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

1 Answer
Johnson Z.
Dec 16, 2015

vertex#=(5/18, -25/36)#

Explanation:

Start by expanding the brackets and simplifying the expression.

#y=5x^2-x-1+(2x-1)^2#
#y=5x^2-x-1+(4x^2-4x+1)#
#y=9x^2-5x#

Take your simplified equation and complete the square.

#y=9x^2-5x#
#y=9(x^2-5/9x+((5/9)/2)^2-((5/9)/2)^2)#
#y=9(x^2-5/9x+(5/18)^2-(5/18)^2)#
#y=9(x^2-5/9x+25/324-25/324)#
#y=9(x^2-5/9x+25/324)-(25/324*9)#
#y=9(x-5/18)^2-(25/color(red)cancelcolor(black)324^36*color(red)cancelcolor(black)9)#
#y=9(x-5/18)^2-25/36#

Recall that the general equation of a quadratic equation written in vertex form is:

#y=a(x-h)^2+k#

where:
#h=#x-coordinate of the vertex
#k=#y-coordinate of the vertex

So in this case, the vertex is #(5/18,-25/36)#.

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