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Homework Help: Curve Sketching

  1. Nov 25, 2008 #1
    1. The problem statement, all variables and given/known data
    This is for my calculus project and we have to pick an equation and go through the process of finding first and second derivatives, asymptotes, symmetry, and max/min to sketch the curve.

    2. Relevant equations
    I chose (x^2-3x+4)/(x^2+2x-1) as my equation.

    3. The attempt at a solution
    I found the first derivative and solved it for find the max or min. I got 1+square root of 2 and 1-square root of 2, but shouldn't there be only one max or min (based on the graph - I used my graphing calculator and this http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html to check)?

    And then the second problem is.. when I calculated to find the asymptote, my horizontal asymptote is y=1. If I look at the graph, There is certainly an asymptote at y=1 when X<0 but not when it's x>0. How does that happen? I'm so stuck right now.
  2. jcsd
  3. Nov 25, 2008 #2


    Staff: Mentor

    The horizontal asymptote is for x < 0 and x >0. You're not looking far enough to the right with your calculator.

    Your work at finding the derivative is correct, and there are two points at which the derivative is 0. You might need to change the size of the window you're using to view the graph so that you can see more of it.
  4. Nov 25, 2008 #3


    User Avatar
    Science Advisor

    Check your graph again with a larger range for x. y drops below 1 for x= 1 but then rises again toward y= 1 as asymptote for x positive.
  5. Nov 25, 2008 #4
    Hmm okay because I know when x=1, y=1, so does that mean I have to note that the asymptote y=1 is for when X<0 and X>0?

    Thanks for helping!
  6. Nov 25, 2008 #5


    Staff: Mentor

    A horizontal asymptote is relevant to behavior in the limit, as x gets large or as x gets very negative. It's not relevant to small values of x, such as 1.

    For your function, the dominant terms in the numerator and denominator are x^2, which means that the graph of the function approaches y = 1 as x gets large and as x gets very negative.
  7. Dec 9, 2008 #6
    Sorry for bumping this thread but I have another question.
    Is there a point of inflection? I got my second derivative and set it equal to 0 and I got this x^3-3x^2-3x-3=0.
    How do you factor it to prove that there is or isn't a point of inflection.
    Last edited: Dec 9, 2008
  8. Dec 9, 2008 #7


    Staff: Mentor

    I'm pretty sure there's not an inflection point for your function. There are a couple of points where the concavity changes sign, but these occur at the vertical asymptotes, so these values of x aren't in the domain of this function.

    You didn't show the work you did to get the equation x^3 -3x^2 -3x -3 = 0, and I suspect you made a mistake in it. This equation has a root around 4.95 or so, but there's nothing going on in the graph of your function that would suggest that concavity changes from negative to positive there.

    For question 2 (BTW you should write the interval as [-1, 0], not the other way around), assuming your equation is correct, you're not likely to find any solutions by factoring it. (On the other hand, your equation might not be correct.) I'm assuming you're trying to find the local maximum, which is around -0.5. To get this you'll most likely need to use some sort of approximating method rather than an algebraic method that would give an exact result.
  9. Dec 9, 2008 #8
    Haha yeah. I realized I put it the other way.. that's why I couldn't factor it, which is why I deleted it from my post ^^ Silly me.

    Hmm okay I will go check my work on my second derivative. Thank you so far!
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