=6(x-3/4)^2-3/8`

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 the method of"`
`color(blue)"completing the square"`

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

`rArry=6(x^2-3/2x)+3`

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

`x^2-3/2x`

`rArry=6(x^2+2(-3/4)xcolor(red)(+9/16)color(red)(-9/16))+3`

`rArry=6(x-3/4)^2-27/8+3`

`rArry=6(x-3/4)^2-3/8larrcolor(red)"in vertex form"`