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Simple Transformations Questions

  1. Jan 19, 2007 #1
    Sorry I don't know how to show the root sign on the forums so I am just gonna use /

    1. The problem statement, all variables and given/known data
    A stretch is applied to the graph of y=/x to produce the graph of y=/9x . Relative to the x and y axis, this stretch may be described as either a........ (then it lists a, b, c, or d answers)

    The answer is:
    horizontal stretch by a factor of 1 over 9 about the y-axis or a vertical stretch by a factor of 3 about the x-axis.

    2. The attempt at a solution
    I understand that the root of 9 is 3, so it makes sense to me that the vertical stretch is 3. But what I don't understand is why is the horizontal stretch is 1 over 9. Should I not find the root of that and then it be 1 over 3? The only thing I can think of is the the 9 is like this /(9x) and I have to take the 9 out and then I still have a nine outside. /9 = 3 then 3^2 = 9 again. So it comes out to be 1 over 9.

    I also assume, although I'm not sure if I'm correct is the reason that it can be a vertical or horizontal stretch is because its not defined in the equation as being a vertical or horizontal stretch.
  2. jcsd
  3. Jan 19, 2007 #2


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    Obviously [itex]y= \sqrt{9x}[/itex] is the same thing as [itex]y= 3\sqrt{x}[/itex]. Since your base equation is [itex]y'= \sqrt{x'}[/itex] (I put the primes on to distinguish the "old" equation from the new), thinking of this a "change in y" it is [itex]y= 3y'= 3(\sqrt{x})[/itex]: the "old value of y" is multiplied by 3 so it is a "stretch" of 3 of the vertical y-axis. However, thinking of this as a change only in x, [itex]y= \sqrt{x'}= \sqrt{9x}[/itex]: x'= 9x or x= (1/9)x'. The "old value" of x' is multiplied by 1/9, not 1/3. Changes in x occur before the "base" function (here [itex]\sqrt{x}[/itex]) is applied, changes in y occur after.

    (Click on the equations to see the Tex code.)
  4. Jan 20, 2007 #3
    Ok thanks for clarifying that for me.
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