Answer:
See below
Step-by-step explanation:
I assume you mean [tex]f(x)=\frac{x^2-2x}{x^3+9x^2-10x}[/tex]:
Holes: Since [tex]f(x)=\frac{x^2-2x}{x^3+9x^2-10x}[/tex] reduces to [tex]f(x)=\frac{x(x-2)}{x(x^2+9x-10)}[/tex], then there is a hole at [tex]x=0[/tex] as [tex]x[/tex] exists in both the numerator and denominator (however, its limit as x approaches 0 is 1/5).
Vertical Asymptotes: If we further reduce [tex]f(x)=\frac{x(x-2)}{x(x^2+9x-10)}[/tex] to [tex]f(x)=\frac{x-2}{(x+10)(x-1)}[/tex], then we see that there are vertical asymptotes at [tex]x=-10[/tex] and [tex]x=1[/tex]
Horizontal Asymptotes: As the degree of the numerator is less than the degree of the numerator ([tex]2<3[/tex]), then there is a horizontal asymptote at [tex]y=0[/tex]
