Question 143907
# 1




{{{y=3 x^2-5 x+9}}} Start with the given equation



{{{y-9=3 x^2-5 x}}}  Subtract {{{9}}} from both sides



{{{y-9=3(x^2+(-5/3)x)}}} Factor out the leading coefficient {{{3}}}



Take half of the x coefficient {{{-5/3}}} to get {{{-5/6}}} (ie {{{(1/2)(-5/3)=-5/6}}}).


Now square {{{-5/6}}} to get {{{25/36}}} (ie {{{(-5/6)^2=(-5/6)(-5/6)=25/36}}})





{{{y-9=3(x^2+(-5/3)x+25/36-25/36)}}} Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of {{{25/36}}} does not change the equation




{{{y-9=3((x-5/6)^2-25/36)}}} Now factor {{{x^2+(-5/3)x+25/36}}} to get {{{(x-5/6)^2}}}



{{{y-9=3(x-5/6)^2-3(25/36)}}} Distribute



{{{y-9=3(x-5/6)^2-25/12}}} Multiply



{{{y=3(x-5/6)^2-25/12+9}}} Now add {{{9}}} to both sides to isolate y



{{{y=3(x-5/6)^2+83/12}}} Combine like terms



Now the quadratic is in vertex form




{{{x=3(y-5/6)^2+83/12}}} Switch x and y



{{{x-83/12=3(y-5/6)^2}}} Add {{{83/12}}} to both sides



{{{(x-83/12)/3=(y-5/6)^2}}} Divide both sides by 3



{{{(x)/3-(83/12)/3=(y-5/6)^2}}} Break up the fraction



{{{(x)/3-83/36=(y-5/6)^2}}} Divide and reduce



{{{0+-sqrt((x)/3-83/36)=y-5/6}}} Take the square root of both sides



{{{0+-sqrt((x)/3-83/36)+5/6=y}}} Add {{{5/6}}} to both sides



So the inverse is {{{y=0+-sqrt((x)/3-83/36)+5/6}}}




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# 2

{{{x=(1/2)y+4}}} Start with the given equation



{{{x-4=(1/2)y}}} Subtract 4 from both sides



{{{2(x-4)=y}}} Multiply both sides by 2



{{{2x-8=y}}} Distribute



So the inverse is {{{y=2x-8}}}