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Simple reflection and translation question

  1. Mar 11, 2013 #1
    1. The problem statement, all variables and given/known data

    If the graph of 1/x is reflected in the y-axis, translated 3/4 units to the right and 2 units up, the resulting graph would have the equation...

    A) 1/(x-2) + 0.75
    B) 3/(4x) +2
    C) 2 - 1/(3/4x)
    D) -1/(4x-3) + 2
    E) 1/(3-4x) + 2

    2. The attempt at a solution

    So the general formula of a hyperbola is y = a/(x-b) + c (in this context anyway, I know there's another form)

    reflection in the y axis -> y=f(-x)
    translated 3/4 right -> b=3/4
    translated 2 up -> c=2

    So if I put it all together I get
    y = 1/(-x-b) + c
    y = 1/(-x-0.75) + 2
    = 1/-(x+0.75) + 2
    = -1/(x+0.75) +2

    The correct answer was D)
  2. jcsd
  3. Mar 11, 2013 #2


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    Homework Helper

    I don't get any of your suggested answers, and your answer is also wrong.

    When you do the reflection about the y-axis, ##y = f(x)## becomes ##y = f(-x) = g(x)## (say).

    When you do the translation to the RIGHT by 0.75, ##y = g(x)## becomes ##y = g(x-0.75)## (note the minus sign here).

    Now ##g(x - 0.75) = g(-(0.75 - x)) = f(0.75 - x)## since ##g(-x) = f(x)##.

    When you finally do the vertical translation upward, you get ##y = f(0.75 - x) + 2##.

    So the final answer is ##y = f(0.75 - x) + 2 = \frac{1}{0.75-x} + 2 = \frac{4}{3-4x} + 2## after simplification.

    If there wasn't a 4 in the numerator in any of the suggested answers, the question is likely wrong.
  4. Mar 11, 2013 #3


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    Homework Helper

    D is not quite correct. It should be y=-1/(x-3/4)+2 or y=-4/(4x-3)+2.
    There is a mistake in your derivation:
    If you shift a function y=f(x) to the right by A the new function is y=f(x-A). (Think: the new function has its asymptote where it goes to infinity at x=3/4)

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