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Math Help - Math Transformations Of A Function

  1. #1
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    Math Transformations Of A Function

    #1 If the point (10,6) is on the graph of y = f(x), what point is on the graph of y = f(-2x-4)? Answer: (-7,6)

    My work:
    y = f(-2(x+2))
    ... Unsure how to complete this and go on.

    #2 The zeros of the function y = f(x) are -4, 1, and 2. Determine the zeros of the new function y = -f(x-1).

    My work:
    (-4,0)
    (1,0)
    (2,0)

    -4+1 = -3
    -1+1 = 0
    2+1 = 3
    (-3) x -1 = 3
    0 x -1 = 0
    3 x -1 = -3

    #3 The maximum value of y = f(x) is 6. Find the maximum value of y = (1/3)f((1/2)x).


    My work: (1/3)f((1/2)(6)) = 1. Not right though.
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  2. #2
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    This should have been posted in the Pre-Calculus subforum.

    Quote Originally Posted by AlphaRock View Post
    #3 The maximum value of y = f(x) is 6. Find the maximum value of y = (1/3)f((1/2)x).

    My work: (1/3)f((1/2)(6)) = 1. Not right though.
    Don't plug in 6 for x!! 6 is a y value.


    Let's compare f(x) to f\left(\frac{1}{2}x\right). The latter is a horizontal stretch by a factor of 2. The minimum and maximum y values, if any, do not change. The maximum value of the latter function would still be y = 6.


    (Look at a concrete example: y = \sin x and y = \sin\left(\frac{1}{2}x\right). In both cases, the minimum and maximum values are -1 and 1, respectively.)


    Okay, now let's compare f\left(\frac{1}{2}x\right) to \frac{1}{3}f\left(\frac{1}{2}x\right). This time, the latter has a vertical shrink by a factor of 1/3. The minimum and maximum y values, if any, this time, will change. The maximum value of the latter function is now y = 2 (just multiply 6 by 1/3).


    (Look at another concrete example: y = \sin x vs. y = \frac{1}{3}\sin x. The min & max values for the original function are -1 and 1, but the min & max values for the transformed function are -1/3 and 1/3.)


    Does that help?


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    Last edited by yeongil; June 17th 2009 at 12:06 AM.
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  3. #3
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    Quote Originally Posted by AlphaRock View Post
    #2 The zeros of the function y = f(x) are -4, 1, and 2. Determine the zeros of the new function y = -f(x-1).

    My work:
    (-4,0)
    (1,0)
    (2,0)

    -4+1 = -3
    -1+1 = 0
    2+1 = 3
    (-3) x -1 = 3
    0 x -1 = 0
    3 x -1 = -3
    Don't do the red -- it's wrong (even though you have the same numbers (-3, 0, & 3) as you had before multiplying by -1).

    You are correct to add 1 to each of the zeroes of the function, because when you go from f(x) to f(x - 1) you have a horizontal shift to the right by 1 unit.

    However, when you go from f(x - 1) to -f(x - 1), you have reflection on the x-axis. The zeroes of the function do not change. (If you have a graphing calculator, try graphing y = x^2 - 6 and y = -(x^2 - 6), and you'll see that the x-intercepts for both graphs are the same.) What is happening is that the y-values are being multiplied by -1, not the x-values. And at the x-intercepts, 0 times -1 is still 0.

    So the x-intercepts for -f(x - 1) are -3, 0, and 3.


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  4. #4
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    Quote Originally Posted by AlphaRock View Post
    #1 If the point (10,6) is on the graph of y = f(x), what point is on the graph of y = f(-2x-4)? Answer: (-7,6)

    My work:
    y = f(-2(x+2))
    ... Unsure how to complete this and go on.
    You only know f(10) so in order to be able to calculate f(-2x-4), you must have -2x-4= 10. Solve that for x. And, of course, the only y value you know is "6" so if -2x-4= 10, f(-2x-4)= f(10)= 6.

    #2 The zeros of the function y = f(x) are -4, 1, and 2. Determine the zeros of the new function y = -f(x-1).
    (-4,0)
    (1,0)
    (2,0)

    My work:
    -4+1 = -3
    -1+1 = 0
    2+1 = 3
    You know that f(-4)= 0, f(1)= 0, and f(2)= 0. Again, you must solve x-1= -4, x- 1= 0, and x-1= 2. Because -0= 0, the "-" before f is irrelevant.
    Yes, saying -4+ 1= 3, -1+1= 0, 2+1= 3 is the same as saying 3-x = -4, -1-0= -1 and 3-1= 2 so, in a backward way, you are solving x-1= -4, x-1= 0, and x- 1= 2.
    (-3) x -1 = 3
    0 x -1 = 0
    3 x -1 = -3
    I don't understand why you are multiplying by -1. The "-" outside f is multiplying a "y" value (y= f(x)) and has nothing to do with x values. As I said before -0= 0.

    #3 The maximum value of y = f(x) is 6. Find the maximum value of y = (1/3)f((1/2)x).[/B]
    My work: (1/3)f((1/2)(6)) = 1. Not right though.
    You replaced "x" by 6! But 6 is a value of "y", not x!
    No matter what x is, the f value cannot be larger than 6 so (1/3)f(anything) cannot be larger than (1/3)(6)= 2.
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