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Graphical linear shifting

  1. May 11, 2005 #1
    I am confused of shifting a linear equation.
    Let f(x)=ax+b
    And g(x) is identical to f(x+2)+5
    For example, we create a specific condition, g(x)=f(x) and (1,2) is a point on f(x) [Does this implies that (1,2) is also a point on g(x)?]
    Next step is to find f(x): By using the given conditions, f(x)= -5x/2+9/2

    The contradiction appears: g(x)=f(x+2)+5
    That's mean shifting the whole curve of f(x) to left parallel to x-axis by 2 units, then by shifting it upwards by 5 units, we get g(x).
    My answer to the previous question ( typed in bold ) is yes but I am not certain with my answer. If I am correct, then the point hasn't moved away.
    However, it's clear to know that the shifting must move the point upward DUE TO A VECTOR NATURE.
    My contradiction is here, anyone helps me solve it?
  2. jcsd
  3. May 11, 2005 #2
    You havent bolded anything, but I'll try to help you out.

    If g(x) is f(x+a)+b, then it isnt necessary for them to intersect at all. It may happen though. I think this is what your question was:


    [tex] f(x) = \frac{-5}{2}x +\frac{9}{2} [/tex] then

    [tex] g(x) = f(x+2)+5 = \left(\frac{-5}{2}(x+2) + \frac{9}{2}\right) + 5 [/tex]

    The large brackets surround f(x+2).
  4. May 11, 2005 #3


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    You say:
    "Let f(x)=ax+b
    And g(x) is identical to f(x+2)+5"

    Given that, what does "For example, we create a specific condition,
    g(x)=f(x)" mean?
  5. May 12, 2005 #4
    [Does this implies that (1,2) is also a point on g(x)?]
    This is the bolded question. Sorry, that's not the solution.
    Once the function f(x) is changed to g(x)=f(x+2)+5,
    that means it shifts left by 2 and upwards by 5.
    Obviously, the new point is not the original point as it has moved upward along the original line.
    This is a general case for all linear equations.
    However, as I mentioned, if I let f(x)=g(x), moreover, (1,2) lies on the equation,...........then f(1)=2 ---> g(1)=2
    That means the point didn't move.
    How to explain the italic sentence?
    My solution: The line is shifted 2 units leftwards and then 2 units rightwards for that specific equation.
  6. May 12, 2005 #5
    (1,2) is a point on f(x), and
    [tex]g(x) = \left(\frac{-5}{2}(x+2) + \frac{9}{2}\right) + 5 [/tex]

    [tex] g(1) = 2 [/tex]

    We're all dandy up til here.

    I really dont know what you mean by this part, but if you notice, f(x) = g(x) as it is. Simplify g(x), you'll find that it already = f(x), the two lines are incident and equivalent. Thus any point on f(x) is a point on g(x).

    If you give me a specific question, I can address it for you, other than that, I really dont know what you are lookin for.
  7. May 13, 2005 #6
    Is the graph shifted? If it is shifted, can you tell me how the shift is?
  8. May 13, 2005 #7
    The graph is not shifted. It is just written in a form where it appears so.
  9. May 13, 2005 #8
    But g(x)=f(x+2)+5? If the condition [g(x)=f(x)] is not told, everyone will shift the graph.
    I'm really confused here, please explain it to me.
  10. May 13, 2005 #9
    But the condition was told.

    I have to admit, I really don't understand the original problem statement so I could be totally off base but this is what I think is happening.

    I think the original problem statement is this:


    Find constants a and b such that f(x) = g(x) and f(1)=2.

    And you did this.

    So, without the condition [f(x)=g(x)] you could not solve the problem and because of the condition the linear shift does not matter. You can think of it as engineering the constants a and b in such a way that the particular linear shift specified by g(x) doesn't produce outputs different from f(x).

    Does this seem right to anybody else?
  11. May 13, 2005 #10
    egsmith, thats exactly what is going on here. I just dont know how to explain the general "g(x) f(x-a)+b then shift f(x) to the right a and up b" in this context. The general rule applies and the shift occurs, but the final result is equivalent to the original result.
  12. May 13, 2005 #11

    matt grime

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    what on earth is the problem?

    if f(x)=ax+b, and g(x)=f(x+2)+5 then g(x)=a(x+2)+b+5 = ax+2a+b+5, now just get on with what ever it is you want to do with it and stop mucking around.
  13. May 13, 2005 #12
    He doesnt understand why there isnt a shift.
  14. May 13, 2005 #13
    Only whozum knows what my problem is.
    Why isn't there a shift? I've proved it mathematically with evidence. I really need to be "solved".
  15. May 13, 2005 #14
    Look at the constant 'a' that you solved for and look at g(x). You sould see some similarities. Basically g(x) shifts f(x) on top of itself so that you can't tell the difference between f(x) and the function after the shift by design.

    There is a shift but it doesn't matter.
  16. May 14, 2005 #15
    So is the original point (1,2) arrives (1,2) after the shift?
  17. May 14, 2005 #16
    Yes, f(1)=2 g(1)=2 by design and as a consequence of the given condition f(x)=g(x).
    Maybe this would help you.
    Graph f(x) with a=-5/2 and b=9/2
    On the same graph, perhaps in a different color, graph f(x+2)
    In a third color, graph f(x+2)+5.
    Hopefully this will show you that the shift does exist. It is just that you end up back to where you started because this is what the problem asked you to do ( when it said let f(x)=g(x) ).
  18. May 15, 2005 #17
    But, if the point arrive at the original point, then, the shift is not just horizontal or vertical.
  19. May 15, 2005 #18


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    If I'm interpreting your question properly, then no. If you shift (1,2) left 2 units and up 5 then it ends up at (-1,7), which you'll see is still on the line.

    This shouldn't be suprising, for any line y=mx+b where m is non zero, a horizontal shift left by k units yields the line y=m(x+k)+b=y=mx+(mk+b). Notice this ls just the original line shifted up mk units. In particular, if m is non zero then any horizontal shift is equivalent to some vertical shift.

    In the case at hand, knowing that going 2 left and 5 up gets the line back to where we started tells us that m=-5/2, since it means a shift left by 2 gives the same line as a shift down by 5 (so k=2, mk=-5 in the last paragraph). The extra condition f(1)=2 is what gave the b constant of 9/2.
  20. May 16, 2005 #19
    But, the prove is :
    Just let f(u+2)=f(v) first
    Note: if u=v and f(u+2)=f(v), then the points are at the same position.-----1
    But, there was actually two shifts of the original function graph-----------2
    The problem is here, if there was really a shift and shift 2 units left and 5 units upwards. With given the equation as I mentioned before, the point obviously
    arrives at a higher position.
    so,1 and 2, one of them must be wrong. If both are correct, then the shift is not just horizontally and vertically.
    Someone please tell me where my mistake is.
  21. May 16, 2005 #20

    matt grime

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    You haven't proved there is a problem. You only think there is a problem.

    We have functions f and g such that f(x+2)+5=g, and such that f=g. The problem is that you think that in order to find, say, g(1) you need to shift f of 1 along 2 and up 5. That isn't true, you need to shove f(3) up 5, that is you shift x along two THEN move f(x) up 5. Which isn't the same thing as what you're claiming.
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