Asymptotes

Vertical Asymptote

Vertical asymptotes occur where the denominator of a rational function is zero, but the numerator is non-zero. In other words, A vertical asymptote in a rational function occurs where the function approaches infinity (positive or negative) as 𝑥 approaches certain values. These asymptotes represent points where the function becomes undefined due to a zero in the denominator that cannot be canceled out.To find vertical asymptotes, set the denominator equal to zero and solve for $ x $.

Definition:

For a rational function $$ R(x) = \frac{P(x)}{Q(x)} $$ Where $P(x)$, and $Q(x)$ are polynomials.
A vertical asymptote occurs at values of 𝑥 for which the denominator $Q(x) = 0$ (making the function undefined) provided that these values do not also make the numerator zero (after simplifying the function).

In other words, a vertical asymptote of a rational function is a vertical line 𝑥 = 𝑎 where the function's value increases or decreases without bound as 𝑥 approaches 𝑎 from the left or right. This occurs when the denominator of the function equals zero at 𝑥 = 𝑎, and the numerator does not also equal zero at that point.

To locate the vertical asymptotes of a rational function:

Factor the numerator and the denominator as much as possible.
Simplify the function by canceling any common factors between the numerator and the denominator. If a factor ( 𝑥 − 𝑎 ) exists in both the numerator and the denominator, canceling it will result in a "hole" at 𝑥 = 𝑎, not a vertical asymptote.
Set the simplified denominator equal to zero and solve for 𝑥. The values of 𝑥 that make the denominator zero after simplification are the locations of the vertical asymptotes.

Note: If a factor in the denominator is canceled by the numerator, the function has a hole at that point, not a vertical asymptote.

Example 1:

Consider the function: $$ f(x) = \frac{1}{x - 2} $$

Numerator: $P(x) = 1 $
Denominator: $Q(x) = x - 2 $

Since there are no common factors between the numerator and the denominator, we proceed by setting $Q(x) = 0$: $$ x - 2 = 0 $$ $$ x = 2 $$ So, 𝑥 = 2 is a vertical asymptote. As 𝑥 approaches 2 from either side, 𝑓( 𝑥 ) becomes very large (positive or negative), meaning the function approaches infinity or negative infinity near this point. so there is a vertical asymptote at $ x = 2 $.

Example 2: $ f(x) = \frac{x + 3}{x^2 - 4} $

The denominator $ x^2 - 4 = 0 $ at $ x = \pm 2 $, so there are vertical asymptotes at $ x = 2 $ and $ x = -2 $.

Example 3: $ f(x) = \frac{x^2 - 1}{x^2 - x - 2} $

The denominator $ x^2 - x - 2 = 0 $ factors to $ (x - 2)(x + 1) = 0 $, giving vertical asymptotes at $ x = 2 $ and $ x = -1 $.

Example 4:

Consider the function: $$ f(x) = \frac{1}{(x + 1)(x - 3)} $$ Here,

Numerator: $P(x) = 1 $
Denominator: $Q(x) = (x + 1)(x - 3) $

Since there are no common factors to cancel, we set each factor in the denominator to zero to find the values of 𝑥 that cause the function to become undefined:

x + 1 = 0, gives, x = - 1
x - 3 = 0, gives, x = 3

Thus, there are vertical asymptotes at 𝑥 = − 1 and 𝑥 = 3.
Vertical asymptotes: $$ x = - 1 \quad and \quad x = 3 $$

Example 5:

Consider the function: $$ f(x) = \frac{(x - 1)(x + 2)}{(x - 1)(x - 3)} $$ Here,

Numerator: $P(x) = (x - 1)(x + 2) $
Denominator: $Q(x) = (x - 1)(x - 3) $

Notice that ( 𝑥 − 1 ) is a common factor in both the numerator and the denominator. We can simplify the function by canceling ( 𝑥 − 1 ): $$ f(x) = \frac{(x + 2)}{(x - 3)} \quad for \quad x \ne 1; $$ After canceling, we have a hole at 𝑥 = 1, not a vertical asymptote. We are left with the denominator 𝑥 − 3, so we set it to zero: $$ x - 3 = 0 $$ $$ x = 3 $$ Thus, there is a vertical asymptote at 𝑥 = 3, and a hole at 𝑥 = 1.
Vertical asymptote: $$ x = 3 $$ Hole: $$ x = 1 $$

Graphical Representation

Vertical asymptotes are represented on a graph as dashed vertical lines at the values of 𝑥 where the function is undefined due to division by zero. The function's graph approaches these lines but never touches or crosses them.

Key Points

Vertical asymptotes occur at values of 𝑥 that make the denominator zero, provided the numerator is not also zero at those points.
If both the numerator and denominator are zero at a point, the function has a hole there, not a vertical asymptote.
The graph of a rational function will approach vertical asymptotes infinitely close but will not intersect them.

Summary of Procedure for Finding Vertical Asymptotes

Factor the numerator and denominator fully.
Simplify by canceling any common factors, noting that any canceled factors indicate a hole, not a vertical asymptote.
Set the simplified denominator to zero and solve for 𝑥 to locate the vertical asymptotes.

Vertical asymptotes represent the values of 𝑥 where the function becomes unbounded (approaches - ∞ or ∞), and they are shown as dashed vertical lines on a graph to indicate where the function is undefined.

Horizontal Asymptote

A horizontal asymptote of a rational function represents a horizontal line that the function approaches as 𝑥 goes to positive or negative infinity. Unlike vertical asymptotes, horizontal asymptotes describe the behavior of the function at extreme values of 𝑥 (either very large or very small).

In other workds, a horizontal asymptotes describe the behavior of a function as $ x \to \infty $ or $ x \to -\infty $. It describes the end behavior of the function, indicating the value that $f(x)$ approaches as $x$ becomes very large or very small. The existence and position of horizontal asymptotes depend on the degrees of the polynomials in the numerator and denominator. They tell us the long-term trend of the function.

Definition:

For a rational function $$ R(x) = \frac{P(x)}{Q(x)} $$ where, $P(x)$, and $Q(x)$ are polynomials, the horizontal asymptote depends on the degrees of $P(x)$, and $Q(x)$. The degree of a polynomial is the highest power of 𝑥 in the expression.

Rules for Finding Horizontal Asymptotes

If the degree of $P(x)$ < degree of $Q(x)$:
The horizontal asymptote is $$ y=0. $$ As 𝑥 grows large, the function approaches zero because the denominator grows faster than the numerator.
Example:
Consider the function, $$ R(x) = \frac{2x + 3}{x^2 + 1} $$ Here,
- Degree of: 1 since $P(x) = 2x + 3 $
- Degree of : 2 since $Q(x) = x^2 + 1 $
Since the degree of 𝑃(𝑥) is less than the degree of 𝑄(𝑥), the horizontal asymptote is: $$ y = 0 $$ As 𝑥 → ∞ or 𝑥 → − ∞, the function 𝑓(𝑥) approaches zero.
Horizontal asymptote : y = 0
If the degree of $P(x)$ = degree of $Q(x)$:
The horizontal asymptote is $$ y = \frac{leading \quad coefficient \quad of \quad Q(x)}{leading \quad coefficient \quad of \quad P(x)} . $$ As 𝑥 grows large, the terms with the highest powers in the numerator and denominator dominate, so the function approaches the ratio of their leading coefficients.
Example:
Consider the function, $$ R(x) = \frac{3x^2 - 5x + 2}{6x^2 + x - 4} $$ Here,
- Degree of $P(x)$: 2 since $P(x) = 3x^2 - 5x + 2 $
- Degree of $Q(x)$: 2 since $Q(x) = 6x^2 + x - 4 $
Since the degrees of $P(x)$ and $Q(x)$ are the same, we find the horizontal asymptote by taking the ratio of the leading coefficients: $$ y = \frac{3}{6} = \frac{1}{2} $$ As 𝑥 → ∞ or 𝑥 → − ∞, 𝑓(𝑥) approaches $\frac{1}{2}$.
Horizontal asymptote $$ y = \frac{1}{2} $$
If the degree of $P(x)$ > degree of $Q(x)$:
There is no horizontal asymptote.
In this case, as 𝑥 grows large, the function increases or decreases without bound, approaching infinity or negative infinity instead of leveling off.
However, if the degree of the numerator is exactly one more than the degree of the denominator, the function has an oblique (slant) asymptote instead.
Example:
Consider the function: $$ f(x) = \frac{x^3 + x + 1}{2x^2 - 4} $$ Here,
- Degree of $P(x)$: 3 since $P(x) = x^3 + x + 1 $
- Degree of $Q(x)$: 2 since $Q(x) = 2x^2 - 4 $
Since the degree of 𝑃(𝑥) is greater than the degree of 𝑄(𝑥), there is no horizontal asymptote.
Horizontal asymptote: None
But the degree of numerator is one greater than denominator, so it has slant asymptote, which is obtained by actual division as follows: $$ \frac{x^3 + x + 1}{2x^2 - 4} = \frac{1}{2}x + \frac{3 x + 1}{2x^2 - 4} $$ Hence the slant asymptote is $ y = \frac{1}{2}x $
Therefore, as 𝑥 → ∞ or 𝑥 → − ∞, 𝑓(𝑥) will increase or decrease without bound parallel to $ y = \frac{1}{2}x $.

Example 1: $ f(x) = \frac{2x^2 + 3}{x^2 + 1} $

Since the degrees of $ P(x) $ and $ Q(x) $ are equal, the horizontal asymptote is at $ y = \frac{2}{1} = 2 $.

Example 2: $ f(x) = \frac{3x + 1}{2x^2 + 4} $

Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $.

Example 3: $ f(x) = \frac{4x^3}{2x^2 + 5} $

Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote but it has the slant asymptote $y = 2x$.

Summary of Procedure for Finding Horizontal Asymptotes

Compare the degrees: of the numerator 𝑃( 𝑥) and the denominator 𝑄( 𝑥).
Apply the rules:
- If the degree of 𝑃(𝑥) < degree of 𝑄(𝑥), the asymptote is 𝑦 = 0.
- If the degree of 𝑃(𝑥) = degree of 𝑄(𝑥), the asymptote is $$ y = \frac{leading \quad coefficient \quad of \quad Q(x)}{leading \quad coefficient \quad of \quad P(x)} . $$
- If the degree of 𝑃(𝑥) > degree of 𝑄(𝑥), there is no horizontal asymptote.

Horizontal asymptotes indicate the end behavior of the function but do not imply that the function cannot cross them. For rational functions, it’s common for the function to approach but never reach these lines as 𝑥 approaches infinity.

Graphical Representation

In a graph, horizontal asymptotes are depicted as dashed horizontal lines that the curve approaches as 𝑥 moves towards positive or negative infinity. Unlike vertical asymptotes, a function's graph may cross its horizontal asymptote.

🧠 Key Points

Horizontal asymptotes describe the end behavior of rational functions.
hey are determined by comparing the degrees of the numerator and denominator polynomials.
A function may cross its horizontal asymptote, but it will approach the asymptote as 𝑥 tends to infinity or negative infinity.
If the degree of the numerator is exactly one more than that of the denominator, the function has an oblique asymptote instead of a horizontal one.

Slant or Oblique Asymptote:

A slant (or oblique) asymptote in a rational function occurs when the degree of the numerator polynomial 𝑃(𝑥) is exactly one more than the degree of the denominator polynomial 𝑄( 𝑥). Slant asymptotes represent a diagonal line that the function approaches as 𝑥 goes to positive or negative infinity.
Definition:
For a rational function $f(x) = \frac{P(x)}{Q(x)}$, where
Degree of $P(x)$ = Degree of $Q(x)$ + 1. The function will have a slant asymptote instead of a horizontal asymptote. The equation of this asymptote is found by performing polynomial long division (or synthetic division if applicable) on 𝑃( 𝑥 ) divided by 𝑄( 𝑥 ). The quotient obtained (ignoring any remainder) represents the equation of the slant asymptote.

How to Locate Slant Asymptotes:

**Ensure that the degree of 𝑃(𝑥) is one more than the degree of ( Q(x)) **: If this is not true, the function does not have a slant asymptote. If it is true, proceed with division.
Divide the numerator by the denominator using polynomial long division (or synthetic division, if applicable). This will yield a quotient and possibly a remainder.
Ignore the remainder. The quotient obtained (in the form 𝑦 = 𝑚𝑥+𝑏 is the equation of the slant asymptote. As 𝑥 → ∞ or 𝑥 → −∞, the effect of the remainder becomes negligible, and the function approaches the line given by this quotient.

Example of Finding Slant Asymptotes

Example 1: Rational Function with a Slant Asymptote:
Consider the function: $$ f(x) = \frac{x^2 + 3x + 2}{x - 1} $$ Here,

Degree of 𝑃( 𝑥) = $x^2 + 3x +2 $ is 2
Degree of 𝑄( 𝑥 ) = $x - 1$ is 1

Since the degree of 𝑃(𝑥) is one more than the degree of 𝑄(𝑥), there is a slant asymptote.

Divide $x^2 + 3x +2 $ by $x - 1$
- First, divide $x^2$ by 𝑥, which gives 𝑥.
- Multiply $x$ by $x - 1$ to get $x^2 - x$
- Subtract $x^2 - x$ from $x^2 + 3x +2 $ which leaves $4x + 2$
- Divide $4x$ by $x$ which gives 4
- Multiply 4 by $x - 1$ to get $4x - 4$
- Subtract $4x - 4 $ from $4x + 2$ which leaves 6
The result of the division is: $$ f(x) = x + 4 + \frac{6}{x - 1} $$ Ignoring the remainder $\frac{6}{x - 1}$ the slant asymptote is: $$ y = x + 4 $$ Slant asymptote: \[y = x + 4\]

As 𝑥 → - ∞ or 𝑥 → ∞, the remainder $\frac{6}{x - 1}$ approaches zero, and $f(x)$ approaches the line $y = x + 4$

Example 2: Rational Function with a Slant Asymptote

Consider the function: $$ f(x) = \frac{2x^3 + 5x^2 - 4x + 3}{x^2 - 1} $$ Here,

Degree of 𝑃( 𝑥) = $2x^3 + 5x^2 - 4x + 3 $ is 3
Degree of 𝑄( 𝑥 ) = $x^2 - 1$ is 2

Since the degree of 𝑃(𝑥) is one more than the degree of 𝑄(𝑥), there is a slant asymptote.

Divide $2x^3 + 5x^2 - 4x + 3 $ by $x^2 - 1$
- First, divide $2x^3$ by $x^2$ which gives 2𝑥.
- Multiply $2x$ by $x^2 - 1$ to get $2x^3 - 2x$
- Subtract $2x^3 - 2x$ from $2x^3 + 5x^2 - 4x + 3 $ which leaves $5x^2 - 2x + 3$
- Divide $5x^2$ by $x^2$ which gives +5
- Multiply 5 by $x^2 - 1$ to get $5x^2 - 5$
- Subtract $5x^2 - 5 $ from $5x^2 - 2x + 3$ which leaves $- 2x + 8$
The result of the division is: $$ f(x) = 2x + 5 + \frac{- 2x + 8}{x^2 - 1} $$ Ignoring the remainder $\frac{- 2x + 8}{x^2 - 1}$ the slant asymptote is: $$ y = 2x + 5 $$ Slant asymptote: \[y = 2x + 5\]

As 𝑥 → - ∞ or 𝑥 → ∞, the remainder $\frac{- 2x + 8}{x^2 - 1}$ approaches zero, and $f(x)$ approaches the line $y = 2x + 5$

Example 3: Rational Function with a Slant Asymptote

Find the slant asymptote of: \[\frac{x^2 + 3x + 2}{x + 1} \]

Degrees: Numerator degree = 2; Denominator degree = 1.
Since the numerator's degree is one more than the denominator's, a slant asymptote exists.
Long Division:
Divide $x^2 + 3x + 2$ by $x + 1$ \[x^2 + 3x + 2 = (x + 2)(x + 1)\] Therefore, \[\frac{x^2 + 3x + 2}{x + 1} = \frac{(x + 2)(x + 1)}{x + 1}\] \[= x + 2\]
Slant Asymptote: $y = x + 2$

Example 4: Rational Function with a Slant Asymptote

Find the slant asymptote of: \[\frac{2x^2 + 3x + 5}{x} \]

Degrees: Numerator degree = 2; Denominator degree = 1.
Long Division:
Divide $2x^2 + 3x + 5$ by $x $ \[2x^2 + 3x + 5 = x(2x + 3) + 5\] Therefore, \[\frac{2x^2 + 3x + 5}{x} = \frac{ x(2x + 3) + 5}{x}\] \[= 2x + 3 + \frac{5}{x}\]
Slant Asymptote: $y = 2x + 3$

Example 5: Rational Function with a Slant Asymptote

Find the slant asymptote of: \[\frac{x^3 + x^2 + 1}{x^2 + 1} \]

Degrees: Numerator degree = 3; Denominator degree = 2.
Long Division:
Divide $x^3 + x^2 + 1$ by $x^2 + 1 $ \[\frac{x^3 + x^2 + 1}{x^2 + 1} = x + \frac{1}{x^2 + 1}\]
Slant Asymptote: $y = x $

📊 Graphical Representation

In a graph, slant asymptotes are represented as dashed diagonal lines that the curve approaches as 𝑥 moves towards positive or negative infinity. Unlike vertical asymptotes, a function's graph may cross its slant asymptote.

🧠 Key Points

Slant asymptotes occur when the degree of the numerator is exactly one more than the degree of the denominator.
They can be found by performing polynomial long division.
Ignore the remainder. The quotient (in the form 𝑦 = 𝑚𝑥 + 𝑏) represents the equation of the slant asymptote.
The graph of a rational function may cross its slant asymptote, but it will approach the asymptote as 𝑥 tends to infinity or negative infinity.
Understanding slant asymptotes is essential for analyzing the end behavior of rational functions and for sketching their graphs accurately.

Slant asymptotes show the diagonal trend of the function at large values of 𝑥, indicating how the function grows or declines over time

Find the vertical and horizontal asymptotes of the following function:

Vertical Asymptote(s):

Horizontal Asymptote:

Asymptotes

Vertical Asymptote

Definition:

Example 1:

Example 2: \( f(x) = \frac{x + 3}{x^2 - 4} \)

Example 3: \( f(x) = \frac{x^2 - 1}{x^2 - x - 2} \)

Example 4:

Example 5:

Graphical Representation

Key Points

Summary of Procedure for Finding Vertical Asymptotes

Horizontal Asymptote

Definition:

Rules for Finding Horizontal Asymptotes

Example:

Example:

Example:

Example 1: \( f(x) = \frac{2x^2 + 3}{x^2 + 1} \)

Example 2: \( f(x) = \frac{3x + 1}{2x^2 + 4} \)

Example 3: \( f(x) = \frac{4x^3}{2x^2 + 5} \)

Summary of Procedure for Finding Horizontal Asymptotes

Graphical Representation

🧠 Key Points

Slant or Oblique Asymptote:

Example of Finding Slant Asymptotes

Example 2: Rational Function with a Slant Asymptote

Example 3: Rational Function with a Slant Asymptote

Example 4: Rational Function with a Slant Asymptote

Example 5: Rational Function with a Slant Asymptote

📊 Graphical Representation

🧠 Key Points

Find the vertical and horizontal asymptotes of the following function: