1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Simple Transformations Questions
Loading...