What is the vertex form of #y=-1/3(x-2)(2x+5) #?

1 Answer

The Vertex Form is #(x--1/4)^2=-3/2*(y-27/8)#

Explanation:

We start from the given
#y=-1/3(x-2)(2x+5)#
Expand first
#y=-1/3(2x^2-4x+5x-10)#
simplify
#y=-1/3(2x^2+x-10)#
insert a #1=2/2# to make factoring of 2 clear

#y=-1/3(2x^2+2/2x-10)#

now, factor out the 2

#y=-2/3(x^2+x/2-5)#
complete the square now by adding #1/16# and subtracting #1/16# inside the grouping symbol

#y=-2/3(x^2+x/2+1/16-1/16-5)#

the first 3 terms inside the grouping symbol is now a Perfect Square Trinomial so that the equation becomes

#y=-2/3((x+1/4)^2-81/16)#
Distribute the #-2/3# inside the grouping symbol
#y=-2/3(x+1/4)^2-2/3(-81/16)#

#y=-2/3(x--1/4)^2+27/8#
let us simplify now to the Vertex Form

#y-27/8=-2/3(x--1/4)^2#

Finally

#(x--1/4)^2=-3/2(y-27/8)#

graph{(x--1/4)^2=-3/2(y-27/8)[-20,20,-10,10]}

God bless...I hope the explanation is useful..