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In the image you have the function and its graph. I don't know how is the graph gotten? I calculated that the zeros are: $ x = 0, x = -3$ and the function isn't defined for: $ x = 4, x = -2$.

But in the solution I see whole three curves for that method, how do I get them? And the asymptote, from where do I get that?

dfgj fghjk
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  • Here is the correct graph: http://img11.hostingpics.net/pics/562985grzezggz.png (Blue parts: Real Parts - Orange pars: Imaginary parts) –  Oct 20 '13 at 09:45
  • @Adobe How did you get it? – dfgj fghjk Oct 20 '13 at 09:46
  • @dfgjfghk Copy Paste all of this into your browser: http://www.wolframalpha.com/input/?i=\sqrt{\frac{x^3+%2B+3x^2}{x+^2+-+2x+-+8}} –  Oct 20 '13 at 09:47
  • @Adobe, you are linking to an image of the function, interpreted as having complex values. The OP can safely ignore the plot of the imaginary part of the function (the red curve in Alpha). – Sammy Black Oct 20 '13 at 10:05
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    @SammyBlack He can deactivate it. –  Oct 20 '13 at 10:07
  • The image in the original post is great, provided that you know how to interpret the different parts of it. For instance, the diagonal black line is the asymptote of the rational function $f(x)$, but it is not part of the graph of the function! – Sammy Black Oct 20 '13 at 10:07
  • Here is the graph for the real part only: http://img11.hostingpics.net/pics/552442grzezggz.png –  Oct 20 '13 at 10:09
  • Yes, that is a graph of the function, which does not show the important features of the function as well as the image originally posted! – Sammy Black Oct 20 '13 at 10:11
  • @SammyBlack What about this one? http://img11.hostingpics.net/pics/262453graph20131020051633.png –  Oct 20 '13 at 10:17

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First, let's talk about the rational function $$ f(x) = \frac{x^3 + 3x^2}{x^2 - 2x - 8} = \frac{x^2(x + 3)}{(x + 2)(x - 4)}. $$ It has

  • vertical asymptotes at $x = -2$ and $x = 4$ (these are the gaps in the domain that you mention),
  • zero at $x = 0$ (of multiplicity 2, meaning that locally the graph looks like a parabola)
  • zero at $x = -3$
  • slant asymptote of $y = x + 5$.

Since there are two real numbers where the function is undefined, the graph is made of three disconnected curves: on the intervals $(-\infty, -2)$, $(-2, 4)$, and $(4, \infty)$. You have a pretty good sketch of $y = f(x)$ in black.

If $g(u) = \sqrt{u}$, then you are interested in the composition $$ (g \circ f)(x) = g \big( f(x) \big) = \sqrt{\frac{x^3 + 3x^2}{x^2 - 2x - 8}}. $$ What is the effect of the square root? First of all, we only consider real square roots of non-negative numbers, so the composite function $g \circ f$ is only defined where $f(x) \ge 0$, specifically on the intervals $[-3, -2)$ and $(4, \infty)$ and, curiously, at the single point $x = 0$.

As far as the shape of your red curve goes, we can observe the qualitative phenomenon that square roots tend to push positive numbers towards $1$. When the black curve is below $y = 1$, then the corresponding points on the red curve are above it. When the black curve is above $y = 1$, then the corresponding points on the red curve are below it.

Regarding the end behavior of $g \circ f$ (what happens as $x \to \infty$), it approaches the curve $y = \sqrt{x + 5}$ asymptotically, as you can see in this image.

Plot of function and asymptotic curve.

Sammy Black
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