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Sketching of lon curve.

  1. Nov 28, 2008 #1
    1. The problem statement, all variables and given/known data
    the qns is : sketch the graph of y=3ln(x+2) , showing clearing the asymptote and the x-intercept.

    im wondering wat is asymptote and how we find it.. :confused:

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 28, 2008 #2
    An asymptote is a vertical or horizontal line (sometimes oblique, but at your level at the moment, I don't think they will be)drawn through the x or y value where the function goes to positive or negative infinity. To determine them you have to evaluate the function to a limit of infinity or to a value that will cause the limit to be infinity.

    so at what value for x would 3ln(x+2) go to +-infinity? What do you know about the limits/boundaries of the ln function?
  4. Nov 28, 2008 #3
    erm it shud be still infinite rite?
  5. Nov 28, 2008 #4
    when will ln be infinite? ie. ln(what)= +-infinity?
  6. Nov 28, 2008 #5
    erm.. i dont know . cos i just started learning limit too
  7. Nov 28, 2008 #6
    have you got a calculator that can do ln? You'll need something to be able to work out what the points are. Put some values in and ln them and see what happens. Eg. what is ln(1), ln(2), ln(1000), ln(1000000000000), ln(0), ln(-1), ln(-2), ln(-100000). try to get more familar with the ln function. Let me know if you get any interesting answers.
  8. Nov 28, 2008 #7
    yah i noe what is happening.
    as in lnx whereby x cannot be smaller or equal to 0 and cannot be 1 thus x cannot be negative values. and when x increases the f(x) will increase.
  9. Nov 28, 2008 #8
    so, start picking some values of x and start working out the corresponding y values and plot the graph. So if x cannot be negative then there is a boundary where the function cannot cross at x=0 or for any negative number. This boundary is your asymptote. This obviouslt has to apply to your problem y=3ln(x+2), at x=0 3ln(x+2)=3ln(2) which is a point. but when x=-2 then 3ln(x+2)=3ln(-2+2)=3ln(0)... etc

    Ln(1) does exist, it is equal to 0. When does ln(x)=1? any ideas? hint: think about what ln means: ln(x) =loge(x). remember that the log1010=1.
  10. Nov 28, 2008 #9
    yah. i noe.. how to plot also. i noe it must be exponent constant for tat x however i wonder how i get the asymptote..
  11. Nov 28, 2008 #10
    erm.. when x approaches -1 ?
  12. Nov 28, 2008 #11
    ln(0) is your asymptote. y=3ln(x+2) therefore when x=-2 y=3ln(0)=infinity. So your asymptote is a vertical line drawn through the x-axis at x=-2.
  13. Nov 28, 2008 #12
  14. Nov 28, 2008 #13
    i tot is ln(1) dhen it will be infinity thus x+2=1 and x=-1 ?:confused:
  15. Nov 28, 2008 #14
    check it on your calculator
  16. Nov 28, 2008 #15
    hahas.. sorry cos my calculator just shows maths error.
    yah tats rite!
    however how i find the horizontal asymptotes
  17. Nov 28, 2008 #16


    Staff: Mentor

    [tex]\lim_{x \rightarrow 0^+} ln (x ) = -\infty[/tex]
    You were off by quite a lot--about as much as it's possible to be off.
  18. Nov 28, 2008 #17
    erm. so it will be [tex]-\infty[/tex] when x approaches 0 from the right..
    but what can we conclude leis? i just just starting learning limits by myself
  19. Nov 29, 2008 #18


    Staff: Mentor

    What's with "erm"? You seem to start your posts with this.

    "but what can we conclude leis?" - What is leis? I'm sure you're not talking about the flower strands they give visitors to Hawaii.

    In your original post you were asked to graph y = 3 ln(x + 2). Part of what you needed to do was find vertical asymptotes. This function is very similar to y = ln x, but it has been translated two units to the left. Any vertical asymptote for y = ln x will also be shifted two units to the left. There are no horizontal asymptotes.
  20. Nov 29, 2008 #19
    thanks for correcting my english, well i did it because of uncertainty..
    once again thank for assisting
  21. Dec 1, 2008 #20
    My bad, I was dumbing down a little.
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