Question 162179
{{{x^2-(1/2)x+1}}} Start with the given expression.



Take half of the {{{x}}} coefficient {{{-1/2}}} to get {{{-1/4}}}. In other words, {{{(1/2)(-1/2)=-1/4}}}.



Now square {{{-1/4}}} to get {{{1/16}}}. In other words, {{{(-1/4)^2=(-1/4)(-1/4)=1/16}}}



{{{x^2-(1/2)x+highlight(1/16-1/16)+1}}} Now add <font size=4><b>and</b></font> subtract {{{1/16}}}. Make sure to place this after the "x" term. Notice how {{{1/16-1/16=0}}}. So the expression is not changed.



{{{(x^2-(1/2)x+1/16)-1/16+1}}} Group the first three terms.



{{{(x-1/4)^2-1/16+1}}} Factor {{{x^2-(1/2)x+1/16}}} to get {{{(x- 1/4)^2}}}.



{{{(x-1/4)^2+15/16}}} Combine like terms.



So after completing the square, {{{x^2-(1/2)x+1}}} transforms to {{{(x-1/4)^2+15/16}}}. So {{{x^2-1/2x+1=(x-1/4)^2+15/16}}}.



So {{{g(x)=x^2-(1/2)x +1}}} is equivalent to {{{g(x)=(x-1/4)^2+15/16}}}.



Notice how {{{g(x)=(x-1/4)^2+15/16}}} is in vertex form {{{y=a(x-h)^2+k}}} where the vertex is (h,k)



Since {{{h=1/4}}} and {{{k=15/16}}}, this means that the vertex is *[Tex \LARGE \left(\frac{1}{4},\frac{15}{16}\right)]