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Combining Transformations; Completing the Square

  1. Oct 7, 2012 #1
    Hello PF!
    1. The problem statement, all variables and given/known data
    The graph of the function y = 2x2 + x +1 is stretched vertically about the x-axis by a factor of 2, stretched horizontally about the y-axis by a factor of 1/3 and translated 2 units right and 4 units down. Write the equation of the transformed function

    2. Relevant equations
    y= af(1/b(x-h))+k


    3. The attempt at a solution
    First I completed the square of y = 2x2 + x +1:
    2x2 + x +1 = y
    2(x2+1/2x+1/16-1/16)+1=y
    2(x+1/4)2+14/16=y

    Then, using mapping notation I calculated what the new x and y coordinates would be on the transformed function:
    (x,y) → (1/3x+2,2y-4)
    Therefore, the point (-2,7) → (4/3,10)

    Using the aforementioned equation I transformed the function:
    a=2
    b=1/3
    h=2
    k=-4

    y=2(2(3(x-7/4)2)-25/8)

    However, when I input the new function into my calculator I received the values (-2,7) → (4/3,53/9) which is demonstrated to be incorrect. Is there a way to solve this problem without putting the function in the form y= af(1/b(x-h))+k? If possible, can anyone show me how to derive the correct equation for the transformed function after completing the square of the function? Thanks!
     
  2. jcsd
  3. Oct 9, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Almost. 1- 1/16= 15/16, not 14/16.

    where did the "1/3" come from? If you are comparing y= 2(x+ 1/4)2+ 15/16 to y= x2 then x changes to x+ 1/4 and y changes to 2y+ 15/16.

     
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