What is the vertex of # y= -3x^2-5x-(3x-2)^2#?

2 Answers
Dec 30, 2017

The vertex is #(7/(24), -143/48)#.

Explanation:

First expand #(3x-2)^2=9x^2-12x+4#.

Substituting that in we have:

#y=-3x^2-5x-(9x^2-12x+4)#

Distribute the negative:

#y=-3x^2-5x-9x^2+12x-4#

Collect like terms:

#y=-12x^2+7x-4#

The vertex is #(h,k)# where #h=-b/(2a)# and #k# is the value of #y# when #h# is substituted.

#h=-(7)/(2(-12))=7/(24)#.

#k=-12(7/(24))^2+7(7/(24))-4=-143/48# (I used a calculator...)

The vertex is #(7/(24), -143/48)#.

Dec 30, 2017

#(7/24,-143/48)#

Explanation:

#"we require to express in standard form"#

#rArry=-3x^2-5x-(9x^2-12x+4)#

#color(white)(rArry)=-3x^2-5x-9x^2+12x-4#

#color(white)(rArry)=-12x^2+7x-4larrcolor(blue)"in standard form"#

#"given the equation of a parabola in standard form then"#
#"the x-coordinate of the vertex is"#

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

#"here "a=-12,b=7,c=-4#

#rArrx_(color(red)"vertex")=-7/(-24)=7/24#

#"substitute this value into the equation for y"#

#y=-12(7/24)^2+7(7/24)-4=-143/48#

#rArrcolor(magenta)"vertex "=(7/24,-143/48)#