**Vertical asymptotes:** we must look for points in which the function is not defined. In this case, the point is #x=2#, since it causes the denominator to vanish:

#f(2)=\frac{2*2^2+5*2}{2-2}=\frac{18}{0}#

which is undefined.

**Horizontal asymptotes:** we can look for horizontal asymptotes, since the domain is unbounded. If they exist, they are given by #lim_{x\to\pm\infty}f(x)#:

#\lim_{x\to\pm\infty}\frac{2x^2+5x}{x-2}=\frac{x(2x+5)}{x(1-\frac{2}{x})} = \frac{2x+5}{1-\frac{2}{x}}\to\frac{\pm\infty}1 = \pm \infty#

**Oblique asymptotes:** we can look for oblique asymptotes, since the function diverges. Here's the theory: if #y=f(x)# approaches some line #y=mx+q# as #x\to\pm\infty#, then we have

#f(x)-(mx+q) \to 0 \iff f(x)\to mx+q#

If we divide both sides by #x#, we have

#f(x)/(x) \to (mx)/x+q/x \to m#

since #q/x\to 0# because #x\to\pm\infty#

So, if #f(x)/x# has a finite limit, that would be the slope of the line it is approaching. Once #m# is known, we simply have

#f(x)\to mx+q \iff f(x)-mx \to q#.

Let's do it!

#m = \lim_{x\to\pm\infty}f(x)/(x)= \lim_{x\to\pm\infty}\frac{2x^2+5x}{x^2-2x}=2#

#q = \lim_{x\to\pm\infty}f(x)-mx = \lim_{x\to\pm\infty}\frac{2x^2+5x}{x-2}-2x#

#=\lim_{x\to\pm\infty}\frac{9x}{x-2}=9#