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Math Help - plotting a graph

  1. #1
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    plotting a graph

    hello, how would I plot a graph y = 2/x for -2<x<2 ?
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  2. #2
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    Quote Originally Posted by sillylilly View Post
    hello, how would I plot a graph y = 2/x for -2<x<2 ?
    Plot the extrema

    f(-2) = -1

    f(2) = 1

    f(0) = unde fined


    If you know the shape of the graph of y=\frac{1}{x} you'll find it's much the same. Plot the two defined points above and draw the standard graph through
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  3. #3
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    yes thats what ive got, so is it really supposed to be
    (-2, -1) (-1, -2) (0,0) (1,2) (2,1) ?

    when i plot this it looks like a z shape and at no point does it cross the other equation y = e^x , which is why i thought there must be something wrong.
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  4. #4
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    y = \frac{2}{x}
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    yeah thats the graph shape i get, its just i need to plot the other equation on the same graph, so i can find the place where they cross over, and they dont cross anywhere. the other equation is y = e^x and the line ive got already for that looks like its right, so where am i going wrong, or what am i missing ?
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  6. #6
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    Quote Originally Posted by sillylilly View Post
    yeah thats the graph shape i get, its just i need to plot the other equation on the same graph, so i can find the place where they cross over, and they dont cross anywhere. the other equation is y = e^x and the line ive got already for that looks like its right, so where am i going wrong, or what am i missing ?
    You need to draw your graphs more carfully. y = 2/x and y = e^x intersect at x = 0.8526 (correct to four decimal places).

    (The exact value is actually x = W(2) where W is the Lambert W-function.)
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  7. #7
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    Quote Originally Posted by sillylilly View Post
    yes thats what ive got, so is it really supposed to be
    (-2, -1) (-1, -2) (0,0) (1,2) (2,1) ?

    when i plot this it looks like a z shape and at no point does it cross the other equation y = e^x , which is why i thought there must be something wrong.
    (0,0) is not a point on either graph.

    On the attached graph y=e^x is in green and y=\frac{2}{x} is in blue.

    As you can see they intersect and it's close enough to Mr F's answer so his is highly likely to be correct (and not just because he's Mr F )
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  8. #8
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    The problem did not say "use a graphing utility to produce the plot".

    Whenever you want to sketch a curve you should start out by finding its x- and y-intercepts by setting y and x equal to 0 respectively. For y=\frac{2}{x} you can't find any finite value, so you know the graph doesn't cut the x & y intercept.
    Then you should determine its asymptotes, You can calculate the vertical asymptote by inspecting where the graph would be undefined (i.e., y=\infty), obviously in this case the denominator is zero at x=0 so you have a vertical asymptote there. In order to find the horizontal asymptote you need to compute the limits of your function as x \rightarrow \pm \infty, if either limit has a finite value L, then the line y=L is a horizontal asymptote. \lim_{x \rightarrow \pm \infty} \frac{2}{x} = 0, so your horizontal asymptote is at y=0.

    Using the values of y' you can then check where the graph is increasing (+ve) or decreasing (-ve). Values of y'' tells you where the graph is concave up or concave down. You have enough information to plot the graph.

    As far as y=e^x is concerned, you know that e^x is the inverse of ln x, so its graph can be obtained from that of ln x by reflection in the line y=x. Therefore its y-int is y=1 (as x=1 was the x-int of ln x), and the range of e^x will be the set of all positive real numbers.
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