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

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 #

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Check by expanding brackets:

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

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

#"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"#