Question

What is the vertex form of `y=6x^2 - 27x - 15`?

Answer 1

`y=6(x-9/4)^2-363/8`

Explanation:

For a more detailed example of method see:

https://socratic.org/s/aFtwtRb4

`y=6x^2-27x-15`

`y=6(x-27/(2xx6))^2+k-15`

`y=6(x-9/4)^2+k-15`

..............................................................
'Gets rid' of the introduced error.

Set `" "6(-9/4)^2+k=0`

`6xx81/16+k=0`

`k=-243/8`
.........................................................

`y=6(x-9/4)^2-243/8-15`

`y=6(x-9/4)^2-363/8`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check by expanding brackets:

`y=6(x^2-18/4x+81/16)-363/8`

`y=6x^2-27x-15`

Answer 2

`y=6(x-9/4)^2-363/8`

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 constant.

`"for a parabola in standard form " y=ax^2+bx+c`

`x_(color(red)"vertex")=-b/(2a)`

`y=6x^2-27x-15" is in this form"`

`"with " a=6,b=-27" and " c=-15`

`rArrx_(color(red)"vertex")=-(-27)/12=27/12=9/4`

`"substitute this value into the function for y-coordinate"`

`rArry_(color(red)"vertex")=(6xx81/16)-(27xx9/4)-15`

`color(white)(xxxxxxx)=243/8-486/8-120/8`

`color(white)(xxxxxxx)=-363/8`

`rArr(h,k)=(9/4,-363/8)`

`rArry=6(x-9/4)^2-363/8larrcolor(red)" in vertex form"`