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Math Help - CURVE SKETCHING question

  1. #1
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    Unhappy CURVE SKETCHING question

    for the function f(x) = (x-4)(x+1)(x-1)

    the domain and range is it x cannot be 4, -1, or 1??

    what is "real zeros"??

    and how do u figure the y-intercept...do u just plug 0 into all the X's?

    what is the end behaviour?

    how do we figure out the symmetry

    and the number of turning points..is it 2?

    and how do u figure out the approximate coordinates of any local maximum points or local minimum points


    THANKS SO MUCH!!

    and if u don't mind..can u show me a very brief sketch if its possible?
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  2. #2
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    Quote Originally Posted by calchelp View Post
    for the function f(x) = (x-4)(x+1)(x-1)

    the domain and range is it x cannot be 4, -1, or 1??
    The domain is all of the real numbers \mathbb{R}=(-\infty,\infty) as it
    is defined everywhere.

    The range is also all of the reals since this is a cubic with real coeficients
    every equation of the form f(x)=r for any r \in \mathbb{R} has at least one real root.

    what is "real zeros"??
    a "real zero", is a real number r such that f(r)=0

    and how do u figure the y-intercept...do u just plug 0 into all the X's?
    yes

    what is the end behaviour?
    This is how the function behaves as x \to \inftyand x \to -\infty. This function is a cubic so as x \to \infty,\ f(x) \to \infty and as x \to -\infty,\ f(x) \to -\infty

    how do we figure out the symmetry
    Don't know what you mean here.

    and the number of turning points..is it 2?
    There must be at least one turning point between each pair of adjacent roots
    and there can be no more than n-1 turning points for a polynomial function of degree n, so yes there are 2 turning points.

    and how do u figure out the approximate coordinates of any local maximum points or local minimum points
    Depends on what you mean by approximate. I would start with them at the mid-points between consecutive roots.

    RonL
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  3. #3
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    Follow Up Questions On Curve Sketching!!!!!

    i still dont understand the real zeros and end behaviour thing, could you show me how its done


    and i meant to ask if thers any symmetry, and how do you know?


    also, how do u find the local max and min for this graph
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by calchelp View Post
    i still dont understand the real zeros and end behaviour thing, could you show me how its done
    The zeros of a function are the solutions of f(x)=0. In this case
    you have a cubic in factorised form, f(x)=(x-4)(x+1)(x-1), so the
    roots are 4, -1, and 1, as these values each make one of the factors
    zero.

    The end behaviour is derived by looking for what happens when |x| becomes
    large. Here we have a cubic which is of the form:

    f(x) = x^3 + some quadratic

    Now when |x| becomes large |x^3| dominates and we can ignore the
    quadratic and so for large |x|:

    f(x) ~ x^3

    which determines the end behaviour.

    and i meant to ask if thers any symmetry, and how do you know?
    There is a point about which a cubic has a symmetry, but finding it
    is too fiddly for me to do here. Essentialy your cubic is a translated
    copy of the cubic y=x(x+a)(x-a) for some a, which has 180 degree
    rotational symmetry about (0,0), so our cubic has such a symmetry
    about the image of (0,0) under the same translation.

    also, how do u find the local max and min for this graph
    Have you done any calculus? Then the local max and min are solutions
    of f'(x)=0.

    RonL
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  5. #5
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    Wink

    wow thanks for all that help . much needed
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