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  1. #1
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    Graphing helpppp...

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

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


    $\displaystyle 2.) x^2(y^2-4) = 4$

    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:

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

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

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

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

    $\displaystyle 0 = x(x^2 - 4) = x(x + 2)(x - 2)$

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

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

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

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

    Domain:
    $\displaystyle y = \frac{1}{8}(x^3-4x)$
    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 $\displaystyle (-\infty, \infty)$.

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

    Asymptotes:
    Horizontal asymptotes: These occur when $\displaystyle \lim_{x \to \pm \infty} y(x) = constant$. In this case $\displaystyle \lim_{x \to \pm \infty} y(x) \to \pm \infty$ 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 $\displaystyle \lim_{x \to \pm \infty} y(x) = (constant)x$. In this case $\displaystyle \lim_{x \to \pm \infty} y(x) \to \frac{1}{8}x^3$ 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 :
    $\displaystyle y = -\frac{1}{8}(-x^3+4x)$
    and
    $\displaystyle y = \frac{1}{8}(x^3-4x)$
    since its symmetrical w/ respect to the origin ryt?
    Last edited by ^_^Engineer_Adam^_^; Nov 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 :
    $\displaystyle y = -\frac{1}{8}(-x^3+4x)$
    and
    $\displaystyle y = \frac{1}{8}(x^3-4x)$
    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 :
    $\displaystyle y = -\frac{1}{8}(-x^3+4x)$
    and
    $\displaystyle y = \frac{1}{8}(x^3-4x)$
    since its symmetrical w/ respect to the origin ryt?
    note that: $\displaystyle -\frac{1}{8}(-x^3+4x) = \frac{1}{8}(x^3-4x)$
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  6. #6
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    Isee thanks!
    Can anyone please help in number 2?

    $\displaystyle x^2 (y^2 - 4) = 4$
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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?

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

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

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

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

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

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