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Math Graphing Problem

  1. Dec 12, 2004 #1
    Hey.. im having problems finding the x-intercepts and the maximum and minimi points using my calculator.. the question is as follows:

    Graph each function. IDentify the x-intercepts and the pointsw here the local maximums and local minimums occur.

    F(x) = x^4-4x^3-x^2+12x-2

    on my ti 83 plus, I went to calc chose Minimum ... then went to trace and for left bound, i chose the place where it turns while on rightbound i chose the point 0" however im not getting a proper answer...

    can anyone please help me do this thing??

    thanks
     
  2. jcsd
  3. Dec 12, 2004 #2
    Hmmm... I don't quite understand what value you chose for your left bound value. Could you be more specific?
    By the way, just to make sure, sometimes the way you enter the function is very important. Just to make sure it's sometimes a good idea to place things like x^4 in parentheses.
     
  4. Dec 12, 2004 #3
    i chose a leftbound value of X= -1.06383 and Y = -9.800981 for finding minimmum and rightbound value of x=.21276596 and y= .4714437

    i really dunno how to do this.. and whichi points to pick... do u know a tutorial for doing this ??

    thanks
     
  5. Dec 12, 2004 #4
    Ah ok I see what your problem is here. First off a minimum on a graph is the lowermost point where the graph dips down, right? Or, as you said in your first post, "where it turns." Your problem when you calculated your leftbound is you were pretty much on top of your minimum value point so the calculator didn't really know what to do because its range didn't include the minimum.
    I don't know of any tutorials for this but I can walk you through an example. First off, clear your equations under "y=", enter only "y=x^2", and hit the graph button. (This graph has only one minimum so it's easy to work with.) Now if you want to find your minimum you go a certain distance for your leftbound so you're not on top of it, maybe x=-2 or that range, and you do the same for rightbound so choose something around x=2. The point of this step is you're basically telling the calculator "look between these two points for the value" so you basically want to be sure the minimum's on that range. A bigger range here is definetely better then too small!
    Next the calculator's going to ask you for a guess. The point of this step is later on you're going to have graphs with maybe ten max and min points so the calculator is going to calculate what the closest minimum value is to that point. So for this step trace to the closest you can get to the minimum value, which in the case of y=x^2 will be at x=0. When you press enter you should get your minimum value to equal zero.
    Hope this helps!
     
  6. Dec 13, 2004 #5
    firstly, thanks for ur help.

    However, i tried what u said ( x=-2.1 and x=2.1) .... and for the guess option, i chose the POINT x=.212, y=.045 and then hit enter but the following is what i get as my minimum:

    x= 1.16 x 10^-7 and y=1.362xx10^-14

    :( dunno where im going wrong..
     
  7. Dec 14, 2004 #6
    Hmmm that didn't happen to me... does this happen fairly consistantly or is it just for this particular problem? I can't really think of what it might be, however, you might want to show a friend/ the teacher what it is you're doing and see if they can help you figure it out.
     
  8. Dec 14, 2004 #7

    Tjl

    User Avatar

    You need to understand how the functions on your calculator work, in order to understand how to do this. You are correct in using the max. and min. functions, but you are working too hard to find the parameters. When you input the right and left bounds using the maximum function, the located point will be the point with the highest y component of the values included. So all you need to do is visually place the bounds, so that the highest point is contained.

    Relative Maximum = (1.144, 6.143)
    Relative Minimum = (-.939, -10.061)

    There is another minimum because the function is degree 4.

    If you have any other problems, just post :)
     
  9. Dec 14, 2004 #8
    ok say you have a reguar parabola and youre trying to find the minimum, which is the vertex, on the left bound you have to press the left key untill you are to the left side of the vertex, and on the right bound press the right key, untill you are to the vertex, hit enter , it should say guess, hit enter again and that should be your answer

    for a downward pointing parabola, it should be max instead of min, now that equation should have many little parabola looking things, so just apply them to that

    and for the zeros you can either go to the table, or you can hit second trace and pick the zero option.

    for the z
     
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