How to find holes and asymptotes?

To find holes in a rational function, we set the common factor present between the numerator and denominator equal to zero and solve for x. The resulting value of is the x-coordinate of the hole. To find the vertical asymptote of a rational function, set the denominator equal to zero and solve for x. To find the horizontal asymptote and oblique asymptote, refer to the degree of the numerator and denominator.

A rational function has holes when common factors exist between the numerator and denominator. To determine the coordinates of a hole, set this common factor equal to zero and solve for x. Then, plug in this x value to the simplified form of the rational function to get the y-coordinate...

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A rational function has holes when common factors exist between the numerator and denominator. To determine the coordinates of a hole, set this common factor equal to zero and solve for x. Then, plug in this x value to the simplified form of the rational function to get the y-coordinate of the hole.

Consider the example below.

Example: find a hole (if any) of the rational function below.

`y = (x^2+5x+6)/(x+3)`

To determine if this function has a hole, factor the polynomial present.

`y=((x+2)(x+3))/(x+3)`

Since there is a common factor between the numerator and denominator, this rational function has a hole. The common factor is x + 3.

To solve for the x-coordinate of the hole, set the common factor equal to zero.

`x + 3= 3`

`x = -3`

To solve for the y-coordinate of the hole, simplify the rational function. To do so, cancel the common factor. It simplifies to the following:

`y=x+2`

And plug-in x = -3.

y = -3 + 2

y=-1

Thus, the rational function has a hole at point (-3, -1).

A rational function has a vertical asymptote when there are x values that will make the denominator zero.

To solve for the vertical asymptote, factor the polynomial present. If there is any common factor between the numerator and denominator, cancel it. When the function is in simplified form, set the denominator equal to zero, and solve for x.

Consider the following example.

Example: find the vertical asymptote of the rational function below.

`y=(x^2+7x+12)/(x^2+3x-4)`

Factor the polynomials present.

`y = ((x+3)(x+4))/((x+4)(x-1))`

Since there is a common factor between the numerator and denominator, cancel it. So this simplifies to the following:

`y=(x+3)/(x-1)`

Then, set the denominator equal to zero and solve for x.

x - 1 = 0

x=1

Therefore, this function has a vertical asymptote at x=1.

To determine if a rational function has horizontal asymptotes, consider these three cases. Let N be the degree of the numerator and D be the degree of the denominator.

(i) N < D

The rational function has a horizontal asymptote. It is y = 0.

(ii) N = D

The rational function has horizontal asymptote. It is `y=a/b,` where a is the leading coefficient of the numerator and b is the leading coefficient of the denominator.

(iii) N > D

The rational function has no horizontal asymptote.

Consider the following example.

Example: if any, find the horizontal asymptote of the rational function below.

`y=(x^2-4)/(x^2+1)`

The degree of the numerator is 2, and the degree of the denominator is 2. Since they are the same, this function has a horizontal asymptote. It will be the quotient of the leading coefficients.

The leading coefficient of the numerator is 1, and the leading coefficient of the denominator is 1. Dividing them, the result is as follows:

`y=1/1`

`y=1`

Therefore, the horizontal asymptote of this function is y=1.

A rational function has an oblique asymptote if the degree of the numerator is greater than denominator by one only.

To find the oblique asymptote, divide the numerator by the denominator. The linear portion of the quotient is the oblique asymptote.

Consider the following example.

Example: if any, find the oblique asymptote of the rational function below.

`y=(3x^3+x^2-7x+5)/(x^2+3x-4)`

The degree of the numerator is 3, and the degree of the denominator is 2. Since N > D by one only, it has an oblique asymptote.

To determine the oblique asymptote, divide the numerator by the denominator.

`y=(3x^3+x^2-7x+5)/(x^2+3x-4) = 3x-8 + (29x-27)/(x^2+3x-4)`

Taking only the linear portion of the quotient, the oblique asymptote is:

`y=3x - 8`

Therefore, the oblique asymptote of this function is y=3x-8.

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