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Homework Help: Curve Sketching

  1. Apr 20, 2006 #1

    Could someone please explain (with steps) that how can i sketch the graphs of y = (1-x)x^1/2


    y^2 = (1-x)x^1/2


    y= x/(x^2-1)

    and y = |(x-1)/(x-2)| (modulus)

    I need to know these for an exam so please be a bit more explanatory in your post.
    Thanks a lot in advance.

  2. jcsd
  3. Apr 20, 2006 #2
    The simplest method (not necessarily the most expedient or most elegant) is simply to plot points and connect the dots. That is, pick a value for x, find the corresponding value(s) for y. Plot them on the graph, then connect the dots. That's basically what a graphing calculator does, btw :tongue:

    For rational functions, there are ways to be much more elegant. Finding horizontal/oblique and vertical asymptotes saves much work (From this point on I'm going to ignore oblique asymptotes, because they only exist when a horizontal asymptote is missing, and the process for dealing with them is virtually the same). Determining whether and where your curve crosses the horizontal asymptote can give you alot of information. Once you find the vertical asymptotes and the points where the curve crosses the horizontal asymptote, you can test each region of your domain for whether the curve is above or below the horizontal asymptote. Since you can't cross the vertical asymptotes or the horizontal asymptote besides the point(s) you found, the rest of the curve should be fairly easy to sketch.

    Another thing that can make your sketch easier is finding the zeros. What values of x can you plug in that will make y=0? If you have a rational function, those zeros can do double duty, helping you determine whether you are above or below (or crossing) your horizontal asymptote.

    When you have square roots involved, you will need to find the domain. Aside from finding the zeros, I can't think of anything else you can use to simplify the sketching process in that case.

    If all else fails and you can't remember any special technique, you can always go back to plotting points, as I mentioned at the beginning of this post.
  4. Apr 20, 2006 #3
    gnuplot would make it

    :biggrin: By the way, how do you know the OP needs to understand analytical expansions ? :biggrin:
  5. Apr 20, 2006 #4
    Some more on this would be more useful. Like how to find the assymptotes etc. And please do tell me the solution to the particular problems mentioned in the original post.
    Thanks anyways
  6. Apr 20, 2006 #5
    Hmm. I forgot to mention holes when discussing rational functions. A hole appears when you have a point missing in your domain that can be patched. E.g. y=(x-3)/(x-3) is a horizontal line at y=1 except when x=3, where there is a hole. Vertical asymptotes are where you have a point missing in your domain, but it can't be patched - the curve tends toward infinity (positive and/or negative) at x near the missing point in the domain.

    In English, a hole appears when you have the same factor in both the numerator and denominator, and the factor in the numerator has at least the power of the factor in the denominator. A vertical asymptote appears when you have a factor in the denominator that can't be cancelled out by one in the numerator.

    Horizontal and oblique asymptotes indicate the behavior of the function as x gets very large in both the negative and positive directions. If you divide the numerator by the denominator (you do remember algebraic long division, right?), the quotient (throwing away the remainder) will be an asymptote if it is 0, a constant, or linear. If your quotient is something like x2-3x+7, you won't have a horizontal asymptote or an oblique asymptote.

    I'm avoiding solving those particular problems for you before seeing your attempts at them. Work on them and if you still need help, we can go from there.
    Last edited: Apr 20, 2006
  7. Apr 20, 2006 #6
    By holes you mean the points at which the function is not defined right ?

    But then as in y = (1-x)x^1/2 the domain would consist only for x belonging to positive real values.
    I still dint get how do you find a horizontal assymptotes. I mean for a hyperbola we already know that the equation of the assymptotes differs from the original eqn by a constant k and since it is a pair of straight lines we can equate the discriminant to zero. But for other type of curves i dint get your method for finding the assymptotes.

    An example would be really helpful

    Meanwhile thanks a lot for the help
  8. Apr 20, 2006 #7
    i tried out the 1st one .. y = (1-x)x^1/2

    passes through origin (starts from there )

    critical point at x=1/3(maxima)

    for x>0 2nd derivative is always <0 thus the curve is convex from 0 to 1/3

    it becomes 0 again for x =1 .

    Now after this how do i plot the remaining curve for x >1 (where y become s <0)
    Can you tell me a smaller approach .
    And do you know any site where there is a graph plotter so i could check my answer.
  9. Apr 20, 2006 #8
    got your point buddy after trying out some examples . Still any short methods you might want to share , i would be more than pleased to know.

    I invite you to join my website at http://www.alchemy-education.com/Board [Broken]
    Last edited by a moderator: May 2, 2017
  10. Apr 21, 2006 #9
    The vertical assymptotes are easy to find. ( all those values of x where function is not defined )

    Is there any proper method besides trial and error for finding horizontal assymptotes.
  11. Apr 21, 2006 #10
    Really sorry for so many posts but there was one problem

    analysing the graph of y^2 = (1-x)x^1/2 we find that f(x) forms a peak at x =0 and a minima (saw it through a graph plotter) .. But i tried double differentiating and it seems that it is not defined for x=0 and same is the case for the 3rd derivative. So shouldnt =0 be just an inflection point // How do i figure differentiability in this case at x=0
  12. Apr 21, 2006 #11
    in regard to horizontal asymptotes ... how is the x axis an assymptote for y= x/(x^2-1) for all x>1 or x<-1
  13. Apr 22, 2006 #12
    Finding horizontal asymptotes is simply a matter of dividing the numerator by the denominator. If the degree of the numerator < degree of the denominator, the quotient will always be zero. The remainder doesn't matter when finding the asymptote. That's why the x-axis was an asymptote for y= x/(x2-1). If the degree of the numerator = degree of the denominator, ignore everything except their leading coefficients to get the quotient. That will be your horizontal asymptote.

    If the degree of the numerator is one more than the degree of the denominator, you have an oblique asymptote, and as far as I know, there is no easy way to quickly get the quotient in that general case. You have to actually do your algebraic long division.

    y2=(1-x)x1/2 is not a function, because of the y2. Since y2 must be non-negative (if we consider only the real plane), either (1-x) and x1/2 are both positive, both negative, or one of them is zero. x1/2 is non-negative, so the both negative case can be ruled out. After a few more steps, you can see this limits your domain to [0,1], with y=0 on either end of the domain.

    If you are going to apply calculus, I would suggest you break the relation into two funtions: y=((1-x)x1/2)1/2 and y=-((1-x)x1/2)1/2. Then apply calculus on each of the functions. You can differentiate without splitting them up like that, but you have to be careful because y can be positive or negative for the same x. If you do that, and If I'm not mistaken, the slope will be vertical at x=0 and x=1 for both functions. The only new features you'd need calculus to find are the extrema and inflection points (if they exist in 0<x<1).
  14. Apr 22, 2006 #13
    How will you sketch y = (cos x) /x
  15. Apr 24, 2006 #14
    I would start out by (lightly) sketching y=1/x and y=cos x. I might also include y=-1/x. Where the maxima of cos x are, stretch (or compress, and possibly invert) them until they hit 1/x. The reason for this is because the maxima of cos x are where y=1. Similarly, the minima of cos x are where y=-1. Stretch the minima of cos x to hit -1/x.

    If you had y=g(x) cos x, for an arbitrary function g(x), the same sort of pattern applies. Stretch the maxima of cos x to hit g(x) and the minima to -g(x). Ever heard of AM radio? AM stands for Amplitude Modulated, which is exactly what this is.

    And, of course, the zeros of cos x will stay at zero, unless g(x)=infinity at any of those points. Then you should resort to calculus (e.g. L'Hopital's rule) to determine whether it's a vertical asymptote or if you can get the location of the hole.
    Last edited: Apr 24, 2006
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