Answer
The function doesn't have a horizontal or oblique asymptote.
$x=0$ is a vertical asymptote
Work Step by Step
$F(x)=\frac{x^4-16}{x^2-2x},$
The degree of the leading coefficient of the numerator is, $n=4$. the degree of the leading coefficient the denominator is, $m=2$.
Thus, $n\geq m+2,$
Therefore, The function doesn't have a horizontal or oblique asymptote.
Factorizing the denominator:
$\frac{(x-2)(x+2)(x^2+4)}{x(x-2)}=\frac{(x+2)(x^2+4)}{x}$
Thus, $x=0$ is a vertical asymptote.
