Answer:
True. The slant asymptote is [tex]y=x-2[/tex]
Step-by-step explanation:
A slant asymptote is the equation of a line of the form y = mx + b
Where m is the slope and b is the cut point with the y axis.
To find these asymptotes, you must make the following limit:
[tex]m = \lim_{x \to \infty}\frac{f(x)}{x}[/tex]
[tex]b = \lim_{x \to \infty}[f(x) -mx][/tex]
With the given function f(x) we make the first limit to find the slope of the line:
[tex]m= \lim_{x \to \infty}\frac{\frac{x^3-2x^2+x-4}{x^2+1}}{x}[/tex]
[tex]m= \lim_{x \to \infty}\frac{x^3-2x^2+x-4}{x^3+x}[/tex]
We divide each term between [tex]x ^ 3[/tex]
[tex]m= \lim_{x \to \infty}\frac{1-0 + 0 -0}{1+0}[/tex]
[tex]m = 1[/tex]
Now we solve the second limit
[tex]b = \lim_{x \to \infty}[\frac{x^3-2x^2+x-4}{x^2+1}-x]\\\\b = \lim_{x \to \infty}[\frac{x^3-2x^2+x-4 -[x^3+x]}{x^2+1}]\\\\b = \lim_{x \to \infty}[\frac{-2x^2-4}{x^2+1}]\\\\b = \lim_{x \to \infty}[\frac{-2\frac{x^2}{x^2}-\frac{4}{x^2}}{\frac{x^2}{x^2}+\frac{1}{x^2}}]\\\\\\b = -2[/tex]
Finally: The slant asymptote is [tex]y = x-2[/tex]
