Results are out! Find what you need...fast. Get quick advice or join the chat
Hey! Sign in to get help with your study questionsNew here? Join for free to post

Basic c2 graphs question

Announcements Posted on
Applying to Uni? Let Universities come to you. Click here to get your perfect place 20-10-2014
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    Say you are given the equation y = x(x^2-1) to sketch the roots are (-1,0) (0,0) (1,0)

    How would you sketch it I mean in terms of which side would you start of with how do you know where the max min points are.

    Also does anybody know what the graph of y=-srt of x look like?
    • 46 followers
    Offline

    ReputationRep:
    (Original post by IShouldBeRevising_)
    Say you are given the equation y = x(x^2-1) to sketch the roots are (-1,0) (0,0) (1,0)

    How would you sketch it I mean in terms of which side would you start of with how do you know where the max min points are.

    Also does anybody know what the graph of y=-srt of x look like?
    The graph of  y= - \sqrt{x} looks like this:



    For the graph of,  y=x(x^2-1)

    You know the values where it crosses the x-axis, just consider all the regions, by subbing in some values to deduce the shape of the graph.

    e.g. It crosses at (1,0), see what is the value of y when x=2, if it is positive then the graph will move upward and if it is negative, then the graph will move downwards.
    • 46 followers
    Offline

    ReputationRep:
    (Original post by IShouldBeRevising_)
    Say you are given the equation y = x(x^2-1) to sketch the roots are (-1,0) (0,0) (1,0)

    How would you sketch it I mean in terms of which side would you start of with how do you know where the max min points are.

    Also does anybody know what the graph of y=-srt of x look like?
    I forgot to answer your other question about max/min points.

    Max/min occurs at stationary points where the gradient is equal to zero.

     y=x(x^2-1) = x^3-x

    Differentiate the above expression, and set dy/dx=0 to find the stationary points. Find the 2nd derivative to check whether the points are max or min.
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    (Original post by raheem94)
    I forgot to answer your other question about max/min points.

    Max/min occurs at stationary points where the gradient is equal to zero.

     y=x(x^2-1) = x^3-x

    Differentiate the above expression, and set dy/dx=0 to find the stationary points. Find the 2nd derivative to check whether the points are max or min.

    (Original post by raheem94)
    The graph of  y= - \sqrt{x} looks like this:



    For the graph of,  y=x(x^2-1)

    You know the values where it crosses the x-axis, just consider all the regions, by subbing in some values to deduce the shape of the graph.

    e.g. It crosses at (1,0), see what is the value of y when x=2, if it is positive then the graph will move upward and if it is negative, then the graph will move downwards.
    Thanks well explained... I get it now

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: May 5, 2012
New on TSR

Submitting your UCAS application

How long did it take for yours to be processed?

Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.