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Algebra II stuff

  1. Feb 25, 2004 #1
    I have a test Friday, and there are a few things that I don't understand. Here are a few examples so you know the context

    1. Direct/Inverse Variation - Identify as Direct or Inverse Variation
    A. x|-2| 4| 6|
    y| 4|-8|-12|

    B. x| -2|-1| 3|
    y|-1/2|-1|1/3|

    2. Translation - Write an equation for the translation of y=2/x with the give asymptotes
    x = 2; y = 3


    Could somebody explain these to me?
     
  2. jcsd
  3. Feb 25, 2004 #2

    Integral

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    The goal of the first question is to identify the relationship between the given x and y values. For the first one you have

    y = -2x

    Can you see the relationship in the second set of numbers?

    Sorry, no time for the 2nd question now.
     
  4. Feb 25, 2004 #3
    I saw the relationship between the numbers, but how do I know whether it's Direct Variation or Inverse Variation?
     
  5. Feb 25, 2004 #4

    Integral

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    A direct relationship is defined by y = kx where k is a constant and an inverse relationship is y = k/x.

    Does that help? :smile:
     
  6. Feb 26, 2004 #5

    HallsofIvy

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    It helps a heckuva lot to know what Direct Variation and Inverse Variation mean! It would be a good idea to actually right down the definitions while you are doing these problems.

    In x|-2| 4| 6|
    y| 4|-8|-12|

    I see that when x changes from -2 to 4 (x is multiplied by -2) y changes from 4 to -8: y is also multiplied by -2. As confirmation, when x changes from -2 to 6 (multiplied by -3), y changes from 4 to -12: also multiplied by -3. x and y change in the same way: "direct" variation

    For the second problem,
    x| -2 |-1| 3 |
    y|-1/2|-1|1/3|

    I see that when x changes from -2 to -1 (x is multiplied by 1/2), y changes from -1/2 to -1 (y is multiplied by 2: 1 over (1/2)). As confirmation, I see that when x changes from -1 to 3 (x is multiplied by -3), y changes from -1 to 1/3 (y is multiplied by -1/3). In each case y is multiplied by the reciprocal or "inverse" of the number x is multiplied by. That is "inverse" variation.

    y= 2/x has vertical asymptote x= 0 for x very close to 0, the fraction "blows up". It has horizontal asymptote y= 0 since for x very large (positive or negative) y will be very close to 0. (Those are thing you would be expected to know before doing a problem like this.)

    You want to "translate" the function y= 2/x so that it will have asymptotes x= 2 and y= 3. You "translate" a graph by adding or subtracting a number to or from x and y. To translate right or left (move the vertical asymptote from x= 0 to x= 2) add or subtract something from x. Specifically, you want something like y= 2/(x-a). Remember that x=0 was an asymptote of y= 2/x because that made the denominator 0. What value of a makes the denominator of y= 2/(x-a) equal to 0?
    To move the horizontal asyptote from y= 0 to y= 3 add or subtract something from the y value: You now have y= 2/(x-a) (you found the correct value of a above) and y still will be close to 0 for x very large. What value of b makes y= 2/(x-a)+ b equal to 3 when x is very large?
     
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