Question

Given the function f(x) x^2-2x/x^3 +9x^2-10x Find any holes, vertical asymptotes, and horizontal asymptotes there may be.

Answer 1

Answer:

See below

Step-by-step explanation:

I assume you mean $f(x)=\frac{x^2-2x}{x^3+9x^2-10x}$:

Holes: Since $f(x)=\frac{x^2-2x}{x^3+9x^2-10x}$ reduces to $f(x)=\frac{x(x-2)}{x(x^2+9x-10)}$, then there is a hole at $x=0$ as $x$ exists in both the numerator and denominator (however, its limit as x approaches 0 is 1/5).

Vertical Asymptotes: If we further reduce $f(x)=\frac{x(x-2)}{x(x^2+9x-10)}$ to $f(x)=\frac{x-2}{(x+10)(x-1)}$, then we see that there are vertical asymptotes at $x=-10$ and $x=1$

Horizontal Asymptotes: As the degree of the numerator is less than the degree of the numerator ($2$), then there is a horizontal asymptote at $y=0$