EQUATION OF HORIZONTAL ASYMPTOTE

Horizontal Asymptotes :

A horizontal asymptote is a horizontal line that is not part of a graph of a function but guides it for x – values “far” to the right and/or “far” to the left. The graph may cross it but eventually, for large enough or small enough values of x (approaching ±∞), the graph would get closer and closer to the asymptote without touching it.

A horizontal asymptote is a special case of a slant asymptote.

Let

deg N(x) = the degree of a numerator

and

deg D(x) = the degree of a denominator

Find the horizontal asymptote of the graph of each rational function.

Problem 1 :

y = 2/(x – 6)

Solution :

 y = 2/(x – 6)

Degree of numerator = 0

Degree of denominator = 1

degree of numerator < degree of denominator

So, equation of the horizontal asymptote is y = 0 which is the x – axis.

Problem 2 :

y = (x + 2)/(x – 4)

Solution :

y = (x + 2)/(x – 4)

Degree of numerator = 1

Degree of denominator = 1

degree of numerator = degree of denominator

y = leading coefficient of N(x)/leading coefficient of D(x).

So, equation of the horizontal asymptote is y = 1.

Problem 3 :

y = (x + 3)/2(x + 4)

Solution :

Given, y = (x + 3)/2(x + 4)

Degree of numerator = 1

Degree of denominator = 1

degree of numerator = degree of denominator

y = leading coefficient of N(x)/leading coefficient of D(x).

So, equation of the horizontal asymptote is y = 1/2.

Problem 4 :

y = (2x² + 3)/(x² – 6)

Solution :

y = (2x² + 3)/(x² – 6)

Degree of numerator = 2

Degree of denominator = 2

degree of numerator = degree of denominator

y = leading coefficient of N(x)/leading coefficient of D(x).

So, equation of the horizontal asymptote is y = 2.

Problem 5 :

y = (3x - 12)/(x² – 2)

Solution :

Given, y = (3x - 12)/(x² – 2)

Degree of numerator = 0

Degree of denominator = 2

degree of numerator < degree of denominator

So, equation of the horizontal asymptote is y = 0 which is the x – axis.

Problem 6 :

y = (3x³ – 4x + 2)/(2x³ + 3)

Solution :

y = (3x³ – 4x + 2)/(2x³ + 3)

Degree of numerator = 3

 Degree of denominator = 3

degree of numerator = degree of denominator

y = leading coefficient of N(x)/leading coefficient of D(x).

y = 3/2

So, equation of the horizontal asymptote is y = 1.5.

For each function, determine the equations of any vertical asymptotes, the locations of any holes, and the existence of any horizontal or oblique asymptotes.

Problem 7 :

y = x/(x + 4)

Solution :

y = x/(x + 4)

Highest exponent of the numerator = 1, highest exponent of the denominator = 1

Equation of horizontal asymptote = 1/1

Then y = 1 is the horizontal asymptote.

Equation of vertical asymptote is at x = -4.

Since there is no common factor in the numerator and in the denominator, there is no hole.

Problem 8 :

y = 1/(x - 5) (x + 3)

Solution :

y = 1/(x - 5) (x + 3)

There is no hole. Vertical asymptotes are at x = 5 and x = -3.

Highest exponent of the numerator = 0

highest exponent of the denominator = 2

N < D

So, x-axis or y = 0 is the horizontal asymptote.

Problem 9 :

y = (x + 4) / (x² - 16)

Solution :

y = (x + 4) / (x² - 16)

Factoring the denominator, we get

y = (x + 4) / (x + 4)(x - 4)

Common factor is x + 4, then there is a hole at x = -4.

Vertical asymptote is at x = 4

Highest exponent of the numerator = 1

Highest exponent of the denominator = 2

Equation of horizontal asymptote y = 1/1

y = 1

Problem 10 :

Consider the function

f(x) = 3/(x - 2)

a) State the equation of the vertical asymptote.

b) Use a table of values to determine the behaviour(s) of the function near its vertical asymptote.

c) State the equation of the horizontal asymptote. 

d) Use a table of values to determine the end behaviours of the function near its horizontal asymptote.

e) Determine the domain and range.

f ) Determine the positive and negative intervals.

g) Sketch the graph.

Solution :

f(x) = 3/(x - 2)

a) The vertical asymptote is at x = 2

b) The intervals are (-∞, 2) and (2, ∞)

• When x ∈ (-∞, 2), f(x) will be negative. That is, when x < 2, f(x) is negative.
• When x ∈ (2, ∞), f(x) will be positive. That is, when x > 2, f(x) is positive.

y-intercept is -3/2.

c) Highest exponent of the numerator = 0, highest exponent of the denominator = 1

Equation of horizontal asymptote is x-axis or y = 0.

d) End behavior :

• x --> -∞ then f(x) --> 0
• x --> ∞ then f(x) --> 0

e) Domain is all real numbers except x = 2

Range is all real values except y = 0

f) 

• x ∈ (-∞, 2), f(x) will be negative
• x ∈ (2, ∞), f(x) will be positive