Question 227665
I'll do the first one to get you started.


a) 




{{{4x^2+3x+1}}} Start with the given expression.



{{{4(x^2+(3/4)x+1/4)}}} Factor out the {{{x^2}}} coefficient {{{4}}}. This step is very important: the {{{x^2}}} coefficient <font size=4><b>must</b></font> be equal to 1.



Take half of the {{{x}}} coefficient {{{3/4}}} to get {{{3/8}}}. In other words, {{{(1/2)(3/4)=3/8}}}.



Now square {{{3/8}}} to get {{{9/64}}}. In other words, {{{(3/8)^2=(3/8)(3/8)=9/64}}}



{{{4(x^2+(3/4)x+highlight(9/64-9/64)+1/4)}}} Now add <font size=4><b>and</b></font> subtract {{{9/64}}} inside the parenthesis. Make sure to place this after the "x" term. Notice how {{{9/64-9/64=0}}}. So the expression is not changed.



{{{4((x^2+(3/4)x+9/64)-9/64+1/4)}}} Group the first three terms.



{{{4((x+3/8)^2-9/64+1/4)}}} Factor {{{x^2+(3/4)x+9/64}}} to get {{{(x+3/8)^2}}}.



{{{4((x+3/8)^2+7/64)}}} Combine like terms.



{{{4(x+3/8)^2+4(7/64)}}} Distribute.



{{{4(x+3/8)^2+7/16}}} Multiply.



So after completing the square, {{{4x^2+3x+1}}} transforms to {{{4(x+3/8)^2+7/16}}}. So {{{4x^2+3x+1=4(x+3/8)^2+7/16}}}.



So {{{f(x)=4x^2+3x+1}}} is equivalent to {{{f(x)=4(x+3/8)^2+7/16}}}.



So the equation {{{y=4(x+3/8)^2+7/16}}} is now in vertex form {{{y=a(x-h)^2+k}}} where {{{a=4}}}, {{{h=-3/8}}}, and {{{k=7/16}}}