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How does changing the value of $\dfrac{p}{q}$ affect the drawing of the graph (domain/range/shape, etc.) How do you calculate asymptotes?

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Below is a question dealing with this type of function. Can someone please refer to the explanations on how to go about solving these questions? I have no idea because I don't understand these graphs

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  • $\begingroup$ Do you know $\sqrt[m]{A^n}=A^{n/m}$? $\endgroup$
    – mathlove
    Jan 4 '14 at 2:52
  • $\begingroup$ yes :) i know this. $\endgroup$
    – confused
    Jan 4 '14 at 2:54
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HINT :

The graph of $y=\sqrt[m]{(x-a)^n}$ is a graph such that the graph of $y=\sqrt[m]{x^n}$ is parallel-shifted by $a$ in the $x$ direction.

Since $y=f(x)$ has a maximal domain of $[-a, \infty)$...

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  • $\begingroup$ well no because i dont understand how the n/m makes a difference to the domain. i mean multiple options have x-a. $\endgroup$
    – confused
    Jan 4 '14 at 3:37
  • $\begingroup$ see your chart. you'll find the difference of domains. $\endgroup$
    – mathlove
    Jan 4 '14 at 3:39
  • $\begingroup$ $x^{1/3}$ and $x^{2/3}$ has the domain $x\in\mathbb R.$ But your function does not have the domain $x\in\mathbb R.$ This means that your function is not in the form of $(x\pm a)^{1/3}, (x\pm a)^{2/3}.$ $\endgroup$
    – mathlove
    Jan 4 '14 at 3:45
  • $\begingroup$ yup i got it - so its (x+a)^(3/2)right... thanks $\endgroup$
    – confused
    Jan 4 '14 at 3:48
  • $\begingroup$ Yes! that's right. You are welcome! $\endgroup$
    – mathlove
    Jan 4 '14 at 3:48

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