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?
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
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.