Question

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


True or False?

Answer 1

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

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)(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)((x^3-2x^2+x-4)/(x^2+1))/(x)`

`m= \lim_(x \to \infty)(x^3-2x^2+x-4)/(x^3+x)`

We divide each term between
`x ^ 3`

`m= \lim_(x \to \infty)(1-0 + 0 -0)/(1+0)`

`m = 1`

Now we solve the second limit

`b = \lim_(x \to \infty)[(x^3-2x^2+x-4)/(x^2+1)-x]\\\\b = \lim_(x \to \infty)[(x^3-2x^2+x-4 -[x^3+x])/(x^2+1)]\\\\b = \lim_(x \to \infty)[(-2x^2-4)/(x^2+1)]\\\\b = \lim_(x \to \infty)[(-2(x^2)/(x^2)-(4)/(x^2))/((x^2)/(x^2)+(1)/(x^2))]\\\\\\b = -2`

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

Answer 2

Option 1. T

Explanation:

The given function is f(x)=
`(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)-((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.