Question

The slant asymptote for x^3-2x^2+x-4/x^2+1 is y=x-2True or False?

Answer 1

Answer:

True. The slant asymptote is $y=x-2$

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:

$m = \lim_{x \to \infty}\frac{f(x)}{x}$

$b = \lim_{x \to \infty}[f(x) -mx]$

With the given function f(x) we make the first limit to find the slope of the line:

$m= \lim_{x \to \infty}\frac{\frac{x^3-2x^2+x-4}{x^2+1}}{x}$

$m= \lim_{x \to \infty}\frac{x^3-2x^2+x-4}{x^3+x}$

We divide each term between $x ^ 3$

$m= \lim_{x \to \infty}\frac{1-0 + 0 -0}{1+0}$

$m = 1$

Now we solve the second limit

$b = \lim_{x \to \infty}[\frac{x^3-2x^2+x-4}{x^2+1}-x]\\\\b = \lim_{x \to \infty}[\frac{x^3-2x^2+x-4 -[x^3+x]}{x^2+1}]\\\\b = \lim_{x \to \infty}[\frac{-2x^2-4}{x^2+1}]\\\\b = \lim_{x \to \infty}[\frac{-2\frac{x^2}{x^2}-\frac{4}{x^2}}{\frac{x^2}{x^2}+\frac{1}{x^2}}]\\\\\\b = -2$

Finally: The slant asymptote is $y = x-2$

Answer 2

Answer:

Option 1. T

Step-by-step explanation:

The given function is f(x)=$\frac{x^{3}-2x^{2}+x-4}{x^{2}+1}$

and the statement states that slant asymptote of the the function is y = x-2

We have to evaluate the statement is true or false.

Now to find the slant asymtote we will divide the numerator by denominator.

We get the function in the form of $f(x)= (x-2)-(\frac{2}{x^{2}+1})$

By ignoring the fraction part we can say that the expression which represents the slant asymptote is (x -2).

Therefore the given statement is true.