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Math Help - Graphing helpppp...

  1. #1
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    Graphing helpppp...

    Find the intercepts, symmetries, domain, range and (vertical and horizontal) asymptotes. Also Sketch the graph:

    1.) y = \frac{1}{8}(x^3-4x)


    2.) x^2(y^2-4) = 4

    Thanks!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the intercepts, symmetries, domain, range and (vertical and horizontal) asymptotes. Also Sketch the graph:

    1.) y = \frac{1}{8}(x^3-4x)

    Thanks!
    Intercepts:
    y-intecept: This is the point (0,y) on the graph, so:
    y = \frac{1}{8}(0^3-4 \cdot 0) = 0

    So the y-intercept is (0,0).

    x-intercepts: These are the points (x,0) on the graph, so:
    0 = \frac{1}{8}(x^3-4x)

    0 = x(x^2 - 4) = x(x + 2)(x - 2)

    So the x-intercepts are (0,0), (0, -2), and (0, 2).

    Symmetries:
    Parity: What is y(-x)?
    y(-x) = \frac{1}{8}((-x)^3-4(-x)) = \frac{1}{8}(-x^3 + 4x) = -y(x)

    Since y(-x) = -y(x) we know that y is an odd function.

    Reflections:
    Reflecting y over the x-axis takes y -> -y:
    -y = \frac{1}{8}(x^3-4x) --> y = -\frac{1}{8}(x^3-4x)
    Since this is not equal to the original y, there is no reflection symmetry.

    Reflecting over the y-axis takes x -> -x
    As this is the same as the parity symmetry, we already know that y(-x) is not equal to y(x). So there is no reflection symmetry.

    Inversion:
    This takes (x,y) -> (-x,-y):
    -y = \frac{1}{8}((-x)^3-4(-x)) = -\frac{1}{8}(-x^3 + 4x)
    This gives the original function back again, so y(x) is symmetric under inversion.

    Domain:
    y = \frac{1}{8}(x^3-4x)
    We look for values of x that makes the function undefined. There are no fractions, square roots, log functions, etc. so the domain is unrestricted. Thus the domain of y is (-\infty, \infty).

    Range:
    We look for all possible values the function can take on its domain. The function is continuous. Note that y -> -\infty for x -> -\infty and y -> \infty for x -> \infty. So the range of y is (-\infty, \infty).

    Asymptotes:
    Horizontal asymptotes: These occur when \lim_{x \to \pm \infty} y(x) = constant. In this case \lim_{x \to \pm \infty} y(x) \to \pm \infty so there are no horizontal asymptotes.

    Vertical asymptotes: These occur when the denominator of a function becomes 0. There are no denominators in this function so there are no vertical asymptotes.

    Slant asymptotes: These occur when \lim_{x \to \pm \infty} y(x) = (constant)x. In this case \lim_{x \to \pm \infty} y(x) \to \frac{1}{8}x^3 so there are no slant asymptotes.

    You should be able to sketch this yourself. However for completeness I have attached one at that bottom of this post.

    -Dan
    Attached Thumbnails Attached Thumbnails Graphing helpppp...-function.jpg  
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  3. #3
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    So the x-intercepts are (0,0), (0, -2), and (0, 2).
    isnt that (0,0), (-2,0) & (2,0)?

    And thank you very much for the help mr top quark!

    So you need to find the coordinates of the equation :
    y = -\frac{1}{8}(-x^3+4x)
    and
    y = \frac{1}{8}(x^3-4x)
    since its symmetrical w/ respect to the origin ryt?
    Last edited by ^_^Engineer_Adam^_^; November 15th 2006 at 02:59 PM.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    isnt that (0,0), (-2,0) & (2,0)?

    And thank you very much for the help mr top quark!

    So you need to find the coordinates of the equation :
    y = -\frac{1}{8}(-x^3+4x)
    and
    y = \frac{1}{8}(x^3-4x)
    since its symmetrical w/ respect to the origin ryt?
    Yeah, I guess I wrote it backward, didn't I?

    I'm not sure what you are trying to confirm for the second part. Coordinates for what?

    -Dan
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  5. #5
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    isnt that (0,0), (-2,0) & (2,0)?

    And thank you very much for the help mr top quark!

    So you need to find the coordinates of the equation :
    y = -\frac{1}{8}(-x^3+4x)
    and
    y = \frac{1}{8}(x^3-4x)
    since its symmetrical w/ respect to the origin ryt?
    note that: -\frac{1}{8}(-x^3+4x) = \frac{1}{8}(x^3-4x)
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  6. #6
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    Isee thanks!
    Can anyone please help in number 2?

    x^2 (y^2 - 4) = 4
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  7. #7
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Isee thanks!
    Can anyone please help in number 2?

    x^2 (y^2 - 4) = 4
    First observe that |y| \ge 2, as otherwise x is imaginary.

    also as y \to \pm \infty, \ x \to 0.

    x \to \infty,\ y \to \pm 2, and

    x \to -\infty,\ y \to \pm 2.

    Now differentiating wrt x shows that there are no maxima or minima for finite x.

    This should be enough to allow us to sketch the curve:
    Attached Thumbnails Attached Thumbnails Graphing helpppp...-gash.jpg  
    Last edited by CaptainBlack; November 17th 2006 at 04:06 AM.
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