Results 1 to 7 of 7

Math Help - curve sketching

  1. #1
    Junior Member Lonehwolf's Avatar
    Joined
    Nov 2008
    Posts
    55

    curve sketching

    Been trying to get past this but it just keeps building on each other. It's for tomorrow:

    f(x)=\frac{x}{(x+1)(x+2)}

    a) Find any vertical and horizontal asys
    b) where axis are crossed
    c) show curve has no turning points
    d) sketch curve, labeling it properly.
    e) sketch f(-x) on the graph.

    So vertical asympotes are as simple as LCM equal to 0 and substitue. Vertical?!?

    Axis cross should be x = 0 and y = 0. It is (0,0) so it should be solved like that.

    Curve's turning points. I want to see if differentiating \frac{dy}{dx} should be enough, or do i need to go down to \frac{d^2y}{dx^2} and make it =0. If the answer is imaginary it's proved, correct?

    sketching curve should be done using the information gained from the previous questions and finding out which side the curve goes when approaching the asymptotes (asy+0.0001), (asy - 0.00001) to get a quick estimate.

    f(-x) means a reflection in the x axis? Thanks for the posts
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Lonehwolf View Post
    Been trying to get past this but it just keeps building on each other. It's for tomorrow:

    f(x)=\frac{x}{(x+1)(x+2)}

    a) Find any vertical and horizontal asys
    b) where axis are crossed
    c) show curve has no turning points
    d) sketch curve, labeling it properly.
    e) sketch f(-x) on the graph.

    So vertical asympotes are as simple as LCM equal to 0 and substitue. Vertical?!?
    that's right. so the vertical asymptotes are x = -1 and x = -2

    for the horizontal asymptotes, find \lim_{x \to \infty}f(x) and \lim_{x \to -\infty}f(x). if you get a finite answer for either (in this case, they will both give the same answer), that's your horizontal asymptotes. it is a line y = something

    Axis cross should be x = 0 and y = 0. It is (0,0) so it should be solved like that.
    don't know what you mean. this is finding the x- (and y-) intercept

    Curve's turning points. I want to see if differentiating \frac{dy}{dx} should be enough, or do i need to go down to \frac{d^2y}{dx^2} and make it =0. If the answer is imaginary it's proved, correct?
    for turning points, you only need the first derivative, and set it to zero.

    sketching curve should be done using the information gained from the previous questions
    yup, is your graph correct?


    and finding out which side the curve goes when approaching the asymptotes (asy+0.0001), (asy - 0.00001) to get a quick estimate.
    huh? oh, ok, i think i got what you're saying. um, yeah, estimates are good when graphing.

    f(-x) means a reflection in the x axis? Thanks for the posts
    nope, you reflect in the y-axis
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member Lonehwolf's Avatar
    Joined
    Nov 2008
    Posts
    55
    Quote Originally Posted by Jhevon View Post
    that's right. so the vertical asymptotes are x = -1 and x = -2

    for the horizontal asymptotes, find \lim_{x \to \infty}f(x) and \lim_{x \to -\infty}f(x). if you get a finite answer for either (in this case, they will both give the same answer), that's your horizontal asymptotes. it is a line y = something

    don't know what you mean. this is finding the x- (and y-) intercept

    for turning points, you only need the first derivative, and set it to zero.

    yup, is your graph correct?


    huh? oh, ok, i think i got what you're saying. um, yeah, estimates are good when graphing.

    nope, you reflect in the y-axis
    Can't work the horizontal asymptotes either way. Not sure how you work with \lim_{x \to \infty}f(x) and \lim_{x \to -\infty}f(x).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Lonehwolf View Post
    Can't work the horizontal asymptotes either way. Not sure how you work with \lim_{x \to \infty}f(x) and \lim_{x \to -\infty}f(x).
    do you know what limits are? this is calculus since you are talking about derivatives, so you should know about limits (in fact, the derivative is a limit!). you have a first degree polynomial divided by a second degree one, as x gets larger in either direction, the limit goes to zero. hence, y = 0 is the horizontal asymptote
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member Lonehwolf's Avatar
    Joined
    Nov 2008
    Posts
    55
    Quote Originally Posted by Jhevon View Post
    do you know what limits are? this is calculus since you are talking about derivatives, so you should know about limits (in fact, the derivative is a limit!). you have a first degree polynomial divided by a second degree one, as x gets larger in either direction, the limit goes to zero. hence, y = 0 is the horizontal asymptote
    Uhh, I do know about limits, but i can't recall those wicked symbols :

    How are they used, I must have been sick during that lesson!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Lonehwolf View Post
    Uhh, I do know about limits, but i can't recall those wicked symbols :

    How are they used, I must have been sick during that lesson!
    \lim_{x \to \infty} f(x) reads "the limit as x goes (or tends) to \infty of f(x)". you pretty much evaluate what will happen if x gets huge. if x gets really big, beyond measure, the function here, will get close to zero.

    a similar interpretation holds for \lim_{x \to - \infty}f(x)

    the same symbol is used in the definition of the derivative, so you must have been sick during several lessons
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member Lonehwolf's Avatar
    Joined
    Nov 2008
    Posts
    55
    Quote Originally Posted by Jhevon View Post
    \lim_{x \to \infty} f(x) reads "the limit as x goes (or tends) to \infty of f(x)". you pretty much evaluate what will happen if x gets huge. if x gets really big, beyond measure, the function here, will get close to zero.

    a similar interpretation holds for \lim_{x \to - \infty}f(x)

    the same symbol is used in the definition of the derivative, so you must have been sick during several lessons
    Thanks for the explanation, I believe this issue is solved, now i gotta start a clean sheet and burn that whole mess of a paper i had.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. curve sketching
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 8th 2010, 02:32 AM
  2. Curve Sketching
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 26th 2010, 07:20 PM
  3. Curve Sketching
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 3rd 2010, 02:48 AM
  4. Curve Sketching
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 19th 2009, 03:09 PM
  5. Curve Sketching again~
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 18th 2008, 02:55 PM

Search Tags


/mathhelpforum @mathhelpforum