Slant Asymptote Formula

A rational function is a polynomial ratio in which the denominator polynomial should not be equal to zero. It is a function that is the polynomial ratio. A rational function is any function of one variable, x, that can be expressed as f(x) = p(x)/q(x), where p(x) and q(x) are polynomials such that q(x) ≠ 0. There are three sorts of asymptotes for a rational function, that is, horizontal, vertical, and slant asymptotes.

Slant Asymptote

A slant asymptote is a hypothetical slant line that seems to touch a portion of the graph. A rational function has a slant asymptote only when the degree of the numerator (a) is exactly one more than the degree of the denominator (b). In other words, the deciding condition is, a + 1 = b. For example, a slant asymptote exists for the function f(x) = x + 1 as the degree of the numerator is 1, which is one greater than that of the denominator. The general equation of slant asymptote of a rational function is of the form Q = mx + c, which is called quotient function produced by long dividing the numerator by the denominator.

Formula

For a rational function f(x) of the form g(x)/h(x), the slant asymptote, S(x) is of the form:

S(x) = 

The value of quotient S(x) is calculated using long division method for the dividend g(x) and divisor h(x).

Example: Obtain the slant asymptote for the function: y = (x² – 3x – 10)/(x – 5).

Solution:

We have, f(x) = (x² – 3x – 10)/(x – 5).

Here f(x) has a slant asymptote as the degree of numerator is one more than that of denominator.

Using the slant asymptote formula, we have

As the quotient obtained is x + 2, the slant asymptote for the given function f(x) is,

S(x) = x + 2

Sample Problems

Problem 1. Obtain the slant asymptote for the function: y = (x² – 2x – 24)/(x + 4).

Solution:

We have, f(x) = (x² – 2x – 24)/(x + 4).

Here f(x) has a slant asymptote as the degree of numerator is one more than that of denominator.

Using the slant asymptote formula, we have

As the quotient obtained is x – 6, the slant asymptote for the given function f(x) is,

S(x) = x – 6

Problem 2. Obtain the slant asymptote for the function: y = (x² – 2x – 8)/(x + 2).

Solution:

We have, f(x) = (x² – 2x – 8)/(x + 2).

Here f(x) has a slant asymptote as the degree of numerator is one more than that of denominator.

Using the slant asymptote formula, we have

As the quotient obtained is x – 4, the slant asymptote for the given function f(x) is,

S(x) = x – 4

Problem 3. Obtain the slant asymptote for the function: y = (x² – 7x + 10)/(x – 2).

Solution:

We have, f(x) = (x² – 7x + 10)/(x – 2).

Here f(x) has a slant asymptote as the degree of numerator is one more than that of denominator.

Using the slant asymptote formula, we have

As the quotient obtained is x – 5, the slant asymptote for the given function f(x) is,

S(x) = x – 5

Problem 4. Obtain the slant asymptote for the function: y = (x² – 3x – 28)/(x – 7).

Solution:

We have, f(x) = (x² – 3x – 28)/(x – 7).

Here f(x) has a slant asymptote as the degree of numerator is one more than that of denominator.

Using the slant asymptote formula, we have

As the quotient obtained is x + 4, the slant asymptote for the given function f(x) is,

S(x) = x + 4

Problem 5. Obtain the slant asymptote for the function: y = (x² – 3x – 18)/(x + 3).

Solution:

We have, f(x) = (x² – 3x – 18)/(x + 3).

Here f(x) has a slant asymptote as the degree of numerator is one more than that of denominator.

Using the slant asymptote formula, we have

As the quotient obtained is x – 6, the slant asymptote for the given function f(x) is,

S(x) = x – 6