What is the vertex form of #y=1/2x^2-1/6x+6/13#?

1 Answer
Jun 11, 2016

#y=1/2(x-1/6)^2+409/936# (assuming I managed the arithmetic correctly)

Explanation:

The general vertex form is
#color(white)("XXX")y=color(green)(m)(x-color(red)(a))^2+color(blue)(b)#
for a parabola with vertex at #(color(red)(a),color(blue)(b))#

Given:
#color(white)("XXX")y=1/2x^2-1/6x+6/13#

#rArr#
#color(white)("XXX")y=1/2(x^2-1/3x)+6/13#

#color(white)("XXX")y=1/2(x^2-1/3x+(1/6)^2)+6/13-1/2*(1/6)^2#

#color(white)("XXX")y=1/2(x-1/6)^2+6/13-1/72#

#color(white)("XXX")y=1/2(x-1/6)^2+(6*72-1*13)/(13*72)#

#color(white)("XXX")y=color(green)(1/2)(x-color(red)(1/6))^2+color(blue)(409/936)#
which is the vertex form with vertex at #(color(red)(1/6),color(blue)(409/936))#

The graph below of the original equation indicates that our answer is at least approximately correct.

graph{1/2x^2-1/6x+6/13 [-0.6244, 1.0606, -0.097, 0.7454]}