Find a limit in order to find an asymptote I need to draw the graph of the function $y=\sqrt[3]{(x^2)(x+9)}$. Suppose $y=k\cdot x + b$ is the equation for the asymptote.
I managed to find $$k=\lim_{x\to\infty}\frac{y(x)}{x}=\lim_{x\to\infty}\sqrt[3]{\frac{x+9}{x}}=1$$
But then I got stuck finding $$b=\lim_{x\to\infty}y(x)-k\cdot x$$
Any ideas on how to deal with it?

You can write $y(x)-x = \frac{y(x)^3-x^3}{y(x)^2+y(x)x+x^2}$. Let us work with this expression
\begin{align*}
y(x)-x &= \frac{x^2(x+9) - x^3}{(x^2(x+9))^{2/3} + x\sqrt[3]{x^2(x+9)} + x^2} \\
&= \frac{9x^2}{(x^2(x+9))^{2/3} + x\sqrt[3]{x^2(x+9)} + x^2} \\
&= \frac{9}{\frac{1}{x^2}(x^2(x+9))^{2/3} + \frac{1}{x}\sqrt[3]{x^2(x+9)} + 1} \\
&= \frac{9}{(\frac{x+9}{x})^{3/2} + \sqrt[3]{\frac{x+9}{x}} + 1 }
\end{align*}
Taking a limit yields $\frac{9}{1+1+1}=3$.
