SOLUTION: Tutors...HELP...Thank you find an equation of the inverse relation y=3x^2-5x+9 and x=1/2y+4

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Question 143907: Tutors...HELP...Thank you
find an equation of the inverse relation y=3x^2-5x+9
and
x=1/2y+4
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
# 1



y=3+x%5E2-5+x%2B9 Start with the given equation


y-9=3+x%5E2-5+x Subtract 9 from both sides


y-9=3%28x%5E2%2B%28-5%2F3%29x%29 Factor out the leading coefficient 3


Take half of the x coefficient -5%2F3 to get -5%2F6 (ie %281%2F2%29%28-5%2F3%29=-5%2F6).

Now square -5%2F6 to get 25%2F36 (ie %28-5%2F6%29%5E2=%28-5%2F6%29%28-5%2F6%29=25%2F36)




y-9=3%28x%5E2%2B%28-5%2F3%29x%2B25%2F36-25%2F36%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 25%2F36 does not change the equation



y-9=3%28%28x-5%2F6%29%5E2-25%2F36%29 Now factor x%5E2%2B%28-5%2F3%29x%2B25%2F36 to get %28x-5%2F6%29%5E2


y-9=3%28x-5%2F6%29%5E2-3%2825%2F36%29 Distribute


y-9=3%28x-5%2F6%29%5E2-25%2F12 Multiply


y=3%28x-5%2F6%29%5E2-25%2F12%2B9 Now add 9 to both sides to isolate y


y=3%28x-5%2F6%29%5E2%2B83%2F12 Combine like terms


Now the quadratic is in vertex form



x=3%28y-5%2F6%29%5E2%2B83%2F12 Switch x and y


x-83%2F12=3%28y-5%2F6%29%5E2 Add 83%2F12 to both sides


%28x-83%2F12%29%2F3=%28y-5%2F6%29%5E2 Divide both sides by 3


%28x%29%2F3-%2883%2F12%29%2F3=%28y-5%2F6%29%5E2 Break up the fraction


%28x%29%2F3-83%2F36=%28y-5%2F6%29%5E2 Divide and reduce


0%2B-sqrt%28%28x%29%2F3-83%2F36%29=y-5%2F6 Take the square root of both sides


0%2B-sqrt%28%28x%29%2F3-83%2F36%29%2B5%2F6=y Add 5%2F6 to both sides


So the inverse is y=0%2B-sqrt%28%28x%29%2F3-83%2F36%29%2B5%2F6






# 2
x=%281%2F2%29y%2B4 Start with the given equation


x-4=%281%2F2%29y Subtract 4 from both sides


2%28x-4%29=y Multiply both sides by 2


2x-8=y Distribute


So the inverse is y=2x-8