Rational Functions and Asymptotic Behavior

Lesson: Rational Expressions and Domain

In algebra, a rational expression is a fraction whose numerator and denominator are polynomials. Some examples are \(\frac{x+3}{x-2}\), \(\frac{2x^2-5}{x^2+1}\), and \(\frac{3}{x^2-9}\).

A big idea with rational expressions is that they are not defined when the denominator equals zero. Since division by zero is impossible, any value that makes the denominator zero must be excluded from the domain.

The domain of an expression is the set of all input values the expression is allowed to use. For rational expressions, finding the domain means finding which values of the variable make the denominator zero, then excluding those values.

This idea is very important because undefined values are connected to discontinuities in rational functions. In later work, these excluded values may become holes or vertical asymptotes in a graph. So before graphing or simplifying, we first check the domain.

Main Idea

To find the domain of a rational expression:

1. Look at the denominator.
2. Set the denominator equal to zero.
3. Solve the equation.
4. Exclude those values from the domain.

In symbols, if a rational expression has the form

$$\frac{P(x)}{Q(x)},$$

then the domain is all real numbers except the values that make \(Q(x)=0\).

Why factoring matters

Sometimes the denominator is easy to solve, such as \(x-4=0\). But often the denominator is a quadratic or another polynomial, so factoring helps us find the restricted values.

For example, if the denominator is

$$x^2-5x+6,$$

we factor it as

$$x^2-5x+6=(x-2)(x-3).$$

Then we set each factor equal to zero:

$$x-2=0 \quad \text{or} \quad x-3=0,$$

so \(x=2\) and \(x=3\) must be excluded from the domain.

Important note about simplifying

Sometimes a factor in the numerator and denominator cancels. Even if it cancels, the original expression still cannot use any value that made the original denominator zero.

For instance, consider

$$\frac{x-3}{x^2-5x+6}.$$

Factoring gives

$$\frac{x-3}{(x-2)(x-3)}.$$

You can simplify this to

$$\frac{1}{x-2},$$

but the original denominator was zero at both \(x=2\) and \(x=3\). So the domain is not just \(x \ne 2\). It is

$$x \ne 2, \; x \ne 3.$$

The value \(x=3\) is still excluded because the original expression was undefined there.

Ways to write the domain

You can describe domain in words or symbols:

• Words: all real numbers except \(2\) and \(3\)
• Symbols: \(x \ne 2\), \(x \ne 3\)
• Interval notation: $$(-\infty,2) \cup (2,3) \cup (3,\infty)$$

Worked Example 1: Simple linear denominator

Find the domain of

$$\frac{5}{x-7}.$$

Step 1: Set the denominator equal to zero.

$$x-7=0$$

Step 2: Solve.

$$x=7$$

Step 3: Exclude this value.

So the domain is all real numbers except \(7\).

Domain: \(x \ne 7\)

Interval notation: $$(-\infty,7) \cup (7,\infty)$$

Worked Example 2: Factor a quadratic denominator

Find the domain of

$$\frac{2x+1}{x^2-9}.$$

Step 1: Factor the denominator.

$$x^2-9=(x-3)(x+3)$$

Step 2: Set each factor equal to zero.

$$x-3=0 \Rightarrow x=3$$

$$x+3=0 \Rightarrow x=-3$$

Step 3: Exclude both values.

So the domain is all real numbers except \(-3\) and \(3\).

Domain: \(x \ne -3,\; x \ne 3\)

Interval notation: $$(-\infty,-3) \cup (-3,3) \cup (3,\infty)$$

Worked Example 3: A factor cancels, but the restriction stays

Find the domain of

$$\frac{x+4}{x^2+x-12}.$$

Step 1: Factor the denominator.

$$x^2+x-12=(x+4)(x-3)$$

So the expression becomes

$$\frac{x+4}{(x+4)(x-3)}.$$

Step 2: Find the values that make the denominator zero.

$$x+4=0 \Rightarrow x=-4$$

$$x-3=0 \Rightarrow x=3$$

Step 3: Exclude both values.

Even though \(x+4\) cancels during simplification, \(x=-4\) is still not allowed because it made the original denominator zero.

The simplified form is

$$\frac{1}{x-3},$$

but the domain of the original expression is

$$x \ne -4, \; x \ne 3.$$

Interval notation: $$(-\infty,-4) \cup (-4,3) \cup (3,\infty)$$

Worked Example 4: Denominator with three factors

Find the domain of

$$\frac{x^2-1}{x(x-5)(x+2)}.$$

Step 1: Look at the denominator.

$$x(x-5)(x+2)$$

Step 2: Set each factor equal to zero.

$$x=0$$

$$x-5=0 \Rightarrow x=5$$

$$x+2=0 \Rightarrow x=-2$$

Step 3: Exclude all three values.

Domain: \(x \ne -2,\; x \ne 0,\; x \ne 5\)

Interval notation: $$(-\infty,-2) \cup (-2,0) \cup (0,5) \cup (5,\infty)$$

Common mistakes to avoid

• Only checking the numerator: The numerator can be zero. That is allowed. The denominator cannot be zero.
• Forgetting to factor completely: If the denominator is not fully factored, you might miss a restricted value.
• Losing restrictions after canceling: Always find restrictions from the original denominator.
• Writing the wrong interval notation: Excluded values create breaks in the intervals.

Quick check: numerator zero vs. denominator zero

Consider

$$\frac{x-2}{x+1}.$$

If \(x=2\), the numerator is zero, and the expression becomes

$$\frac{0}{3}=0,$$

which is perfectly defined.

If \(x=-1\), the denominator is zero, and the expression is undefined.

So the only value excluded from the domain is \(x=-1\).

Connection to rational functions

When a rational expression is used as a function, the excluded domain values show where the function has a break. These breaks help explain asymptotic behavior and discontinuities.

If a denominator factor does not cancel, it often leads to a vertical asymptote. If a factor cancels, it often leads to a hole. In both cases, the excluded value comes from the denominator being zero in the original expression.

Strategy you can always use

1. Write the rational expression clearly.
2. Focus on the denominator only.
3. Factor the denominator completely.
4. Set each denominator factor equal to zero.
5. List those values as excluded from the domain.
6. If needed, write the answer in interval notation.

Brief Summary

A rational expression is undefined wherever its denominator equals zero. To find the domain, factor the denominator, solve for the values that make it zero, and exclude those values.

Be careful when simplifying: canceled factors still create domain restrictions if they were in the original denominator. Always use the original expression to determine the full domain.

Operations on Rational Expressions

Rational expressions are fractions that contain polynomials in the numerator, the denominator, or both. Examples include \(\frac{x+3}{x-2}\) and \(\frac{2x^2-8}{x^2-1}\).

Learning how to add, subtract, multiply, and divide rational expressions is important because these expressions appear often in algebra, rational functions, and equations involving fractions.

In this lesson, you will learn how to:

• simplify rational expressions,
• find restrictions on the variable,
• multiply and divide rational expressions,
• add and subtract rational expressions using a least common denominator,
• simplify complex rational expressions.

1. What is a rational expression?

A rational expression is any expression of the form

$$\frac{P(x)}{Q(x)}$$

where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x)\neq 0\).

This means the denominator can never be zero. So whenever you work with a rational expression, you should first think about restrictions: which values of the variable make the denominator zero?

For example, in \(\frac{x+1}{x-4}\), the denominator is zero when \(x=4\). So \(x\neq 4\).

2. Simplifying rational expressions

Simplifying a rational expression is similar to simplifying a numerical fraction. You factor the numerator and denominator, then cancel any common factors.

It is very important to remember that you can only cancel factors, not terms that are being added or subtracted.

For example,

$$\frac{x^2-9}{x^2+x-12}$$

First factor both parts:

$$x^2-9=(x-3)(x+3)$$

$$x^2+x-12=(x+4)(x-3)$$

So the expression becomes

$$\frac{(x-3)(x+3)}{(x+4)(x-3)}$$

Now cancel the common factor \((x-3)\):

$$\frac{x+3}{x+4}$$

But the original denominator also tells us that \(x\neq 3\) and \(x\neq -4\).

Even though \((x-3)\) canceled, \(x=3\) is still not allowed because it made the original denominator zero.

3. Multiplying rational expressions

To multiply rational expressions:

1. Factor each numerator and denominator completely.
2. Cancel any common factors.
3. Multiply what remains.

This works because multiplying fractions follows the same rule as with numbers:

$$\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$$

Worked Example 1: Multiplying rational expressions

Simplify:

$$\frac{x^2-4}{x^2-5x+6}\cdot \frac{x-3}{x+2}$$

Step 1: Factor everything.

$$x^2-4=(x-2)(x+2)$$

$$x^2-5x+6=(x-2)(x-3)$$

So:

$$\frac{(x-2)(x+2)}{(x-2)(x-3)}\cdot \frac{x-3}{x+2}$$

Step 2: Cancel common factors.

The factors \((x-2)\), \((x-3)\), and \((x+2)\) all cancel:

$$1$$

Step 3: State restrictions.

From the original denominators,

$$x^2-5x+6=(x-2)(x-3),\quad x+2\neq 0$$

So:

$$x\neq 2,\; x\neq 3,\; x\neq -2$$

Answer: \(1\), with restrictions \(x\neq 2,3,-2\).

4. Dividing rational expressions

To divide rational expressions, use the rule:

$$\frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\cdot \frac{d}{c}$$

In words: keep the first fraction, change division to multiplication, and flip the second fraction.

Then factor, cancel common factors, and simplify.

Worked Example 2: Dividing rational expressions

Simplify:

$$\frac{x^2-1}{x^2+2x+1}\div \frac{x-1}{x+1}$$

Step 1: Rewrite as multiplication by the reciprocal.

$$\frac{x^2-1}{x^2+2x+1}\cdot \frac{x+1}{x-1}$$

Step 2: Factor.

$$x^2-1=(x-1)(x+1)$$

$$x^2+2x+1=(x+1)^2$$

So:

$$\frac{(x-1)(x+1)}{(x+1)^2}\cdot \frac{x+1}{x-1}$$

Step 3: Cancel common factors.

Cancel \((x-1)\) and one \((x+1)\) from numerator and denominator:

$$1$$

Step 4: State restrictions.

From the original problem:

• \(x^2+2x+1=(x+1)^2\neq 0\), so \(x\neq -1\)
• Also, the divisor \(\frac{x-1}{x+1}\) cannot be zero, so \(x-1\neq 0\), meaning \(x\neq 1\)

Answer: \(1\), with restrictions \(x\neq -1,1\).

5. Adding and subtracting rational expressions

Adding and subtracting rational expressions is different from multiplying and dividing. You cannot add the numerators and denominators directly.

For example, this is not correct:

$$\frac{1}{x} + \frac{1}{x+1} \neq \frac{2}{2x+1}$$

To add or subtract rational expressions, you need a common denominator. Usually, you use the least common denominator (LCD), which is the smallest denominator that contains all the factors needed.

Steps for adding or subtracting rational expressions:

1. Factor each denominator completely.
2. Find the LCD.
3. Rewrite each expression with the LCD.
4. Add or subtract the numerators.
5. Simplify if possible.
6. State any restrictions.

How to find the LCD

Suppose the denominators are \((x-2)\) and \((x+3)(x-2)\). The LCD must include each factor the greatest number of times it appears. So the LCD is

$$ (x-2)(x+3) $$

If the denominators are \((x-1)^2\) and \((x-1)(x+4)\), then the LCD is

$$ (x-1)^2(x+4) $$

Worked Example 3: Adding rational expressions

Simplify:

$$\frac{3}{x} + \frac{2}{x+1}$$

Step 1: Find the LCD.

The denominators are \(x\) and \(x+1\), so the LCD is

$$x(x+1)$$

Step 2: Rewrite each fraction.

$$\frac{3}{x}=\frac{3(x+1)}{x(x+1)}$$

$$\frac{2}{x+1}=\frac{2x}{x(x+1)}$$

Step 3: Add the numerators.

$$\frac{3(x+1)}{x(x+1)}+\frac{2x}{x(x+1)}=\frac{3(x+1)+2x}{x(x+1)}$$

Expand the numerator:

$$\frac{3x+3+2x}{x(x+1)}=\frac{5x+3}{x(x+1)}$$

Step 4: State restrictions.

\(x\neq 0\) and \(x\neq -1\).

Answer:

$$\frac{5x+3}{x(x+1)}$$

with \(x\neq 0,-1\).

Worked Example 4: Subtracting and simplifying

Simplify:

$$\frac{2}{x^2-1}-\frac{1}{x-1}$$

Step 1: Factor the denominator.

$$x^2-1=(x-1)(x+1)$$

So the problem is:

$$\frac{2}{(x-1)(x+1)}-\frac{1}{x-1}$$

Step 2: Find the LCD.

The LCD is

$$ (x-1)(x+1) $$

Step 3: Rewrite the second fraction.

$$\frac{1}{x-1}=\frac{x+1}{(x-1)(x+1)}$$

Step 4: Subtract the numerators.

$$\frac{2}{(x-1)(x+1)}-\frac{x+1}{(x-1)(x+1)}=\frac{2-(x+1)}{(x-1)(x+1)}$$

Simplify the numerator:

$$\frac{2-x-1}{(x-1)(x+1)}=\frac{1-x}{(x-1)(x+1)}$$

Step 5: Factor the numerator if possible.

$$1-x=-(x-1)$$

So:

$$\frac{-(x-1)}{(x-1)(x+1)}$$

Cancel \((x-1)\):

$$-\frac{1}{x+1}$$

Step 6: State restrictions.

From the original expression, \(x\neq 1\) and \(x\neq -1\).

Answer:

$$-\frac{1}{x+1}$$

with \(x\neq 1,-1\).

6. Complex rational expressions

A complex rational expression is a fraction that contains one or more rational expressions inside it, such as

$$\frac{\frac{1}{x}+\frac{1}{2}}{\frac{3}{x}}$$

To simplify complex rational expressions, there are two common methods:

• simplify the numerator and denominator separately first, or
• multiply the entire expression by the LCD of all the small denominators.

Here we will use the first method.

Worked Example 5: Simplifying a complex rational expression

Simplify:

$$\frac{\frac{1}{x}+\frac{1}{2}}{\frac{3}{x}}$$

Step 1: Simplify the numerator.

The LCD of \(x\) and \(2\) is \(2x\).

$$\frac{1}{x}=\frac{2}{2x}, \qquad \frac{1}{2}=\frac{x}{2x}$$

So:

$$\frac{1}{x}+\frac{1}{2}=\frac{2+x}{2x}$$

Step 2: Rewrite the full expression.

$$\frac{\frac{x+2}{2x}}{\frac{3}{x}}$$

Step 3: Divide by multiplying by the reciprocal.

$$\frac{x+2}{2x}\cdot \frac{x}{3}$$

Step 4: Simplify.

Cancel the common factor \(x\):

$$\frac{x+2}{6}$$

Restrictions: Since \(x\) appears in denominators, \(x\neq 0\).

Answer:

$$\frac{x+2}{6}$$

with \(x\neq 0\).

7. Common mistakes to avoid

• Canceling terms instead of factors.
  For example, in \(\frac{x+2}{x}\), you cannot cancel the \(x\) from \(x+2\). The numerator is a sum, not a product.
• Forgetting to factor first.
  Many expressions only simplify after factoring completely.
• Adding denominators when adding fractions.
  You add or subtract only the numerators once the denominators are the same.
• Forgetting restrictions.
  Even if a factor cancels, values that made the original denominator zero are still not allowed.
• Not flipping the second fraction when dividing.
  Division of rational expressions always becomes multiplication by the reciprocal.

8. Problem-solving checklist

When working with rational expressions, ask yourself:

1. Did I factor everything completely?
2. Did I identify any values that make a denominator zero?
3. If multiplying or dividing, did I cancel only common factors?
4. If adding or subtracting, did I use the LCD?
5. Did I simplify my final answer as much as possible?

Summary

Operations on rational expressions follow the same basic ideas as operations on numerical fractions, but you must be more careful because the denominators contain variables.

To multiply and divide, factor first and cancel common factors. To add and subtract, find the least common denominator, rewrite each expression, and then combine the numerators.

Always remember that any value making an original denominator equal to zero is not allowed. Keeping track of those restrictions is an important part of every problem involving rational expressions.

Solving Rational Equations means solving equations that contain rational expressions, which are fractions with polynomials in the numerator, denominator, or both. A simple example is \(\frac{1}{x} = 3\). A more complicated one might be \(\frac{2}{x-1} + \frac{1}{x+2} = 1\).

These equations are important because the variable can appear in the denominator. That creates an extra challenge: some values of the variable are not allowed, because division by zero is undefined. When solving rational equations, you must always check for these restricted values and make sure your final answers are valid.

The main idea is to clear the fractions by multiplying every term in the equation by the least common denominator (LCD). This turns the rational equation into a simpler equation, often linear or quadratic, which is much easier to solve.

Step 1: Find values that are not allowed.

Any value that makes a denominator equal to zero is excluded from the domain of the equation. These values can never be solutions.

For example, in the equation

$$\frac{3}{x-4} + \frac{1}{x} = 2,$$

the denominators are \(x-4\) and \(x\). So the restricted values are:

$$x \ne 4 \quad \text{and} \quad x \ne 0.$$

It is helpful to write these restrictions before doing any algebra.

Step 2: Find the least common denominator (LCD).

The LCD is the smallest expression that contains all the denominator factors. You use it to eliminate all fractions in one step.

For example:

• The LCD of \(x\) and \(x-2\) is \(x(x-2)\).
• The LCD of \(x^2\) and \(x\) is \(x^2\).
• The LCD of \((x+1)\) and \((x+1)(x-3)\) is \((x+1)(x-3)\).

Step 3: Multiply every term by the LCD.

This is the key step. When you multiply every term by the LCD, the denominators cancel. Be careful to multiply every term on both sides of the equation.

For instance, if you have

$$\frac{1}{x} + \frac{2}{x+3} = 4,$$

the LCD is \(x(x+3)\). Multiplying each term by the LCD gives:

$$x(x+3)\left(\frac{1}{x}\right) + x(x+3)\left(\frac{2}{x+3}\right) = x(x+3)(4).$$

Then simplify each term:

$$x+3 + 2x = 4x(x+3).$$

After the fractions are gone, solve the resulting equation.

Step 4: Solve the new equation.

Once the fractions have been removed, solve using the algebra skills you already know. The equation may become:

• a linear equation,
• a quadratic equation, or
• another polynomial equation that can be simplified.

Step 5: Check for extraneous solutions.

An extraneous solution is a value that appears during the algebra but does not actually work in the original equation. This often happens in rational equations because multiplying by expressions involving the variable can introduce values that make a denominator zero.

So after solving, you must:

1. compare your answers to the restricted values, and
2. substitute them back into the original equation if needed.

If an answer makes any denominator zero, it must be rejected.

Worked Example 1: A basic rational equation

Solve:

$$\frac{2}{x} = 6$$

Step 1: Restriction

Since \(x\) is in the denominator, \(x \ne 0\).

Step 2: LCD

The only denominator is \(x\), so the LCD is \(x\).

Step 3: Multiply both sides by \(x\)

$$x\left(\frac{2}{x}\right) = x(6)$$

$$2 = 6x$$

Step 4: Solve

$$x = \frac{2}{6} = \frac{1}{3}$$

Step 5: Check

\(\frac{1}{3}\) is not a restricted value. Substitute into the original equation:

$$\frac{2}{1/3} = 6$$

$$6 = 6$$

Solution: \(x = \frac{1}{3}\)

Worked Example 2: Different denominators

Solve:

$$\frac{1}{x} + \frac{1}{2} = \frac{3}{x}$$

Step 1: Restriction

\(x \ne 0\)

Step 2: LCD

The denominators are \(x\), \(2\), and \(x\). The LCD is \(2x\).

Step 3: Multiply every term by \(2x\)

$$2x\left(\frac{1}{x}\right) + 2x\left(\frac{1}{2}\right) = 2x\left(\frac{3}{x}\right)$$

$$2 + x = 6$$

Step 4: Solve

$$x = 4$$

Step 5: Check

\(4\) is allowed since it does not make a denominator zero.

Substitute back:

$$\frac{1}{4} + \frac{1}{2} = \frac{3}{4}$$

$$\frac{1}{4} + \frac{2}{4} = \frac{3}{4}$$

$$\frac{3}{4} = \frac{3}{4}$$

Solution: \(x = 4\)

Worked Example 3: Variables in two denominators

Solve:

$$\frac{3}{x-1} - \frac{1}{x+2} = 1$$

Step 1: Restrictions

The denominators cannot be zero:

$$x-1 \ne 0 \Rightarrow x \ne 1$$

$$x+2 \ne 0 \Rightarrow x \ne -2$$

Step 2: LCD

The LCD is \((x-1)(x+2)\).

Step 3: Multiply every term by the LCD

$$ (x-1)(x+2)\left(\frac{3}{x-1}\right) - (x-1)(x+2)\left(\frac{1}{x+2}\right) = (x-1)(x+2)(1) $$

Simplify:

$$3(x+2) - (x-1) = (x-1)(x+2)$$

Expand both sides:

$$3x+6-x+1 = x^2+x-2$$

$$2x+7 = x^2+x-2$$

Move all terms to one side:

$$0 = x^2 - x - 9$$

or

$$x^2 - x - 9 = 0$$

Step 4: Solve the quadratic

Use the quadratic formula:

$$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-9)}}{2(1)}$$

$$x = \frac{1 \pm \sqrt{1+36}}{2}$$

$$x = \frac{1 \pm \sqrt{37}}{2}$$

Step 5: Check restrictions

Neither \(\frac{1+\sqrt{37}}{2}\) nor \(\frac{1-\sqrt{37}}{2}\) equals \(1\) or \(-2\), so both are valid.

Solutions:

$$x = \frac{1+\sqrt{37}}{2} \quad \text{or} \quad x = \frac{1-\sqrt{37}}{2}$$

Worked Example 4: An equation with an extraneous solution

Solve:

$$\frac{x}{x-2} = \frac{4}{x-2}$$

Step 1: Restriction

$$x-2 \ne 0 \Rightarrow x \ne 2$$

Step 2: LCD

The LCD is \(x-2\).

Step 3: Multiply both sides by \(x-2\)

$$ (x-2)\left(\frac{x}{x-2}\right) = (x-2)\left(\frac{4}{x-2}\right) $$

$$x = 4$$

This solution is allowed, so it works.

Now let us look at a case where algebra may suggest a value that must be rejected.

Solve:

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

At first glance, both sides simplify to the same expression whenever \(x \ne 3\). The restriction is:

$$x \ne 3$$

If someone multiplied by \((x-3)^2\), they would get:

$$x-3 = x-3$$

This identity is true for all values of \(x\), but we must still keep the restriction. So the solution is all real numbers except \(3\).

This example shows why restrictions matter. Even if the algebra looks valid for every number, values that make a denominator zero are never allowed.

Common mistakes to avoid

• Forgetting restrictions. Always identify values that make denominators zero before solving.
• Using the wrong LCD. Factor denominators carefully if needed, then choose the smallest expression containing all factors.
• Not multiplying every term. The LCD must be distributed to each term on both sides of the equation.
• Arithmetic errors after clearing fractions. Go slowly when expanding and combining like terms.
• Not checking answers. Some solutions may be extraneous and must be rejected.

Helpful strategy

1. Write the restrictions.
2. Find the LCD.
3. Multiply every term by the LCD.
4. Simplify and solve.
5. Check each solution in the original equation.

If you follow this process carefully, rational equations become much more manageable.

Brief Summary

To solve a rational equation, first identify any values that make denominators zero. Then find the LCD and multiply every term by it to remove the fractions. Solve the resulting equation, and finally check each answer to make sure it is not restricted and does work in the original equation. This final check is essential because rational equations can produce extraneous solutions.

Vertical Asymptotes and Removable Discontinuities

Rational functions are functions written as a fraction of two polynomials. They often have places where the function is not defined, and those points can create important features on the graph.

Two common types of discontinuities in rational functions are vertical asymptotes and removable discontinuities. These may look similar at first because both come from values that make the denominator equal to zero, but they are not the same. The key difference is whether the factor causing the zero cancels.

This lesson will show you how to tell them apart, how to find them from an equation, and what they mean on a graph.

1. What is a rational function?

A rational function has the form

$$f(x)=\frac{P(x)}{Q(x)}$$

where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x)\neq 0\).

The denominator cannot be zero, so any value of \(x\) that makes \(Q(x)=0\) is not in the domain. Those values are the first places we check when looking for discontinuities.

2. What is a discontinuity?

A discontinuity is a break in the graph. In rational functions, discontinuities happen at values of \(x\) that make the denominator zero.

There are two types we will focus on:

• Vertical asymptote: the graph shoots up or down without bound near a certain \(x\)-value.
• Removable discontinuity: the graph has a hole at one point, even though the rest of the graph follows a smooth path.

3. Vertical asymptotes

A vertical asymptote happens when a factor in the denominator makes the denominator zero and does not cancel with the numerator.

If \((x-a)\) stays in the denominator after simplifying, then \(x=a\) is a vertical asymptote.

Near a vertical asymptote, the function values become very large positive or very large negative numbers. On a graph, the curve rises or falls sharply near that line.

Example pattern:

$$f(x)=\frac{1}{x-3}$$

At \(x=3\), the denominator is zero. The factor \((x-3)\) does not cancel, so \(x=3\) is a vertical asymptote.

4. Removable discontinuities

A removable discontinuity happens when a factor in the denominator makes the denominator zero, but that same factor also appears in the numerator and cancels.

After canceling, the simplified function no longer has that factor. The graph behaves like the simplified function, except the original function is still undefined at the canceled value. This creates a hole.

Example pattern:

$$f(x)=\frac{x-3}{x-3}$$

For all \(x\neq 3\), this simplifies to \(f(x)=1\). But the original function is undefined at \(x=3\), so there is a hole at \((3,1)\).

Important: Even if a factor cancels, the original function is still undefined at that value. Canceling helps us identify the graph's shape, but it does not put that value back into the domain.

5. How to tell the difference

To decide whether a denominator zero gives a vertical asymptote or a removable discontinuity, use this process:

1. Factor the numerator and denominator completely.
2. Find values of \(x\) that make the denominator zero.
3. Check whether the factor causing the zero cancels.
4. If it does not cancel, there is a vertical asymptote.
5. If it does cancel, there is a removable discontinuity (a hole).

6. How to find the hole's coordinates

If a factor cancels, the \(x\)-value of the hole comes from the canceled factor.

Then substitute that \(x\)-value into the simplified function to find the \(y\)-value of the hole.

So for a hole:

• The \(x\)-coordinate comes from the canceled factor.
• The \(y\)-coordinate comes from plugging that \(x\)-value into the simplified expression.

Worked Example 1: A basic vertical asymptote

Find any vertical asymptotes or removable discontinuities of

$$f(x)=\frac{2}{x+5}$$

Step 1: Find where the denominator is zero.

$$x+5=0 \Rightarrow x=-5$$

Step 2: Check for cancellation.

The numerator is just \(2\), so there is no factor \((x+5)\) to cancel.

Conclusion: \(x=-5\) is a vertical asymptote.

There is no removable discontinuity.

Worked Example 2: A basic removable discontinuity

Find any vertical asymptotes or removable discontinuities of

$$f(x)=\frac{x-4}{x-4}$$

Step 1: Factor if needed.

This is already factored.

Step 2: Find where the denominator is zero.

$$x-4=0 \Rightarrow x=4$$

Step 3: Check for cancellation.

The factor \((x-4)\) appears in both numerator and denominator, so it cancels:

$$f(x)=1 \quad \text{for } x\neq 4$$

Step 4: Find the hole.

The hole is at \(x=4\). Using the simplified function \(y=1\), the hole is at

$$ (4,1) $$

Conclusion: The function has a removable discontinuity at \((4,1)\), not a vertical asymptote.

Worked Example 3: One hole and one vertical asymptote

Find the vertical asymptotes and removable discontinuities of

$$f(x)=\frac{(x-2)(x+1)}{(x-2)(x-5)}$$

Step 1: Identify factors.

The denominator is zero at

$$x=2 \quad \text{and} \quad x=5$$

Step 2: Cancel common factors.

The factor \((x-2)\) appears in both numerator and denominator, so it cancels:

$$f(x)=\frac{x+1}{x-5} \quad \text{for } x\neq 2,5$$

Step 3: Classify each value.

• \(x=2\): the factor canceled, so this is a removable discontinuity.
• \(x=5\): the factor \((x-5)\) remains in the denominator, so this is a vertical asymptote.

Step 4: Find the hole's coordinates.

Plug \(x=2\) into the simplified function:

$$y=\frac{2+1}{2-5}=\frac{3}{-3}=-1$$

So the hole is at

$$ (2,-1) $$

Conclusion:

• Vertical asymptote: \(x=5\)
• Removable discontinuity: \((2,-1)\)

Worked Example 4: Factoring first

Find the vertical asymptotes and removable discontinuities of

$$f(x)=\frac{x^2-9}{x^2-x-12}$$

Step 1: Factor numerator and denominator.

$$x^2-9=(x-3)(x+3)$$

$$x^2-x-12=(x-4)(x+3)$$

So

$$f(x)=\frac{(x-3)(x+3)}{(x-4)(x+3)}$$

Step 2: Cancel common factors.

The factor \((x+3)\) cancels:

$$f(x)=\frac{x-3}{x-4} \quad \text{for } x\neq -3,4$$

Step 3: Classify the excluded values.

• \(x=-3\): canceled factor, so this is a removable discontinuity.
• \(x=4\): denominator still zero after simplifying, so this is a vertical asymptote.

Step 4: Find the hole's coordinates.

Substitute \(x=-3\) into the simplified function:

$$y=\frac{-3-3}{-3-4}=\frac{-6}{-7}=\frac{6}{7}$$

So the hole is at

$$ \left(-3,\frac{6}{7}\right) $$

Conclusion:

• Vertical asymptote: \(x=4\)
• Removable discontinuity: \(\left(-3,\frac{6}{7}\right)\)

7. Graph meaning

It helps to connect the algebra to the graph.

• If a factor does not cancel, the graph breaks near that \(x\)-value and heads up or down. That is a vertical asymptote.
• If a factor does cancel, the graph looks like the simplified function, but with one missing point. That missing point is the hole.

You can think of a removable discontinuity as a point that could be “fixed” if the function were redefined there. That is why it is called removable.

8. Common mistakes to avoid

• Do not assume every denominator zero is a vertical asymptote. First check whether the factor cancels.
• Do not forget to factor completely. A common factor may be hidden until you factor.
• Do not plug the hole's x-value into the original function. The original function is undefined there. Use the simplified function to find the hole's y-value.
• Do not say the hole disappears completely. The factor cancels algebraically, but the original function still has a missing point.

9. Quick checklist

When you are given a rational function, ask:

1. Can I factor the numerator and denominator?
2. What values make the denominator zero?
3. Which factors cancel?
4. Which zeros stay in the denominator? Those are vertical asymptotes.
5. Which zeros come from canceled factors? Those are removable discontinuities.
6. If there is a hole, what is its \(y\)-value in the simplified function?

10. Summary

Vertical asymptotes and removable discontinuities both begin with values that make the denominator zero. The difference is what happens after factoring and simplifying.

If the factor causing the zero does not cancel, the function has a vertical asymptote. If the factor does cancel, the function has a removable discontinuity, which appears as a hole in the graph.

To solve these problems correctly, always factor first, cancel common factors carefully, and then classify each excluded value.

Horizontal Asymptotes describe what happens to a function as the input gets very large positive or very large negative. They tell us the end behavior of a graph.

For rational functions, which are functions written as one polynomial divided by another polynomial, horizontal asymptotes can often be found by comparing the degrees of the numerator and denominator and, in some cases, their leading coefficients.

This lesson will show you what horizontal asymptotes mean, how to find them, and how to avoid common mistakes.

Reminder: A rational function has the form

$$f(x)=\frac{P(x)}{Q(x)}$$

where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x)\neq 0\).

What is a horizontal asymptote?

A horizontal asymptote is a horizontal line \(y=L\) that the graph of a function approaches as \(x\to \infty\) or as \(x\to -\infty\).

In symbols, if

$$\lim_{x\to \infty} f(x)=L \quad \text{or} \quad \lim_{x\to -\infty} f(x)=L,$$

then \(y=L\) is a horizontal asymptote.

This does not mean the graph has to touch the asymptote. It also does not mean the graph stays below or above it forever. It only means the graph gets closer to that line far to the left or far to the right.

Why do degree and leading coefficient matter?

When \(x\) becomes very large, the highest-power terms in the numerator and denominator have the biggest effect on the value of the function. Lower-power terms matter less and less.

For example, in

$$\frac{3x^4+2x-1}{5x^4+7x^2+9},$$

the terms \(3x^4\) and \(5x^4\) dominate for large values of \(x\). So the function behaves like

$$\frac{3x^4}{5x^4}=\frac{3}{5}.$$

That means the horizontal asymptote is \(y=\frac{3}{5}\).

The 3 main rules for horizontal asymptotes

1. If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote is \(y=0\).
2. If the degree of the numerator is equal to the degree of the denominator, then the horizontal asymptote is the ratio of the leading coefficients.
3. If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote.

Let us write these more clearly.

Suppose

$$f(x)=\frac{a_nx^n+\cdots}{b_mx^m+\cdots}.$$

• If \(n<m\), then the horizontal asymptote is \(y=0\).
• If \(n=m\), then the horizontal asymptote is \(y=\frac{a_n}{b_m}\).
• If \(n>m\), then there is no horizontal asymptote.

Important note: If the numerator’s degree is exactly one more than the denominator’s degree, the function may have a slant asymptote instead. But for this lesson, we focus on whether a horizontal asymptote exists.

How to find a horizontal asymptote step by step

1. Identify the degree of the numerator.
2. Identify the degree of the denominator.
3. Compare the degrees.
4. If the degrees are equal, divide the leading coefficients.
5. State the horizontal asymptote as an equation like \(y=0\) or \(y=2\).

Worked Example 1: Numerator degree less than denominator degree

Find the horizontal asymptote of

$$f(x)=\frac{2x+5}{x^2-4}.$$

Step 1: Find the degree of the numerator.

The numerator is \(2x+5\), which has degree \(1\).

Step 2: Find the degree of the denominator.

The denominator is \(x^2-4\), which has degree \(2\).

Step 3: Compare the degrees.

Since \(1<2\), the numerator degree is less than the denominator degree.

Conclusion: The horizontal asymptote is

$$y=0.$$

This makes sense because for very large \(x\), the function behaves like

$$\frac{2x}{x^2}=\frac{2}{x},$$

and \(\frac{2}{x}\to 0\) as \(x\to \infty\) or \(x\to -\infty\).

Worked Example 2: Degrees are equal

Find the horizontal asymptote of

$$f(x)=\frac{4x^3-7x+1}{2x^3+9x^2-5}.$$

Step 1: Degree of the numerator is \(3\).

Step 2: Degree of the denominator is \(3\).

Step 3: The degrees are equal, so use the ratio of leading coefficients.

The leading coefficient of the numerator is \(4\). The leading coefficient of the denominator is \(2\).

So the horizontal asymptote is

$$y=\frac{4}{2}=2.$$

Conclusion:

$$y=2$$

For large values of \(x\), the function behaves like

$$\frac{4x^3}{2x^3}=2.$$

Worked Example 3: Numerator degree greater than denominator degree

Find the horizontal asymptote of

$$f(x)=\frac{x^4+3x^2-1}{5x-2}.$$

Step 1: Degree of the numerator is \(4\).

Step 2: Degree of the denominator is \(1\).

Step 3: Compare the degrees.

Since \(4>1\), the numerator degree is greater than the denominator degree.

Conclusion: There is no horizontal asymptote.

The function does not level off to a constant value as \(x\) gets very large.

Worked Example 4: Be careful after simplifying

Find the horizontal asymptote of

$$f(x)=\frac{x^2-1}{x^2+x}.$$

Step 1: Factor the expression if possible.

$$x^2-1=(x-1)(x+1)$$

$$x^2+x=x(x+1)$$

So

$$f(x)=\frac{(x-1)(x+1)}{x(x+1)}.$$

For \(x\neq -1\), this simplifies to

$$f(x)=\frac{x-1}{x}.$$

Step 2: Use the simplified form to study end behavior.

The degree of the numerator is \(1\), and the degree of the denominator is \(1\).

The leading coefficients are both \(1\).

So the horizontal asymptote is

$$y=\frac{1}{1}=1.$$

Conclusion:

$$y=1$$

Even though the function has a hole because of the canceled factor, the horizontal asymptote is still based on the end behavior, which is \(y=1\).

Common mistakes to avoid

• Mixing up horizontal and vertical asymptotes. Horizontal asymptotes describe end behavior as \(x\to \pm\infty\). Vertical asymptotes happen at specific \(x\)-values where the denominator becomes zero.
• Forgetting to compare degrees first. Do not just divide the leading coefficients unless the degrees are equal.
• Thinking there must always be a horizontal asymptote. Some rational functions have none.
• Assuming the graph never crosses a horizontal asymptote. A graph can cross a horizontal asymptote. The asymptote only describes what happens far away from the center of the graph.

Can a graph cross a horizontal asymptote?

Yes. A horizontal asymptote is about what the function does far to the left or right, not what it does in the middle.

For example, consider

$$f(x)=\frac{x}{x^2+1}.$$

The numerator has degree \(1\), and the denominator has degree \(2\), so the horizontal asymptote is \(y=0\).

But when \(x=0\),

$$f(0)=0,$$

so the graph actually touches the line \(y=0\). This is allowed.

A quick way to think about it

• If the bottom grows faster than the top, the fraction shrinks to \(0\).
• If the top and bottom grow at the same rate, the fraction settles to the ratio of leading coefficients.
• If the top grows faster than the bottom, the function does not settle to a horizontal line.

Practice Check

Try these on your own:

1. $$f(x)=\frac{3x-2}{x^5+1}$$
2. $$f(x)=\frac{7x^2+4}{x^2-9}$$
3. $$f(x)=\frac{2x^3+1}{x^2+6}$$

Answers:

1. Degree of numerator \(1\), degree of denominator \(5\), so horizontal asymptote: $$y=0$$
2. Degrees are equal, so use leading coefficients \(7\) and \(1\), giving horizontal asymptote: $$y=7$$
3. Numerator degree \(3\), denominator degree \(2\), so there is no horizontal asymptote.

Summary

To find the horizontal asymptote of a rational function, compare the degree of the numerator with the degree of the denominator.

• If numerator degree \(<\) denominator degree, then \(y=0\).
• If numerator degree \(=\) denominator degree, then \(y=\) ratio of leading coefficients.
• If numerator degree \(>\) denominator degree, then there is no horizontal asymptote.

Horizontal asymptotes help you understand how the graph behaves far to the left and right. They are a powerful tool for sketching and analyzing rational functions.

Lesson: Slant (Oblique) Asymptotes

When we graph a rational function, we often want to know what the graph does far to the left and far to the right. Sometimes the graph gets closer and closer to a horizontal line. Sometimes it gets closer to a vertical line near a value where the function is undefined. But there is another important possibility: the graph may get closer to a slanted line.

This slanted line is called a slant asymptote or an oblique asymptote. In this lesson, you will learn what a slant asymptote is, when it happens, and how to find it using polynomial long division.

What is a slant asymptote?

A slant asymptote is a line of the form \(y=mx+b\) that the graph of a rational function approaches as \(x\) becomes very large positive or very large negative.

In other words, for large values of \(x\), the function behaves almost like a line.

Slant asymptotes happen for rational functions when the degree of the numerator is exactly one more than the degree of the denominator.

For example:

Here, the numerator has degree \(2\) and the denominator has degree \(1\). Since \(2\) is exactly one more than \(1\), this function has a slant asymptote.

When does a slant asymptote occur?

• If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is usually \(y=0\).
• If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients.
• If the degree of the numerator is exactly one greater than the degree of the denominator, there is a slant asymptote.

This lesson focuses on that third case.

Why polynomial long division works

To find a slant asymptote, we divide the numerator by the denominator.

After division, the rational function can be written in this form:

If the degree of the numerator is one greater than the degree of the denominator, the quotient will be a linear expression, like \(mx+b\).

So the function becomes:

As \(x\) gets very large, the fraction \(\frac{r(x)}{Q(x)}\) gets very close to \(0\). That means the graph gets closer and closer to the line:

That line is the slant asymptote.

Steps to find a slant asymptote

1. Check the degrees of the numerator and denominator.
2. Make sure the numerator’s degree is exactly one more than the denominator’s degree.
3. Use polynomial long division.
4. Ignore the remainder part when writing the asymptote.
5. The quotient is the equation of the slant asymptote.

Important note

The slant asymptote is based on the quotient only, not the remainder.

Also, the graph may cross a slant asymptote. Unlike some early ideas students have about asymptotes, an asymptote is not always a line the graph can never touch. It is a line the graph approaches for very large positive or negative values of \(x\).

Worked Example 1

Find the slant asymptote of:

Step 1: Check the degrees.

The numerator has degree \(2\). The denominator has degree \(1\). Since \(2=1+1\), a slant asymptote exists.

Step 2: Divide.

We divide \(x^2+5x+6\) by \(x+2\).

First, divide \(x^2\) by \(x\):

Put \(x\) in the quotient.

Now multiply:

Subtract:

Next, divide \(3x\) by \(x\):

Put \(3\) in the quotient.

Multiply:

Subtract:

So:

This means the function is exactly the line \(y=x+3\), except we must remember the original denominator cannot be zero, so \(x=-2\) is not allowed.

Slant asymptote:

Worked Example 2

Find the slant asymptote of:

Step 1: Check the degrees.

The numerator has degree \(2\), and the denominator has degree \(1\). So a slant asymptote exists.

Step 2: Use long division.

Divide \(2x^2+3x-1\) by \(x-4\).

First term:

Put \(2x\) in the quotient.

Multiply:

Subtract:

Next term:

Put \(11\) in the quotient.

Multiply:

Subtract:

So:

As \(x\) becomes very large or very negative, \(\frac{43}{x-4}\) gets closer to \(0\).

So the slant asymptote is:

Worked Example 3

Find the slant asymptote of:

Step 1: Check the degrees.

The numerator has degree \(3\). The denominator has degree \(2\). Since \(3=2+1\), a slant asymptote exists.

Step 2: Divide.

We divide \(x^3-2x^2+0x+4\) by \(x^2+1\).

First term:

Put \(x\) in the quotient.

Multiply:

Subtract:

Next term:

Put \(-2\) in the quotient.

Multiply:

Subtract:

So:

As \(x\) gets very large in either direction, \(\frac{-x+6}{x^2+1}\) approaches \(0\).

So the slant asymptote is:

Worked Example 4

Determine whether the function has a slant asymptote, and if so, find it:

Step 1: Compare degrees.

The numerator has degree \(2\), and the denominator also has degree \(2\).

Since the degrees are equal, this function does not have a slant asymptote.

Instead, it has a horizontal asymptote equal to the ratio of the leading coefficients:

This example is important because not every rational function has a slant asymptote. You should always check the degrees first.

How to write the final answer

After division, suppose you get:

Then the slant asymptote is just:

You do not include the fraction part in the asymptote.

Common mistakes to avoid

• Forgetting to compare degrees first. A slant asymptote only happens when the numerator’s degree is exactly one more than the denominator’s degree.
• Using the remainder in the asymptote. The asymptote comes from the quotient only.
• Making errors in long division. Line up terms carefully and include missing powers, such as writing \(0x\) if needed.
• Confusing vertical and slant asymptotes. Vertical asymptotes come from values that make the denominator zero. Slant asymptotes describe end behavior.

Quick check for understanding

For each function, decide whether a slant asymptote exists:

• \(\frac{x+1}{x^2-4}\) → no, numerator degree is less than denominator degree
• \(\frac{x^2+1}{x-3}\) → yes, numerator degree is one more than denominator degree
• \(\frac{4x^3-2}{2x^3+1}\) → no, degrees are equal
• \(\frac{x^4+1}{x^2+3}\) → no slant asymptote, because the degree difference is more than one

Final idea to remember

A slant asymptote tells you the line that a rational function follows more and more closely at the far ends of the graph.

To find it, use polynomial long division. If the numerator’s degree is exactly one greater than the denominator’s degree, the quotient will be linear, and that linear expression is the slant asymptote.

Brief Summary

A slant asymptote occurs in a rational function when the degree of the numerator is exactly one greater than the degree of the denominator. To find it, divide the numerator by the denominator using polynomial long division. The quotient, not the remainder, gives the slant asymptote. If the quotient is \(mx+b\), then the slant asymptote is \(y=mx+b\).

Graphing Rational Functions means sketching the graph of a function that can be written as a quotient of polynomials:

$$f(x)=\frac{p(x)}{q(x)}, \quad q(x)\neq 0$$

To graph these well, you do not usually plot many random points. Instead, you study the function’s key features: the domain, intercepts, asymptotes, holes, and a few test points. Then you combine that information into an accurate sketch.

This lesson will show you a clear step-by-step method you can use on almost any rational function in 11th Grade maths.

1. What makes rational functions different?

Rational functions are different from polynomials because the denominator can create places where the function is undefined. These undefined values often lead to:

• vertical asymptotes, where the graph rises or falls without bound,
• holes, where one point is missing from an otherwise smooth curve,
• and special end behavior, often described by a horizontal or slant asymptote.

Because of this, graphing rational functions is really about understanding how the graph behaves near these important features.

2. A step-by-step method for graphing

When graphing a rational function, use this order:

1. Factor the numerator and denominator if possible.
2. Find the domain by identifying values that make the denominator zero.
3. Check for common factors. A common factor that cancels creates a hole.
4. Find the x-intercepts by setting the numerator equal to zero, but only after accounting for any cancelled factors.
5. Find the y-intercept by substituting \(x=0\), if allowed.
6. Find any vertical asymptotes from denominator factors that remain after simplification.
7. Find the horizontal or slant asymptote by comparing the degrees of the numerator and denominator.
8. Choose test points in the intervals created by the vertical asymptotes and holes.
9. Sketch the graph, showing the asymptotes, intercepts, and the way each branch behaves.

3. Domain

The domain of a rational function is all real numbers except the values that make the denominator zero.

For example, if

$$f(x)=\frac{x+1}{x-3}$$

then \(x=3\) is not allowed, because the denominator would be zero. So the domain is all real numbers except \(3\).

You can write this as:

$$x\neq 3$$

or in interval notation:

$$(-\infty,3)\cup(3,\infty)$$

4. Holes and removable discontinuities

If the numerator and denominator have a common factor, that factor can cancel. This means the simplified function may look fine at that \(x\)-value, but the original function is still undefined there. That creates a hole.

Example:

$$f(x)=\frac{(x-2)(x+1)}{(x-2)(x-3)}$$

The factor \((x-2)\) cancels, so the simplified function is

$$f(x)=\frac{x+1}{x-3}, \quad x\neq 2,3$$

Because \(x=2\) was cancelled, the graph has a hole at \(x=2\), not a vertical asymptote there.

To find the coordinates of the hole, substitute \(x=2\) into the simplified function:

$$y=\frac{2+1}{2-3}=\frac{3}{-1}=-3$$

So the hole is at \((2,-3)\).

5. Intercepts

x-intercepts occur where the graph crosses the \(x\)-axis, so \(y=0\). For a rational function, this happens when the numerator is zero and the denominator is not zero.

So:

• Set the numerator equal to zero.
• Make sure the factor did not cancel.

y-intercepts occur where \(x=0\). Substitute \(x=0\) into the function, if \(0\) is in the domain.

6. Vertical asymptotes

A vertical asymptote occurs at values of \(x\) that make the denominator zero after simplification.

If a denominator factor cancels, it creates a hole instead of a vertical asymptote.

Near a vertical asymptote, the function values become very large positive or very large negative numbers.

7. Horizontal and slant asymptotes

The end behavior of a rational function depends on the degrees of the numerator and denominator.

• If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y=0\).
• If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.
• If the degree of the numerator is one more than the degree of the denominator, there is a slant asymptote. Find it by division.

For horizontal asymptotes:

$$f(x)=\frac{2x^2+1}{x^2-5} \quad \Rightarrow \quad y=\frac{2}{1}=2$$

since both top and bottom have degree 2.

For a slant asymptote:

$$f(x)=\frac{x^2+1}{x-1}$$

Divide \(x^2+1\) by \(x-1\):

$$\frac{x^2+1}{x-1}=x+1+\frac{2}{x-1}$$

So the slant asymptote is

$$y=x+1$$

8. Test points and branch behavior

After finding vertical asymptotes, divide the number line into intervals. Then test one \(x\)-value in each interval to see whether the function is positive or negative there.

This helps you decide whether each branch lies above or below the \(x\)-axis and how it approaches the asymptotes.

For example, if there is a vertical asymptote at \(x=2\), you might test one point from each interval:

• left of 2, such as \(x=1\),
• right of 2, such as \(x=3\).

These test points make your sketch much more accurate.

Worked Example 1: A basic rational function

Graph:

$$f(x)=\frac{x+1}{x-2}$$

Step 1: Domain

The denominator cannot be zero:

$$x-2=0 \Rightarrow x=2$$

So the domain is:

$$x\neq 2$$

Step 2: Check for holes

There are no common factors, so there are no holes.

Step 3: x-intercept

Set the numerator equal to zero:

$$x+1=0 \Rightarrow x=-1$$

So the \(x\)-intercept is:

$$(-1,0)$$

Step 4: y-intercept

Substitute \(x=0\):

$$f(0)=\frac{0+1}{0-2}=\frac{1}{-2}=-\frac{1}{2}$$

So the \(y\)-intercept is:

$$(0,-\tfrac{1}{2})$$

Step 5: Vertical asymptote

The denominator is zero at \(x=2\), so the vertical asymptote is:

$$x=2$$

Step 6: Horizontal asymptote

The numerator and denominator both have degree 1. The ratio of leading coefficients is \(1/1=1\).

So the horizontal asymptote is:

$$y=1$$

Step 7: Test points

Test \(x=1\):

$$f(1)=\frac{2}{-1}=-2$$

Test \(x=3\):

$$f(3)=\frac{4}{1}=4$$

Sketch description

• One branch is left of \(x=2\), passing through \((-1,0)\) and \((0,-\tfrac12)\).
• As \(x\to 2^-\), the graph falls toward \(-\infty\).
• The other branch is right of \(x=2\).
• As \(x\to 2^+\), the graph rises toward \(+\infty\).
• As \(x\to \pm\infty\), the graph approaches \(y=1\).

Worked Example 2: A graph with a hole

Graph:

$$f(x)=\frac{(x-1)(x+2)}{(x-1)(x-3)}$$

Step 1: Simplify

Cancel the common factor \((x-1)\):

$$f(x)=\frac{x+2}{x-3}, \quad x\neq 1,3$$

Step 2: Domain

The original denominator was zero at \(x=1\) and \(x=3\), so both are excluded.

Step 3: Hole

Since \((x-1)\) cancelled, there is a hole at \(x=1\).

Find its \(y\)-value using the simplified function:

$$y=\frac{1+2}{1-3}=\frac{3}{-2}=-\frac{3}{2}$$

So the hole is at:

$$(1,-\tfrac{3}{2})$$

Step 4: Vertical asymptote

The remaining denominator is zero at \(x=3\), so:

$$x=3$$

Step 5: x-intercept

Set the simplified numerator equal to zero:

$$x+2=0 \Rightarrow x=-2$$

So the \(x\)-intercept is:

$$(-2,0)$$

Step 6: y-intercept

Substitute \(x=0\):

$$f(0)=\frac{2}{-3}=-\frac{2}{3}$$

So the \(y\)-intercept is:

$$(0,-\tfrac{2}{3})$$

Step 7: Horizontal asymptote

The degrees are equal, so use the ratio of leading coefficients:

$$y=1$$

Sketch description

• The graph looks like \(\frac{x+2}{x-3}\), but with a missing point at \((1,-\tfrac32)\).
• There is a vertical asymptote at \(x=3\).
• There is a horizontal asymptote at \(y=1\).
• The graph crosses the \(x\)-axis at \((-2,0)\).

Worked Example 3: A function with a slant asymptote

Graph:

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

Step 1: Domain

The denominator is zero when:

$$x+1=0 \Rightarrow x=-1$$

So the domain is:

$$x\neq -1$$

Step 2: Check for common factors

There is no common factor, so there is no hole.

Step 3: Vertical asymptote

Since \(x=-1\) makes the denominator zero, the vertical asymptote is:

$$x=-1$$

Step 4: x-intercepts

Set the numerator equal to zero:

$$x^2+2x+3=0$$

The discriminant is:

$$b^2-4ac=2^2-4(1)(3)=4-12=-8$$

Since the discriminant is negative, there are no real x-intercepts.

Step 5: y-intercept

Substitute \(x=0\):

$$f(0)=\frac{3}{1}=3$$

So the \(y\)-intercept is:

$$(0,3)$$

Step 6: Slant asymptote

The numerator has degree 2 and the denominator has degree 1, so the degree of the numerator is one more. This means there is a slant asymptote.

Divide:

$$\frac{x^2+2x+3}{x+1}=x+1+\frac{2}{x+1}$$

So the slant asymptote is:

$$y=x+1$$

Step 7: Test points

Test \(x=-2\):

$$f(-2)=\frac{4-4+3}{-1}=-3$$

Test \(x=1\):

$$f(1)=\frac{1+2+3}{2}=3$$

Sketch description

• The graph has two branches separated by \(x=-1\).
• It approaches the slant asymptote \(y=x+1\) as \(x\to \pm\infty\).
• It has no real \(x\)-intercepts.
• It crosses the \(y\)-axis at \((0,3)\).

9. Important patterns to remember

• Cancelled factor \(\Rightarrow\) hole
• Remaining denominator zero \(\Rightarrow\) vertical asymptote
• Numerator zero \(\Rightarrow\) possible \(x\)-intercept
• Substitute \(x=0\) \(\Rightarrow\) \(y\)-intercept
• Compare degrees \(\Rightarrow\) horizontal or slant asymptote

10. Common mistakes

• Forgetting to factor first. If you do not factor, you may miss a hole.
• Calling a cancelled factor a vertical asymptote. Cancelled factors create holes, not asymptotes.
• Using the original numerator for x-intercepts after cancellation. Only non-cancelled zeros of the numerator give x-intercepts.
• Forgetting that the graph may cross a horizontal asymptote. A horizontal asymptote describes end behavior, not a wall the graph can never cross.
• Not using test points. Without them, the branches may be drawn on the wrong side of an asymptote.

11. Quick graphing checklist

1. Factor completely.
2. State the domain.
3. Cancel common factors and identify holes.
4. Find x-intercepts.
5. Find the y-intercept.
6. Find vertical asymptotes.
7. Find horizontal or slant asymptote.
8. Use test points in each interval.
9. Sketch carefully and mark holes with open circles.

Summary

To graph a rational function, start by factoring and identifying where the denominator is zero. Then determine whether those values create holes or vertical asymptotes. Next, find the intercepts and the horizontal or slant asymptote. Finally, use test points to decide where each branch goes, and sketch the graph with all important features clearly labeled.

Rational Inequalities are inequalities that involve a rational expression, which is a fraction made of polynomials. For example,

\[ \frac{x-2}{x+1} > 0 \]

is a rational inequality.

Solving a rational inequality means finding all values of \(x\) that make the inequality true. Unlike solving a rational equation, we are not just looking for points where the two sides are equal. We must find where the expression is positive, negative, or possibly zero.

This topic is important because rational expressions can change sign at special values. They can also be undefined when the denominator is zero, so we must be careful.

Main idea: To solve a rational inequality, find the values where the numerator is zero and where the denominator is zero. These values split the number line into intervals. Then test the sign of the expression in each interval.

Step-by-step method

1. Move everything to one side so the inequality has the form

   \[ \frac{P(x)}{Q(x)} > 0, \quad \frac{P(x)}{Q(x)} < 0, \quad \frac{P(x)}{Q(x)} \ge 0, \quad \text{or} \quad \frac{P(x)}{Q(x)} \le 0. \]

2. Factor the numerator and denominator completely if possible.

3. Find the critical values:

   • Zeros of the numerator: where the expression equals \(0\).

   • Zeros of the denominator: where the expression is undefined.

4. Place these critical values on a number line. They divide the number line into intervals.

5. Use a sign chart by testing one value from each interval, or by analyzing the signs of the factors.

6. Choose the intervals that make the inequality true.

7. Include or exclude endpoints correctly:

   • If the inequality is \(\ge\) or \(\le\), include numerator zeros if they make the expression equal to \(0\).

   • Never include values that make the denominator zero.

Why sign charts work

A rational expression can only change sign at values where a factor becomes zero or undefined. Between those critical values, the sign stays the same. That is why testing one point in each interval is enough.

Important facts to remember

• A rational expression is undefined when the denominator is zero.

• If the numerator is zero and the denominator is not zero, the expression equals \(0\).

• If a factor appears an even number of times, such as \((x-3)^2\), its sign does not change when passing through that value.

• If a factor appears an odd number of times, such as \((x-3)\), its sign does change when passing through that value.

Worked Example 1: A basic rational inequality

Solve

\[ \frac{x-4}{x+2} > 0 \]

Step 1: Find critical values.

• Numerator zero: \(x-4=0\Rightarrow x=4\)

• Denominator zero: \(x+2=0\Rightarrow x=-2\)

These values divide the number line into three intervals:

\[ (-\infty,-2), \quad (-2,4), \quad (4,\infty) \]

Step 2: Test each interval.

Choose a test value from each interval.

• For \(( -\infty,-2 )\), test \(x=-3\):

  \[ \frac{-3-4}{-3+2} = \frac{-7}{-1}=7>0 \]

  So this interval works.

• For \(( -2,4 )\), test \(x=0\):

  \[ \frac{0-4}{0+2}=\frac{-4}{2}=-2<0 \]

  So this interval does not work.

• For \(( 4,\infty )\), test \(x=5\):

  \[ \frac{5-4}{5+2}=\frac{1}{7}>0 \]

  So this interval works.

Step 3: Decide about endpoints.

• \(x=4\) makes the expression \(0\), but the inequality is strict \((>)\), so do not include it.

• \(x=-2\) makes the denominator zero, so do not include it.

Solution:

\[ x\in (-\infty,-2)\cup(4,\infty) \]

Worked Example 2: Including zero in the answer

Solve

\[ \frac{x+1}{x-3} \le 0 \]

Step 1: Find critical values.

• Numerator zero: \(x=-1\)

• Denominator zero: \(x=3\)

Intervals:

\[ (-\infty,-1), \quad (-1,3), \quad (3,\infty) \]

Step 2: Test each interval.

• Test \(x=-2\):

  \[ \frac{-2+1}{-2-3}=\frac{-1}{-5}>0 \]

• Test \(x=0\):

  \[ \frac{0+1}{0-3}=\frac{1}{-3}<0 \]

• Test \(x=4\):

  \[ \frac{4+1}{4-3}=\frac{5}{1}>0 \]

We want values where the expression is less than or equal to zero, so we take the negative interval and also any numerator zero that is allowed.

Step 3: Check endpoints.

• At \(x=-1\), the expression is \(0\), and \(0\le 0\), so include \(-1\).

• At \(x=3\), the expression is undefined, so do not include \(3\).

Solution:

\[ x\in [-1,3) \]

Worked Example 3: Factored form with more than two critical values

Solve

\[ \frac{(x-1)(x+4)}{(x-2)(x+3)} < 0 \]

Step 1: Find critical values.

• Numerator zeros: \(x=1\), \(x=-4\)

• Denominator zeros: \(x=2\), \(x=-3\)

Put them in order:

\[ -4,\,-3,\,1,\,2 \]

Intervals:

\[ (-\infty,-4),\quad (-4,-3),\quad (-3,1),\quad (1,2),\quad (2,\infty) \]

Step 2: Use signs of factors.

We can test points, or we can look at signs factor by factor.

Test values:

• \(x=-5\):

  \[ \frac{(-)(-)}{(-)(-)}=\frac{+}{+}=+ \]

• \(x=-3.5\):

  \[ \frac{(-)(+)}{(-)(-)}=\frac{-}{+}=- \]

• \(x=0\):

  \[ \frac{(-)(+)}{(-)(+)}=\frac{-}{-}=+ \]

• \(x=1.5\):

  \[ \frac{(+)(+)}{(-)(+)}=\frac{+}{-}=- \]

• \(x=3\):

  \[ \frac{(+)(+)}{(+)(+)}=+ \]

We want the expression to be negative, so choose the negative intervals.

Step 3: Endpoints.

• Do not include \(x=-4\) or \(x=1\) because the inequality is strict \((<)\).

• Do not include \(x=-3\) or \(x=2\) because they make the denominator zero.

Solution:

\[ x\in (-4,-3)\cup(1,2) \]

Worked Example 4: Repeated factors

Solve

\[ \frac{(x-2)^2}{(x+1)(x-5)} \ge 0 \]

Step 1: Find critical values.

• Numerator zero: \(x=2\), with multiplicity 2

• Denominator zeros: \(x=-1\), \(x=5\)

Intervals:

\[ (-\infty,-1),\quad (-1,2),\quad (2,5),\quad (5,\infty) \]

Step 2: Think about signs.

The factor \((x-2)^2\) is always nonnegative. It is positive for every \(x\neq 2\), and zero at \(x=2\). Because the exponent is even, the sign does not change as we pass through \(x=2\).

So the sign mostly depends on the denominator \((x+1)(x-5)\).

• For \(x=-2\): denominator is \((-)(-) = +\), so the whole expression is positive.

• For \(x=0\): denominator is \((+)(-) = -\), so the whole expression is negative.

• For \(x=3\): denominator is \((+)(-) = -\), so the whole expression is negative.

• For \(x=6\): denominator is \((+)(+) = +\), so the whole expression is positive.

We want the expression to be greater than or equal to zero.

Step 3: Endpoints.

• At \(x=2\), the numerator is zero and the denominator is not zero, so the expression equals \(0\). Since \(\ge 0\), include \(2\).

• At \(x=-1\) and \(x=5\), the expression is undefined, so do not include them.

Solution:

\[ x\in (-\infty,-1)\cup\{2\}\cup(5,\infty) \]

How to handle rational inequalities with more than one fraction

Sometimes the inequality is written like this:

\[ \frac{1}{x} - \frac{2}{x-1} > 0 \]

First combine the fractions into a single rational expression.

Use a common denominator:

\[ \frac{1}{x} - \frac{2}{x-1} = \frac{x-1}{x(x-1)} - \frac{2x}{x(x-1)} = \frac{x-1-2x}{x(x-1)} = \frac{-x-1}{x(x-1)} \]

So the inequality becomes

\[ \frac{-x-1}{x(x-1)} > 0 \]

Then solve it using the same sign chart method.

Common mistakes to avoid

• Forgetting domain restrictions. If the denominator is zero, that value can never be in the solution.

• Only solving numerator equals zero. In rational inequalities, denominator zeros matter just as much because they split the number line.

• Using the wrong endpoint rules.

  • Include numerator zeros only if the inequality allows equality.

  • Never include denominator zeros.

• Not factoring completely. Factored form makes sign analysis much easier.

• Assuming the sign always changes at every critical value. It may not change at repeated factors with even powers.

Quick checklist for solving rational inequalities

1. Get one rational expression on one side.

2. Factor numerator and denominator.

3. Find all zeros and undefined points.

4. Mark them on a number line.

5. Test the sign in each interval.

6. Select the intervals that satisfy the inequality.

7. Include or exclude endpoints correctly.

Brief Summary

To solve a rational inequality, rewrite it as one fraction, factor completely, and find the critical values from the numerator and denominator. These values divide the number line into intervals. Use a sign chart to determine where the expression is positive or negative. Include numerator zeros only when equality is allowed, and never include denominator zeros.