Question 196648
I'm assuming that you want to complete the square.





{{{(1/2)y^2-3y+13/2}}} Start with the given expression.



{{{(1/2)(y^2-6y+13)}}} Factor out the {{{y^2}}} coefficient {{{1/2}}}. This step is very important: the {{{y^2}}} coefficient <font size=4><b>must</b></font> be equal to 1.



Take half of the {{{y}}} coefficient {{{-6}}} to get {{{-3}}}. In other words, {{{(1/2)(-6)=-3}}}.



Now square {{{-3}}} to get {{{9}}}. In other words, {{{(-3)^2=(-3)(-3)=9}}}



{{{(1/2)(y^2-6y+highlight(9-9)+13)}}} Now add <font size=4><b>and</b></font> subtract {{{9}}} inside the parenthesis. Make sure to place this after the "y" term. Notice how {{{9-9=0}}}. So the expression is not changed.



{{{(1/2)((y^2-6y+9)-9+13)}}} Group the first three terms.



{{{(1/2)((y-3)^2-9+13)}}} Factor {{{y^2-6y+9}}} to get {{{(y-3)^2}}}.



{{{(1/2)((y-3)^2+4)}}} Combine like terms.



{{{(1/2)(y-3)^2+(1/2)(4)}}} Distribute.



{{{(1/2)(y-3)^2+2}}} Multiply.



So after completing the square, {{{1/2y^2-3y+13/2}}} transforms to {{{(1/2)(y-3)^2+2}}}. 



In other words, {{{(1/2)y^2-3y+13/2=(1/2)(y-3)^2+2}}}.