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Math Help - Graphing Functions

  1. #1
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    Graphing Functions

    Hello all, I have two functions that i have trouble getting how it looks like as a graph.

    I'll post my problem for the first graph first and then I'll post my second one after the first one has been solved.

    The first function for the graph: f(x)= (x^2 +x -2)/(x^2)

    I got the horizontal and vertical asymptotes, x=0 is the V.A. and y=1 is the H.A.

    I found the derivaties of the first prime and second prime. Then I evaluated the graph.

    From what it looks like to me. The graph slopes downward in the third quadrant. and the graph goes upward above the H.A. and then decreases again.

    I'm really not sure what the graph really looks like.
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  2. #2
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    Quote Originally Posted by snakeman11689 View Post
    Hello all, I have two functions that i have trouble getting how it looks like as a graph.

    I'll post my problem for the first graph first and then I'll post my second one after the first one has been solved.

    The first function for the graph: f(x)= (x^2 +x -2)/(x^2)

    I got the horizontal and vertical asymptotes, x=0 is the V.A. and y=1 is the H.A.

    I found the derivaties of the first prime and second prime. Then I evaluated the graph.

    From what it looks like to me. The graph slopes downward in the third quadrant. and the graph goes upward above the H.A. and then decreases again.

    I'm really not sure what the graph really looks like.
    f(x)= \frac{x^2 +x -2}{x^2}

    \lim_{x \rightarrow \infty} f(x) = 1

    One is a horizontal asymptote. Likewise, zero is a vertical asymptote. You can now construct a "new grid" based on these lines.

    This function is continuous everywhere except zero, so it must hold values on both sides of the vertical asymptote. In order for it to be a function, it cannot have more than one f(x) value for some given x value. Our conclusion is that this function must be in either the 1st and 3rd quadrants (of our grid crossed by these asymptotes) or the 2nd and 4th quadrants. To find that out, test for positive or negative values.

    To find the curvature, note the function is of the form a+bx+cx^{-1}, so it must be parabolic.

    Lastly, to find the concavity, you can analyze the one-sided limits of f(x) as x approaches -\infty, 1 \text{ (left side)}, 1 \text{ (right side)}, \infty.
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  3. #3
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    Here is what i got, it looks a bit sloppy, but oh well. from the first quadrant it goes up as x approaches 4 and the it goes down as 4 approaches six and then six goes to infinity.

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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by snakeman11689 View Post
    Here is what i got, it looks a bit sloppy, but oh well. from the first quadrant it goes up as x approaches 4 and the it goes down as 4 approaches six and then six goes to infinity.

    Uhm...this is pretty far off, why first of all doesnt your curve left of your VA approach your HA? Secondly both sides of your VA will be below your HA.
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    Uhm...this is pretty far off, why first of all doesnt your curve left of your VA approach your HA? Secondly both sides of your VA will be below your HA.
    I'm confused?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by snakeman11689 View Post
    I'm confused?
    Your whole curve should be below your horizontal asymptote, and secondly your curve should approach your asymptote as |x| gets bigger.
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    Uhm...this is pretty far off, why first of all doesnt your curve left of your VA approach your HA? Secondly both sides of your VA will be below your HA.
    got it, thanks!!
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