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Homework Help: Shifting/expanding a function?

  1. Aug 10, 2010 #1
    1. The problem statement, all variables and given/known data

    The original function is: f(x) = square root of (4 - x^2)

    1) Find the domain and draw the graph of y = -f(2x)
    2) Find the domain and draw the graph of y = f(2x + 1)
    3) Find the domain and draw the graph of y = f(-x)^2 + 1

    2. Relevant equations

    f(x) = square root of (4 - x^2)
    y = -f(2x)
    y = f(2x + 1)
    f(-x)^2 + 1

    3. The attempt at a solution

    I am struggling with a shifting/expanding/reflecting a function problem.

    The function is: f(x) = square root of (4 - x^2)

    1) Find the domain and draw the graph of y = -f(2x)
    I think I need to reflect across the x-axis and dilate by two (horizontal expansion). I wrote the domain as x < 1. I'm not sure how to rewrite the equation.

    2) Find the domain and draw the graph of y = f(2x + 1)
    I thought I should dilate by 2 and shift to the left one unit. Again I don't know how to rewrite the original equation. I put the domain as x < 0 but I am not sure if it is correct.

    3) Find the domain and draw the graph of y = f(-x)^2 + 1
    This part is hard but I thought I should reflect across the y axis and shift up one unit. I don't know how to handle the square part and I don't know how to rewrite the equation. I put the domain as x > 1 but I am not sure if it's right.
     
  2. jcsd
  3. Aug 11, 2010 #2

    vela

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    First, tell us what the domain of f(x), the original function, is.
     
  4. Aug 11, 2010 #3
    I think it's x<2 so that the portion under the square root isn't negative.
     
  5. Aug 11, 2010 #4

    Mentallic

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    Think about the negative values of x as well. Do you know what the shape of this graph is?
     
  6. Aug 11, 2010 #5
    Oh, you are right, it's -2<x<2. The graph looks like a semicircle with radius 2.
     
  7. Aug 11, 2010 #6

    vela

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    The domain also includes 2 and -2, so it should be -2 ≤ x ≤ 2.

    This means whatever gets plugged into f has to be between -2 and 2 inclusive. For instance, if we wanted to know the domain of f(3x), we need to find values of x such that -2 ≤ 3x ≤ 2. Dividing by 3 to get x by itself yields -2/3 ≤ x ≤ 2/3. That would be the domain of f(3x). If we wanted to find the domain of f(4x+3), we start with -2 ≤ 4x+3 ≤ 2, which eventually gets you -5/4 ≤ x ≤ -1/4. Make sense?
     
  8. Aug 11, 2010 #7
    Thanks, Yes that makes sense, but if the function gets stretched, it seems like the domain could be bigger than 2. That's the part I really don't understand, how to shift and expand the function.
     
  9. Aug 11, 2010 #8

    Mentallic

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    For your function [tex]f(x)=\sqrt{4-x^2}[/tex] ::

    f(x)+n will move the graph up n units (or down if n is negative)

    f(x+n) will move the graph left n units (or right if n is negative)

    f(nx) will squish the domain from [itex]-2\leq x\leq 2[/itex] to [itex]-2/n\leq x\leq 2/n[/itex] but the graph will still cut the y-axis at 2 since for [tex]\sqrt{4-(nx)^2}[/tex], f(0) still equals 2.
    If n<0 then it will stretch out the domain. These will all be ellipses.

    n.f(x) will stretch the semi-circle up (ellipse), giving a different y-axis intercept while the domain stays the same. For 0<n<1 it will squish it, and for n<0 it will flip it around the x-axis so you have an upside down ellipse.
     
  10. Aug 11, 2010 #9
    Okay, I understand this but I can't figure out the last one, because not everything is in the parentheses. I have shifted the equation to be sqr root (4 - x^4) + 1. Do I solve for the domain like this -2 < ((-x)^2)+1 < 2 ? It doesn't seem to work.
     
  11. Aug 11, 2010 #10

    vela

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    Mentallic summed the basic rules up nicely.

    When you have a combination, such as f(3x+2), you just have to do it in steps, so f(3x+2) is the graph of f(3x) shifted by 2 to the left and f(3x) is the graph of f(x) shrunk by a factor of 3.

    The third problem doesn't fall completely under those rules for converting the graph because you're squaring f(x). In this case, it's probably easiest to just work out algebraically what y equals and then plot the end result. When drawing the graph, though, keep in mind what the domain of y is.
     
  12. Aug 11, 2010 #11

    vela

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    No, you still find the domain the same way as before. Whatever gets plugged into f has to be between -2 and 2, so you get -2 ≤ -x ≤ 2, which is equivalent to 2 ≥ x ≥ -2.
     
  13. Aug 11, 2010 #12

    vela

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    I just noticed you also did the algebra incorrectly in #3. You should have

    [tex]y=f(-x)^2+1 = [f(-x)]^2+1 = [\sqrt{4-(-x)^2}]^2+1[/tex]

    If the problem actually involves f(-x2) or f[(-x)2], it'll be different.
     
  14. Aug 11, 2010 #13
    That looks to me like what I do have.
    sqr root (4 - x^4) + 1

    It's due in 10 mins so I guess this is what I am sticking with!
     
  15. Aug 11, 2010 #14

    vela

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    You're getting the order of operations wrong. What you have is

    [tex]f[(-x)^2]+1 = \sqrt{4-[(-x)^2]^2}+1 = \sqrt{4-x^4}+1[/tex]

    but f(-x)2+1 means

    [tex][f(-x)]^2+1 = [\sqrt{4-(-x)^2}]^2+1[/tex]

    Note the difference in the placement of the square brackets between the two cases. If the problem meant for you to square -x rather than f(-x), it would have been written f[(-x)2].
     
  16. Aug 12, 2010 #15
    Thanks for all the help, I think I understand now.
     
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