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

1 Answer
Tony B
Apr 10, 2017

Thus the vertex#->(x,y)=(5/4,15/8)#

Explanation:

#color(red)("For a full explanation of completing the square method see:")#
https://socratic.org/s/aDHYWAiE

We need to include the #x # that is outside the brackets

Expanding the brackets we have:
#y=2(x-1)^2" "color(white)(.)+3+x#
#y=2x^2-4x+2+3-x#

#y=2x^2-5x+5#

As the question presents a part vertex form equation it is reasonable to assume that the questioner's intention is for you to continue using vertex form format.

#y=2(x^2-5/2x)+5+k#

Where #k# is a correction constant

#y=2(x-5/4)^2+5+k#

Set #" "2(-5/4)^2+k=0" "=>" "k=-25/8# giving:

#y=2(x-5/4)^2+5-25/8#

#y=2(x-5/4)^2+15/8#

Thus the vertex#->(x,y)=(5/4,15/8)#

Tony B

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