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I am a little bit confused about graphing

  1. Oct 9, 2006 #1
    so if the plot of x(t) is given, then I get confused about how to graph x(4-t/2), I know how to graph x(t+4), x(t/2) and x(-2) ..
    but when they are all put together, I get very confused about the order ...
    any help?
  2. jcsd
  3. Oct 9, 2006 #2
    First look at [tex] x(-\frac{t}{2}) [/tex]. Then [tex] x(-\frac{t}{2} +4) [/tex] will be that graph shifted to the left 4 units. It will be stretched out (depending if it is linear, quadratic, etc..). If it is quadratic, the graph will be stretched by a factor of 4, and will not be affected by the negative sign.
    Last edited: Oct 9, 2006
  4. Oct 9, 2006 #3
    why would it strecthed out by factor of 4?

    the only stretching I see is streching by factor of 2 ... coz I see t/2....
    i see shifting by 4 though ...
    and the order you suggested doesnt seem to work out ...
  5. Oct 9, 2006 #4


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    If it's a QUADRATIC, the t/2 term gets squared.

    The order he suggested is the only way to do it. First you scale it, then you shift it
  6. Oct 9, 2006 #5
    alright, 3 things need to be done, scaling, reversing and shifting ...
    now please tell me the exact order? where does the reversal fit into?
  7. Oct 26, 2006 #6
  8. Oct 26, 2006 #7
    Knowing that something does something is not nearly as important as why something does something. There's a vey easy conceptual way to look at this.

    Well, if you recall what a function is, a function is the graphical representation of how a equation's value changes as a variable changes. So let's look at basic physics equation:

    D = R*t (If you like, you can corrolate it to y = mx. Right now I'm disregarding y-scale changes and translations becuase they should make sense after this explination.)

    Where D = Distance, R = speed, and t = time, the independant variable for this function. Now, when we graph this function, the Y-axis is going to give us the distance traveled at a certain X value, measured in time.

    Now, think about it this way: if we double the time, it is the same thing as doubling the velocity (communative property). Notice the difference in these two equations:

    1) D = R*t
    2) D = R*2*t = 2R*t

    Which means that equation two is going twice as fast as equation one. So, when you graph it, what would you expect would happen? E2 will reach the same points as E1 in half the time, because it is operating twice as fast. This causes the "stretch" (Or scale-change) effect.

    Now, let's look at adding strictly to your X-value. If I take that same equation, but I add five to X, why does it move -5 over? Well, look at it conceptually:

    D = R(t+5)

    What you essentially saying is that if you add 5 to it at all points, you are making it start out "earlier". So, say you give car#1 5 seconds of speed time before you give car#2. This is mathmatically described as:

    Car1) D = R(t+5)
    Car2) D = Rt

    Notice that if you start a car out five seconds longer, it is making it 5 closer to the "end". So this causes it to shift five over to the left, because it is doing things quicker. That is, when t = -5, it is the origin, because that is actually where it started out relative to the other car --five seconds earlier.

    So, you asked what happens when you have (t+4)/2. Well, it is starting out 4 seconds earlier, and it is going half as fast. Therefore, you'll have to move it back four because it "started out earlier", and stretch it out because it's going "slower".

    I hope that explination helps. It took me a while to conceptualize it myself.
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