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TI-83 Question

  1. Nov 28, 2005 #1
    With my buddies TI-83 I've plotted a cubic function but am wondering how I can find the x value for a given y value.
     
  2. jcsd
  3. Nov 28, 2005 #2

    tmc

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    Plot both f(x), the cubic, and g(x)=k, the constant function.

    On top of "Trace", click on "Calc."

    Go through the menu, look for intersect, and follow the instructions.

    (The menu titles might not be exact, I dont have one in front of me)
     
  4. Nov 28, 2005 #3
    I'm not sure if I totally follow but I assumed k = "y value". In my case this was 10^12. I don't really know what the intersect function is asking by "first curve?", "second curve?" and "guess?", but I clicked past them. With lower numbers it seems to work but with 10^12 I get "ERR: NO SIGN CHNG"
     
  5. Nov 28, 2005 #4
    While viewing the graph, click "2nd" then "Trace" to bring up the Calc menu. Select "Value" and enter the x value for which you want the y value. You don't use "Intersection" unless you want the point at which two or more curves intersect and from what you're saying there's only one curve.
     
  6. Nov 28, 2005 #5
    I want to do the opposite of what you described! I need the x value for a given y value. Thanks anyway.
     
    Last edited: Nov 28, 2005
  7. Nov 28, 2005 #6
    Oh, there is no way that I know of using the original graphing software on the TI-83 to do that. I suppose the reason is because there can be multiple x values for a single y value in a function, not the other way around.
     
  8. Nov 28, 2005 #7

    tmc

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    sure he can, simply by doing it my way...

    Basically, he wants to solve for f(x)-k=0. So he can either plot f(x)-k, and look for where it crosses the x-axis (I dont remember if a function does that), or make two curves, f(x) and k, and find the x-coordinate of their intersection. Their intersection is where f(x)=k.

    Another way would be to go into the solver (i dont remember where it is to be honest, hopefully you know...), which solves equations for you.
     
  9. Nov 29, 2005 #8
    Yes, this worked. Thanks!
     
  10. Nov 29, 2005 #9
    I understand where you're coming from now. That's a nice skill. I'll have to try that out sometime. :)
     
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