Question 237368
b) 


{{{f(x)=x^2-kx+9}}} Start with the given function.



{{{f(x)=x^2-4x+9}}} Plug in {{{k=4}}}



Now let's complete the square.



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



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



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



{{{f(x)=(x^2-4x+4)-4+9}}} Group the first three terms.



{{{f(x)=(x-2)^2-4+9}}} Factor {{{x^2-4x+4}}} to get {{{(x-2)^2}}}.



{{{f(x)=(x-2)^2+5}}} Combine like terms.



So after completing the square, {{{x^2-4x+9}}} transforms to {{{(x-2)^2+5}}}. So {{{x^2-4x+9=(x-2)^2+5}}}.



So {{{f(x)=x^2-4x+9}}} is equivalent to {{{f(x)=(x-2)^2+5}}}.



Now {{{f(x)=(x-2)^2+5}}} is in the form {{{f(x)=(x-p)^2+q}}} where p=2 and q=5.