Results 1 to 7 of 7

Math Help - Graph of Basic functions

  1. #1
    Member
    Joined
    Jan 2008
    Posts
    132

    Graph of Basic functions

    Hello! Here's a question that I need help with. Thanks!

    1) Construct a cubic equation that describes a graph that has no stationary points. Describe how you arrived at your equation.

    2) Quadratic equations may describe graphs that do not intersect the x-axis at all. Is it possible to construct a cubic equation that describes a graph that does not intesect the x-axis at all? [My answer is no, but I don't really know how to explain it.] Why/Why not? State and explain an implication of your result using the Remainder-Factor Theorem.

    Thank you in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Tangera View Post
    Hello! Here's a question that I need help with. Thanks!

    1) Construct a cubic equation that describes a graph that has no stationary points. Describe how you arrived at your equation.

    2) Quadratic equations may describe graphs that do not intersect the x-axis at all. Is it possible to construct a cubic equation that describes a graph that does not intesect the x-axis at all? [My answer is no, but I don't really know how to explain it.] Why/Why not? State and explain an implication of your result using the Remainder-Factor Theorem.

    Thank you in advance!
    1) This is an excellent question. Please read through the attachment .... you will find the answer - case 3 is what you want ..... (there are solutions after the questions).
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Tangera View Post
    Hello! Here's a question that I need help with. Thanks!

    1) Construct a cubic equation that describes a graph that has no stationary points. Describe how you arrived at your equation.

    2) Quadratic equations may describe graphs that do not intersect the x-axis at all. Is it possible to construct a cubic equation that describes a graph that does not intesect the x-axis at all? [My answer is no, but I don't really know how to explain it.] Why/Why not? State and explain an implication of your result using the Remainder-Factor Theorem.

    Thank you in advance!
    2) Your answer is correct.

    From the general equation of a cubic it's clear that the range of a cubic is all real number numbers. In other words, -\infty < dx^3 + ax^2 + bx + c < + \infty. In order to go from -oo to +oo (or vice versa) the cubic has to cross the x-axis. The fancy theorem that formalises this is called the intermediate value theorem.

    The implication is that all cubics have at least one real linear factor => at least one real root. Not sure why you'd bring in the remainder-factor theorem .....
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2008
    Posts
    132
    Thank you very much for answering my questions! The document you provided is going to be really useful for me since we are going to be learning more about cubic functions.

    If I am not wrong, perhaps the part of the question that says: " State and explain an implication of your result using the Remainder-Factor Theorem." requires some sort of conclusion that a cubic function can always be expressed/factorized to the following 3 forms?

    [inspired by a part of the document you provided, so thank you!]
    Since there will always be at least 1 root for a cubic equation,
    y= (linear factor)(irreducible quadratic factor) OR
    y= (linear factor)^2 (irreducible quadratic factor)
    y= (linear factor)(linear factor)(linear factor)

    Does that mean that once we know a root of the equation, we can always find the factor too? (It should be so, right?)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Tangera View Post
    Thank you very much for answering my questions! The document you provided is going to be really useful for me since we are going to be learning more about cubic functions.

    If I am not wrong, perhaps the part of the question that says: " State and explain an implication of your result using the Remainder-Factor Theorem." requires some sort of conclusion that a cubic function can always be expressed/factorized to the following 3 forms?

    [inspired by a part of the document you provided, so thank you! Mr F says: You're welcome.]

    Since there will always be at least 1 root for a cubic equation,

    y= (linear factor)(irreducible quadratic factor) Mr F says: Yes.

    y= (linear factor)^2 (irreducible quadratic factor) Mr F says: No. (linear factor)^2 gives a quadratic ..... (quadratic)(irreducible quadratic) = quartic, not cubic.

    y= (linear factor)(linear factor)(linear factor) Mr F says: Yes. And included in this case is (linear factor)^2 (linear factor).


    Does that mean that once we know a root of the equation, we can always find the factor too? Mr F says: Yes. If \alpha is a root then (x - \alpha) is always a factor.

    (It should be so, right? Mr F says: Yes.)
    Some examples to illustrate:

    1. y= (linear factor)(irreducible quadratic factor): One x-intercept.

    y = (x + 2)(x^2 + x + 2) = x^3 + 3x^2 + 4x + 4. Has a NON-Stationary point of inflexion and NO stationary points.

    y = (x - 1)(x^2 + x + 1) = x^3 - 1. Has a stationary point of inflexion - its only stationary point.


    2. y= (linear factor)(linear factor)(linear factor): At least two x-intercepts.

    y = (x - 1)(x + 1)(x + 2). Has two stationary points (max turning point and min turning point) and also a (non-stationary) point of inflection. Has three x-intercepts.

    y = (x - 1)^2 (x + 1). Has two stationary points (max turning point and min turning point) and also a (non-stationary) point of inflection. The min turning point is also an x-intercept (at x = 1). Has two x-intercepts.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jan 2008
    Posts
    132
    Okiedokie. Thank you for correcting my mistake! (I didn't even notice it!) >< Sorry to trouble you again, could you explain the difference between a stationary point of inflexion and a non-stationary point of inflexion?
    Last edited by Tangera; January 28th 2008 at 04:28 AM. Reason: Need to clarify stationary point of inflexion!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Tangera View Post
    Okiedokie. Thank you for correcting my mistake! (I didn't even notice it!) >< Sorry to trouble you again, could you explain the difference between a stationary point of inflexion and a non-stationary point of inflexion?
    There are two sorts of points of inflexions:

    At a stationary point of inflexion the gradient of the tangent is zero (ie. f'(x) = 0).

    At a NON-stationary point of inflexion the gradient of the curve is NOT zero.

    Why is a point of inflexion spcial enough to get its own name:

    At any point of inflexion, the concavity of the curve changes. The meaning of concavity boils down to whether the tangent is above or below the curve. Cocave down: the graph lies below its tangent. Concave up: the graph lies above its tangent. So at a point where the concavity of a curve changes (ie. at a point of inflexion), the tangent switches from being below to above (or vice versa) the graph.

    Points of inflexion can be found by solving f''(x) = 0. But note that solutions to f''(x) = 0 are not always points of inflexion ..... Each solution must be tested.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Very Basic Algebra & Graph help - Profit.
    Posted in the Algebra Forum
    Replies: 4
    Last Post: November 12th 2011, 10:57 AM
  2. Basic 'Logs' and 'Functions'
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: February 27th 2011, 03:44 AM
  3. basic graph theory
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 27th 2009, 03:14 AM
  4. Transformations/Basic Functions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 21st 2009, 06:35 PM
  5. Help - Basic functions
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 14th 2009, 02:49 AM

Search Tags


/mathhelpforum @mathhelpforum