Introduction to Rational Expressions
Hello dear student! Welcome to Unit 2 of your Grade 11 Mathematics. In this unit, we will learn about rational expressions and rational functions. These topics build directly on what you already know about fractions and polynomials. So, let’s begin step by step!
A rational expression is simply a fraction where both the numerator and the denominator are polynomials. Think of it like a regular fraction such as \( \frac{3}{5} \), but instead of just numbers, we have polynomials.
\[
\text{Rational Expression} = \frac{P(x)}{Q(x)}, \quad \text{where } Q(x) \neq 0
\]
For example:
\[
\frac{x+1}{x-3}, \quad \frac{2x^2-5x+3}{x+4}, \quad \frac{x^3+8}{x^2-4}
\]
Notice something important — the denominator can never be zero. Why? Because division by zero is undefined in mathematics. So, for each rational expression, we must identify the values that make the denominator zero. Those values are called restricted values or excluded values.
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Finding Restricted Values
To find restricted values, we set the denominator equal to zero and solve.
Example 1
Find the restricted value(s) of \( \dfrac{x+2}{x-5} \).
Solution:
Set the denominator equal to zero:
Set the denominator equal to zero:
\[ x – 5 = 0 \implies x = 5 \]
So, \( x = 5 \) is the restricted value. The expression is defined for all real numbers except \( x = 5 \).Example 2
Find the restricted value(s) of \( \dfrac{3x+1}{x^2-9} \).
Solution:
Set the denominator equal to zero:
Set the denominator equal to zero:
\[ x^2 – 9 = 0 \implies (x-3)(x+3) = 0 \implies x = 3 \text{ or } x = -3 \]
So, the restricted values are \( x = 3 \) and \( x = -3 \).Can you see the pattern? We always factor the denominator first, then set each factor to zero. This gives us all the restricted values.
Practice 1: Find the restricted value(s) of \( \dfrac{2x-1}{x^2+4x+3} \).
Factor the denominator: \( x^2+4x+3 = (x+1)(x+3) \).
Set each factor to zero: \( x+1=0 \Rightarrow x=-1 \) and \( x+3=0 \Rightarrow x=-3 \).
Restricted values: \( x = -1 \) and \( x = -3 \).
Set each factor to zero: \( x+1=0 \Rightarrow x=-1 \) and \( x+3=0 \Rightarrow x=-3 \).
Restricted values: \( x = -1 \) and \( x = -3 \).
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Simplifying Rational Expressions
Simplifying a rational expression is just like simplifying a regular fraction. We divide the numerator and denominator by their common factors. Remember: we can only cancel factors, never individual terms!
Example 3
Simplify \( \dfrac{x^2-4}{x^2-4x+4} \).
Solution:
Step 1: Factor both numerator and denominator.
Step 1: Factor both numerator and denominator.
\[
\frac{x^2-4}{x^2-4x+4} = \frac{(x-2)(x+2)}{(x-2)(x-2)}
\]
Step 2: Cancel the common factor \( (x-2) \):\[
= \frac{x+2}{x-2}, \quad x \neq 2
\]
Note: We write \( x \neq 2 \) because the original denominator becomes zero when \( x=2 \). Even though the factor canceled, the restriction remains!
Key Exam Note: After simplifying, you MUST carry forward all original restricted values. A common exam mistake is to forget the restrictions from canceled factors. Always check the original denominator!
Example 4
Simplify \( \dfrac{6x^2+9x}{3x^2} \).
Solution:
\[
\frac{6x^2+9x}{3x^2} = \frac{3x(2x+3)}{3x \cdot x} = \frac{2x+3}{x}, \quad x \neq 0
\]
Here, we factor out the GCF from the numerator and cancel \( 3x \).
Practice 2: Simplify \( \dfrac{x^2-9}{x^2+6x+9} \) and state any restrictions.
Factor: numerator \( = (x-3)(x+3) \), denominator \( = (x+3)(x+3) \).
Cancel \( (x+3) \):
\[ \frac{x-3}{x+3}, \quad x \neq -3 \]
The restriction comes from both the original denominator (double root at \( x=-3 \)) and from the simplified form.
Cancel \( (x+3) \):
\[ \frac{x-3}{x+3}, \quad x \neq -3 \]
The restriction comes from both the original denominator (double root at \( x=-3 \)) and from the simplified form.
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Exam-Style Questions — Simplifying Rational Expressions
Q1. Simplify \( \dfrac{x^2-5x+6}{x^2-x-2} \) and state the restriction(s).
Numerator: \( x^2-5x+6 = (x-2)(x-3) \)
Denominator: \( x^2-x-2 = (x-2)(x+1) \)
Cancel \( (x-2) \):
\[ \frac{x-3}{x+1}, \quad x \neq 2, \; x \neq -1 \]
Both restrictions \( x=2 \) and \( x=-1 \) come from the original denominator.
Denominator: \( x^2-x-2 = (x-2)(x+1) \)
Cancel \( (x-2) \):
\[ \frac{x-3}{x+1}, \quad x \neq 2, \; x \neq -1 \]
Both restrictions \( x=2 \) and \( x=-1 \) come from the original denominator.
Q2. Simplify \( \dfrac{4x^2-12x}{8x^3-24x^2} \) and state the restriction(s).
Numerator: \( 4x(x-3) \)
Denominator: \( 8x^2(x-3) \)
Cancel \( 4x(x-3) \):
\[ \frac{1}{2x}, \quad x \neq 0, \; x \neq 3 \]
Note: Even though \( x=3 \) canceled, it remains a restriction from the original expression.
Denominator: \( 8x^2(x-3) \)
Cancel \( 4x(x-3) \):
\[ \frac{1}{2x}, \quad x \neq 0, \; x \neq 3 \]
Note: Even though \( x=3 \) canceled, it remains a restriction from the original expression.
Q3. A student simplified \( \dfrac{x+3}{x-2} \) to \( \dfrac{3}{-2} \) by “canceling the \( x \)’s.” Explain why this is wrong.
You can only cancel factors, not individual terms. Here, \( x+3 \) and \( x-2 \) are not factored — they are sums. The \( x \) in the numerator is not a factor; it is part of the sum \( x+3 \). There is no common factor to cancel. This is one of the most common mistakes in this topic!
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Multiplication of Rational Expressions
Multiplying rational expressions works just like multiplying regular fractions: multiply the numerators together and multiply the denominators together. But to keep things simple, we usually factor first and cancel before multiplying.
\[
\frac{A}{B} \times \frac{C}{D} = \frac{A \cdot C}{B \cdot D}
\]
Example 5
Multiply: \( \dfrac{x+1}{x-3} \times \dfrac{x^2-9}{x+1} \).
Solution:
First, factor where possible: \( x^2-9 = (x-3)(x+3) \).
First, factor where possible: \( x^2-9 = (x-3)(x+3) \).
\[
\frac{x+1}{x-3} \times \frac{(x-3)(x+3)}{x+1}
\]
Cancel \( (x+1) \) and \( (x-3) \):\[
= x+3, \quad x \neq 3, \; x \neq -1
\]
Example 6
Multiply: \( \dfrac{2x^2+4x}{x^2-1} \times \dfrac{x-1}{x+2} \).
Solution:
Factor everything:
Factor everything:
\[
\frac{2x(x+2)}{(x-1)(x+1)} \times \frac{x-1}{x+2}
\]
Cancel \( (x+2) \) and \( (x-1) \):\[
= \frac{2x}{x+1}, \quad x \neq 1, \; x \neq -1, \; x \neq -2
\]
Practice 3: Multiply \( \dfrac{x^2-16}{x+5} \times \dfrac{x+5}{x-4} \).
Factor: \( x^2-16 = (x-4)(x+4) \).
\[ \frac{(x-4)(x+4)}{x+5} \times \frac{x+5}{x-4} \]
Cancel \( (x+5) \) and \( (x-4) \):
\[ = x+4, \quad x \neq 4, \; x \neq -5 \]
\[ \frac{(x-4)(x+4)}{x+5} \times \frac{x+5}{x-4} \]
Cancel \( (x+5) \) and \( (x-4) \):
\[ = x+4, \quad x \neq 4, \; x \neq -5 \]
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Division of Rational Expressions
Division of rational expressions follows the same rule as dividing fractions: multiply by the reciprocal of the divisor.
\[
\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C} = \frac{AD}{BC}
\]
Example 7
Divide: \( \dfrac{x^2-1}{x+3} \div \dfrac{x-1}{x^2-9} \).
Solution:
Step 1: Rewrite as multiplication by the reciprocal.
Step 1: Rewrite as multiplication by the reciprocal.
\[
\frac{x^2-1}{x+3} \times \frac{x^2-9}{x-1}
\]
Step 2: Factor everything.\[
\frac{(x-1)(x+1)}{x+3} \times \frac{(x-3)(x+3)}{x-1}
\]
Step 3: Cancel common factors: \( (x-1) \) and \( (x+3) \).\[
= (x+1)(x-3) = x^2 – 2x – 3, \quad x \neq -3, \; x \neq 1, \; x \neq 3
\]
Exam-Style Questions — Multiplication and Division
Q4. Simplify \( \dfrac{x^2+3x+2}{x^2-4} \div \dfrac{x+1}{x-2} \).
Rewrite as multiplication: \( \frac{x^2+3x+2}{x^2-4} \times \frac{x-2}{x+1} \)
Factor: numerator \( = (x+1)(x+2) \), denominator \( = (x-2)(x+2) \)
\[ \frac{(x+1)(x+2)}{(x-2)(x+2)} \times \frac{x-2}{x+1} \]
Cancel all: \( = 1 \), with restrictions \( x \neq 2, \; x \neq -2, \; x \neq -1 \).
Factor: numerator \( = (x+1)(x+2) \), denominator \( = (x-2)(x+2) \)
\[ \frac{(x+1)(x+2)}{(x-2)(x+2)} \times \frac{x-2}{x+1} \]
Cancel all: \( = 1 \), with restrictions \( x \neq 2, \; x \neq -2, \; x \neq -1 \).
Q5. Simplify \( \left(\dfrac{x^2-25}{x^2+6x+5}\right) \times \left(\dfrac{x+1}{x-5}\right) \div \left(\dfrac{x+5}{x^2+x}\right) \).
Factor everything first:
\( x^2-25 = (x-5)(x+5) \)
\( x^2+6x+5 = (x+1)(x+5) \)
\( x^2+x = x(x+1) \)
The expression becomes:
\[ \frac{(x-5)(x+5)}{(x+1)(x+5)} \times \frac{x+1}{x-5} \div \frac{x+5}{x(x+1)} \]
Simplify the first two fractions (cancel \( (x+5), (x+1), (x-5) \)): gives \( 1 \).
Then: \( 1 \div \frac{x+5}{x(x+1)} = 1 \times \frac{x(x+1)}{x+5} = \frac{x(x+1)}{x+5} \)
Restrictions: \( x \neq -5, -1, 0, 5 \).
\( x^2-25 = (x-5)(x+5) \)
\( x^2+6x+5 = (x+1)(x+5) \)
\( x^2+x = x(x+1) \)
The expression becomes:
\[ \frac{(x-5)(x+5)}{(x+1)(x+5)} \times \frac{x+1}{x-5} \div \frac{x+5}{x(x+1)} \]
Simplify the first two fractions (cancel \( (x+5), (x+1), (x-5) \)): gives \( 1 \).
Then: \( 1 \div \frac{x+5}{x(x+1)} = 1 \times \frac{x(x+1)}{x+5} = \frac{x(x+1)}{x+5} \)
Restrictions: \( x \neq -5, -1, 0, 5 \).
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Addition and Subtraction of Rational Expressions
Adding and subtracting rational expressions is similar to adding and subtracting regular fractions. The key rule is: you need a common denominator.
Case 1: Same Denominator
\[
\frac{A}{C} + \frac{B}{C} = \frac{A+B}{C}, \qquad \frac{A}{C} – \frac{B}{C} = \frac{A-B}{C}
\]
Example 8
Add: \( \dfrac{3}{x+2} + \dfrac{5}{x+2} \).
Solution:
\[
\frac{3+5}{x+2} = \frac{8}{x+2}, \quad x \neq -2
\]
Simple, right? Just add the numerators and keep the denominator.Case 2: Different Denominators
When denominators are different, we must find the Least Common Denominator (LCD). The LCD is the LCM of the denominators.
Example 9
Add: \( \dfrac{2}{x-1} + \dfrac{3}{x+2} \).
Solution:
Step 1: Find the LCD. The denominators are \( x-1 \) and \( x+2 \). They share no common factor, so:
Step 1: Find the LCD. The denominators are \( x-1 \) and \( x+2 \). They share no common factor, so:
\[ \text{LCD} = (x-1)(x+2) \]
Step 2: Rewrite each fraction with the LCD.\[
\frac{2(x+2)}{(x-1)(x+2)} + \frac{3(x-1)}{(x-1)(x+2)}
\]
Step 3: Add the numerators.\[
= \frac{2(x+2)+3(x-1)}{(x-1)(x+2)} = \frac{2x+4+3x-3}{(x-1)(x+2)} = \frac{5x+1}{(x-1)(x+2)}
\]
Restrictions: \( x \neq 1, \; x \neq -2 \).Example 10
Subtract: \( \dfrac{5}{x^2-4} – \dfrac{2}{x+2} \).
Solution:
Factor: \( x^2-4 = (x-2)(x+2) \).
The LCD is \( (x-2)(x+2) \).
Factor: \( x^2-4 = (x-2)(x+2) \).
The LCD is \( (x-2)(x+2) \).
\[
\frac{5}{(x-2)(x+2)} – \frac{2(x-2)}{(x-2)(x+2)}
\]
\[
= \frac{5 – 2(x-2)}{(x-2)(x+2)} = \frac{5-2x+4}{(x-2)(x+2)} = \frac{9-2x}{(x-2)(x+2)}
\]
Restrictions: \( x \neq 2, \; x \neq -2 \).
Practice 4: Add \( \dfrac{4}{x-3} + \dfrac{1}{x+1} \).
LCD \( = (x-3)(x+1) \)
\[ \frac{4(x+1)}{(x-3)(x+1)} + \frac{1(x-3)}{(x-3)(x+1)} \]
\[ = \frac{4x+4+x-3}{(x-3)(x+1)} = \frac{5x+1}{(x-3)(x+1)} \]
Restrictions: \( x \neq 3, \; x \neq -1 \).
\[ \frac{4(x+1)}{(x-3)(x+1)} + \frac{1(x-3)}{(x-3)(x+1)} \]
\[ = \frac{4x+4+x-3}{(x-3)(x+1)} = \frac{5x+1}{(x-3)(x+1)} \]
Restrictions: \( x \neq 3, \; x \neq -1 \).
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Complex Fractions
A complex fraction is a fraction where the numerator and/or the denominator itself contains a fraction. For example:
\[
\frac{\dfrac{1}{x} + \dfrac{1}{y}}{\dfrac{1}{x} – \dfrac{1}{y}}
\]
There are two main methods to simplify complex fractions:
Method 1: Simplify the numerator and denominator separately, then divide.
Method 2: Multiply the numerator and denominator of the big fraction by the LCD of all small fractions.
Example 11
Simplify: \( \dfrac{\dfrac{1}{x} + \dfrac{1}{y}}{\dfrac{1}{x} – \dfrac{1}{y}} \).
Solution (Method 2 — usually faster):
The small fractions have denominators \( x \) and \( y \), so LCD \( = xy \).
Multiply both the top and bottom of the main fraction by \( xy \):
The small fractions have denominators \( x \) and \( y \), so LCD \( = xy \).
Multiply both the top and bottom of the main fraction by \( xy \):
\[
\frac{xy\left(\dfrac{1}{x}+\dfrac{1}{y}\right)}{xy\left(\dfrac{1}{x}-\dfrac{1}{y}\right)} = \frac{y+x}{y-x} = \frac{x+y}{y-x}
\]
Example 12
Simplify: \( \dfrac{\dfrac{2}{x+1}+1}{\dfrac{3}{x+1}-2} \).
Solution:
LCD of the small fractions is \( x+1 \). Multiply top and bottom by \( x+1 \):
LCD of the small fractions is \( x+1 \). Multiply top and bottom by \( x+1 \):
\[
\frac{(x+1)\left(\dfrac{2}{x+1}+1\right)}{(x+1)\left(\dfrac{3}{x+1}-2\right)} = \frac{2+(x+1)}{3-2(x+1)} = \frac{x+3}{3-2x-2} = \frac{x+3}{1-2x}
\]
Restrictions: \( x \neq -1 \) and \( 1-2x \neq 0 \Rightarrow x \neq \frac{1}{2} \).Exam-Style Questions — Addition, Subtraction, Complex Fractions
Q6. Simplify \( \dfrac{x}{x^2-9} + \dfrac{2}{x+3} \).
Factor: \( x^2-9 = (x-3)(x+3) \). LCD \( = (x-3)(x+3) \).
\[ \frac{x}{(x-3)(x+3)} + \frac{2(x-3)}{(x-3)(x+3)} \]
\[ = \frac{x+2x-6}{(x-3)(x+3)} = \frac{3x-6}{(x-3)(x+3)} = \frac{3(x-2)}{(x-3)(x+3)} \]
Restrictions: \( x \neq 3, \; x \neq -3 \).
\[ \frac{x}{(x-3)(x+3)} + \frac{2(x-3)}{(x-3)(x+3)} \]
\[ = \frac{x+2x-6}{(x-3)(x+3)} = \frac{3x-6}{(x-3)(x+3)} = \frac{3(x-2)}{(x-3)(x+3)} \]
Restrictions: \( x \neq 3, \; x \neq -3 \).
Q7. Simplify \( \dfrac{\dfrac{x}{x-2}-\dfrac{2}{x+2}}{\dfrac{x^2}{x^2-4}-1} \).
Factor: \( x^2-4 = (x-2)(x+2) \). LCD of all small fractions \( = (x-2)(x+2) \).
Numerator: \( \frac{x(x+2)-2(x-2)}{(x-2)(x+2)} = \frac{x^2+2x-2x+4}{(x-2)(x+2)} = \frac{x^2+4}{(x-2)(x+2)} \)
Denominator: \( \frac{x^2-(x^2-4)}{(x-2)(x+2)} = \frac{4}{(x-2)(x+2)} \)
Now divide: \( \frac{x^2+4}{(x-2)(x+2)} \div \frac{4}{(x-2)(x+2)} = \frac{x^2+4}{4} \)
Restrictions: \( x \neq 2, \; x \neq -2 \).
Numerator: \( \frac{x(x+2)-2(x-2)}{(x-2)(x+2)} = \frac{x^2+2x-2x+4}{(x-2)(x+2)} = \frac{x^2+4}{(x-2)(x+2)} \)
Denominator: \( \frac{x^2-(x^2-4)}{(x-2)(x+2)} = \frac{4}{(x-2)(x+2)} \)
Now divide: \( \frac{x^2+4}{(x-2)(x+2)} \div \frac{4}{(x-2)(x+2)} = \frac{x^2+4}{4} \)
Restrictions: \( x \neq 2, \; x \neq -2 \).
Q8. Simplify \( \dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x^2}} \).
LCD of small fractions \( = x^2 \). Multiply top and bottom by \( x^2 \):
\[ \frac{x^2\left(1+\dfrac{1}{x}\right)}{x^2\left(1-\dfrac{1}{x^2}\right)} = \frac{x^2+x}{x^2-1} \]
Factor: \( \frac{x(x+1)}{(x-1)(x+1)} = \frac{x}{x-1} \), \( x \neq 0, \; x \neq 1, \; x \neq -1 \).
\[ \frac{x^2\left(1+\dfrac{1}{x}\right)}{x^2\left(1-\dfrac{1}{x^2}\right)} = \frac{x^2+x}{x^2-1} \]
Factor: \( \frac{x(x+1)}{(x-1)(x+1)} = \frac{x}{x-1} \), \( x \neq 0, \; x \neq 1, \; x \neq -1 \).
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Rational Functions
Now let’s move to the second major part of this unit. A rational function is a function of the form:
\[
f(x) = \frac{P(x)}{Q(x)}
\]
where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) \neq 0 \).
So, a rational expression becomes a rational function when we treat it as a function of \( x \). Everything we learned about simplifying, adding, and operations on rational expressions applies to rational functions too.
Domain of a Rational Function
The domain of a rational function is all real numbers except the values that make the denominator zero.
Example 13
Find the domain of \( f(x) = \dfrac{3x+1}{x^2-4x+3} \).
Solution:
Factor the denominator: \( x^2-4x+3 = (x-1)(x-3) \).
Set to zero: \( x=1 \) or \( x=3 \).
Factor the denominator: \( x^2-4x+3 = (x-1)(x-3) \).
Set to zero: \( x=1 \) or \( x=3 \).
\[
\text{Domain} = \{x \in \mathbb{R} \mid x \neq 1, \; x \neq 3\}
\]
In interval notation: \( (-\infty, 1) \cup (1, 3) \cup (3, \infty) \).
Practice 5: Find the domain of \( f(x) = \dfrac{x+5}{x^2+1} \).
The denominator \( x^2+1 = 0 \) has no real solutions (since \( x^2 \geq 0 \), so \( x^2+1 \geq 1 > 0 \)).
Therefore, the domain is all real numbers: \( (-\infty, \infty) \).
This is a special case — not every rational function has a restricted domain!
Therefore, the domain is all real numbers: \( (-\infty, \infty) \).
This is a special case — not every rational function has a restricted domain!
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Graphing Rational Functions — Vertical Asymptotes
A vertical asymptote is a vertical line \( x = a \) that the graph of the function approaches but never touches or crosses. Vertical asymptotes occur at values where the denominator is zero and the factor does not cancel with the numerator.
Key Rule:
If a factor in the denominator does NOT cancel with the numerator → Vertical Asymptote at that value.
If a factor in the denominator DOES cancel with the numerator → Hole (removable discontinuity) at that value.
If a factor in the denominator does NOT cancel with the numerator → Vertical Asymptote at that value.
If a factor in the denominator DOES cancel with the numerator → Hole (removable discontinuity) at that value.
Example 14
Find the vertical asymptotes of \( f(x) = \dfrac{2x}{x^2-9} \).
Solution:
Factor denominator: \( x^2-9 = (x-3)(x+3) \).
Neither \( (x-3) \) nor \( (x+3) \) cancels with the numerator \( 2x \).
Factor denominator: \( x^2-9 = (x-3)(x+3) \).
Neither \( (x-3) \) nor \( (x+3) \) cancels with the numerator \( 2x \).
\[ \text{Vertical asymptotes: } x = 3 \text{ and } x = -3 \]
Example 15
Find the vertical asymptotes and holes of \( f(x) = \dfrac{x^2-4}{x^2-4x+4} \).
Solution:
Numerator: \( x^2-4 = (x-2)(x+2) \)
Denominator: \( x^2-4x+4 = (x-2)(x-2) \)
The factor \( (x-2) \) cancels (one copy), but one copy of \( (x-2) \) remains in the denominator.
Hole: There is no separate hole here because the remaining factor still gives a vertical asymptote at \( x=2 \). The canceled factor simply means the function equals \( \frac{x+2}{x-2} \) everywhere except at \( x=2 \).
Numerator: \( x^2-4 = (x-2)(x+2) \)
Denominator: \( x^2-4x+4 = (x-2)(x-2) \)
The factor \( (x-2) \) cancels (one copy), but one copy of \( (x-2) \) remains in the denominator.
\[ f(x) = \frac{x+2}{x-2}, \quad x \neq 2 \]
Vertical asymptote: \( x = 2 \) (the remaining factor in the denominator).Hole: There is no separate hole here because the remaining factor still gives a vertical asymptote at \( x=2 \). The canceled factor simply means the function equals \( \frac{x+2}{x-2} \) everywhere except at \( x=2 \).
Example 16
Find vertical asymptotes and holes of \( f(x) = \dfrac{(x-1)(x+3)}{(x-1)(x+5)} \).
Solution:
\( (x-1) \) cancels → Hole at \( x = 1 \)
\( (x+5) \) does not cancel → Vertical asymptote at \( x = -5 \)
The simplified function is \( f(x) = \dfrac{x+3}{x+5} \), with a hole at \( x=1 \).
To find the \( y \)-coordinate of the hole, substitute \( x=1 \) into the simplified function: \( y = \dfrac{1+3}{1+5} = \dfrac{4}{6} = \dfrac{2}{3} \).
Hole at \( \left(1, \dfrac{2}{3}\right) \).
\( (x-1) \) cancels → Hole at \( x = 1 \)
\( (x+5) \) does not cancel → Vertical asymptote at \( x = -5 \)
The simplified function is \( f(x) = \dfrac{x+3}{x+5} \), with a hole at \( x=1 \).
To find the \( y \)-coordinate of the hole, substitute \( x=1 \) into the simplified function: \( y = \dfrac{1+3}{1+5} = \dfrac{4}{6} = \dfrac{2}{3} \).
Hole at \( \left(1, \dfrac{2}{3}\right) \).
Key Exam Note: To find the \( y \)-coordinate of a hole, substitute the \( x \)-value into the simplified function (after canceling), NOT the original function (which would be undefined there).
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Horizontal Asymptotes
A horizontal asymptote is a horizontal line \( y = L \) that the graph approaches as \( x \to \infty \) or \( x \to -\infty \).
Rules for Horizontal Asymptotes (for \( f(x) = \dfrac{P(x)}{Q(x)} \)):
1. If degree of \( P \) < degree of \( Q \): Horizontal asymptote at \( y = 0 \) (the \( x \)-axis).
2. If degree of \( P \) = degree of \( Q \): Horizontal asymptote at \( y = \dfrac{\text{leading coefficient of } P}{\text{leading coefficient of } Q} \).
3. If degree of \( P \) > degree of \( Q \): No horizontal asymptote (there may be a slant/oblique asymptote instead).
1. If degree of \( P \) < degree of \( Q \): Horizontal asymptote at \( y = 0 \) (the \( x \)-axis).
2. If degree of \( P \) = degree of \( Q \): Horizontal asymptote at \( y = \dfrac{\text{leading coefficient of } P}{\text{leading coefficient of } Q} \).
3. If degree of \( P \) > degree of \( Q \): No horizontal asymptote (there may be a slant/oblique asymptote instead).
Example 17
Find the horizontal asymptote of \( f(x) = \dfrac{3x+2}{5x^2-1} \).
Solution: Degree of numerator \( = 1 \), degree of denominator \( = 2 \).
Since \( 1 < 2 \), the horizontal asymptote is \( y = 0 \).
Since \( 1 < 2 \), the horizontal asymptote is \( y = 0 \).
Example 18
Find the horizontal asymptote of \( f(x) = \dfrac{4x^2-3x+1}{2x^2+5x-7} \).
Solution: Both numerator and denominator have degree 2.
\[ y = \frac{4}{2} = 2 \]
The horizontal asymptote is \( y = 2 \).Example 19
Find the horizontal asymptote of \( f(x) = \dfrac{x^3+1}{x^2-4} \).
Solution: Degree of numerator \( = 3 \), degree of denominator \( = 2 \).
Since \( 3 > 2 \), there is no horizontal asymptote.
Since \( 3 > 2 \), there is no horizontal asymptote.
Practice 6: Find the horizontal asymptote of \( f(x) = \dfrac{5x-3}{7x+2} \).
Both have degree 1. Leading coefficient ratio: \( \frac{5}{7} \).
Horizontal asymptote: \( y = \frac{5}{7} \).
Horizontal asymptote: \( y = \frac{5}{7} \).
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Oblique (Slant) Asymptotes
When the degree of the numerator is exactly one more than the degree of the denominator, there is an oblique (slant) asymptote. We find it by performing polynomial long division.
Example 20
Find the oblique asymptote of \( f(x) = \dfrac{x^2+3x+1}{x-2} \).
Solution:
Perform long division of \( x^2+3x+1 \) by \( x-2 \):
As \( x \to \pm\infty \), the remainder term \( \frac{11}{x-2} \to 0 \).
Oblique asymptote: \( y = x + 5 \)
Perform long division of \( x^2+3x+1 \) by \( x-2 \):
x + 5
_________
x-2 | x^2 + 3x + 1
x^2 – 2x
———
5x + 1
5x – 10
——–
11
So, \( f(x) = x + 5 + \dfrac{11}{x-2} \).As \( x \to \pm\infty \), the remainder term \( \frac{11}{x-2} \to 0 \).
Oblique asymptote: \( y = x + 5 \)
Key Exam Note: An oblique asymptote exists only when degree(numerator) = degree(denominator) + 1. If the degree difference is 2 or more, there is no oblique asymptote (the end behavior follows a polynomial curve instead).
Putting It All Together — Full Analysis of a Rational Function
Example 21
Fully analyze \( f(x) = \dfrac{x^2-1}{x^2-4x+3} \): find domain, intercepts, asymptotes, and holes.
Solution:
Step 1: Factor
Numerator: \( x^2-1 = (x-1)(x+1) \)
Denominator: \( x^2-4x+3 = (x-1)(x-3) \)
Step 2: Simplify
\( f(x) = \dfrac{(x-1)(x+1)}{(x-1)(x-3)} = \dfrac{x+1}{x-3} \), \( x \neq 1, x \neq 3 \)
Step 3: Domain
\( (-\infty, 1) \cup (1, 3) \cup (3, \infty) \)
Step 4: Hole
\( x=1 \) canceled → Hole at \( x=1 \).
\( y \)-coordinate: \( \dfrac{1+1}{1-3} = \dfrac{2}{-2} = -1 \).
Hole at \( (1, -1) \).
Step 5: Vertical Asymptote
\( x-3 \) remains → \( x = 3 \)
Step 6: Horizontal Asymptote
Both numerator and denominator have degree 1. Ratio of leading coefficients: \( \frac{1}{1} = 1 \).
Horizontal asymptote: \( y = 1 \)
Step 7: Intercepts
\( x \)-intercept: Set numerator of simplified form to zero: \( x+1=0 \Rightarrow x=-1 \). Point: \( (-1, 0) \).
\( y \)-intercept: \( f(0) = \dfrac{0+1}{0-3} = -\dfrac{1}{3} \). Point: \( \left(0, -\dfrac{1}{3}\right) \).
Step 1: Factor
Numerator: \( x^2-1 = (x-1)(x+1) \)
Denominator: \( x^2-4x+3 = (x-1)(x-3) \)
Step 2: Simplify
\( f(x) = \dfrac{(x-1)(x+1)}{(x-1)(x-3)} = \dfrac{x+1}{x-3} \), \( x \neq 1, x \neq 3 \)
Step 3: Domain
\( (-\infty, 1) \cup (1, 3) \cup (3, \infty) \)
Step 4: Hole
\( x=1 \) canceled → Hole at \( x=1 \).
\( y \)-coordinate: \( \dfrac{1+1}{1-3} = \dfrac{2}{-2} = -1 \).
Hole at \( (1, -1) \).
Step 5: Vertical Asymptote
\( x-3 \) remains → \( x = 3 \)
Step 6: Horizontal Asymptote
Both numerator and denominator have degree 1. Ratio of leading coefficients: \( \frac{1}{1} = 1 \).
Horizontal asymptote: \( y = 1 \)
Step 7: Intercepts
\( x \)-intercept: Set numerator of simplified form to zero: \( x+1=0 \Rightarrow x=-1 \). Point: \( (-1, 0) \).
\( y \)-intercept: \( f(0) = \dfrac{0+1}{0-3} = -\dfrac{1}{3} \). Point: \( \left(0, -\dfrac{1}{3}\right) \).
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Exam-Style Questions — Rational Functions, Asymptotes, Holes
Q9. For \( f(x) = \dfrac{x^2-x-6}{x^2-9} \), find: (a) domain, (b) vertical asymptote(s), (c) horizontal asymptote, (d) hole(s) with coordinates.
Factor: numerator \( = (x-3)(x+2) \), denominator \( = (x-3)(x+3) \).
Simplified: \( f(x) = \dfrac{x+2}{x+3} \), \( x \neq 3, x \neq -3 \).
(a) Domain: \( (-\infty,-3) \cup (-3,3) \cup (3,\infty) \)
(b) Vertical asymptote: \( x = -3 \) (remaining factor)
(c) Horizontal asymptote: \( y = \frac{1}{1} = 1 \) (equal degrees)
(d) Hole at \( x=3 \): \( y = \frac{3+2}{3+3} = \frac{5}{6} \). Hole at \( \left(3, \frac{5}{6}\right) \).
Simplified: \( f(x) = \dfrac{x+2}{x+3} \), \( x \neq 3, x \neq -3 \).
(a) Domain: \( (-\infty,-3) \cup (-3,3) \cup (3,\infty) \)
(b) Vertical asymptote: \( x = -3 \) (remaining factor)
(c) Horizontal asymptote: \( y = \frac{1}{1} = 1 \) (equal degrees)
(d) Hole at \( x=3 \): \( y = \frac{3+2}{3+3} = \frac{5}{6} \). Hole at \( \left(3, \frac{5}{6}\right) \).
Q10. Find the oblique asymptote of \( f(x) = \dfrac{2x^2+x-3}{x+2} \).
Divide \( 2x^2+x-3 \) by \( x+2 \):
\( 2x^2 \div x = 2x \). Multiply: \( 2x(x+2) = 2x^2+4x \). Subtract: \( -3x-3 \).
\( -3x \div x = -3 \). Multiply: \( -3(x+2) = -3x-6 \). Subtract: \( 3 \).
Result: \( f(x) = 2x – 3 + \dfrac{3}{x+2} \).
Oblique asymptote: \( y = 2x – 3 \).
\( 2x^2 \div x = 2x \). Multiply: \( 2x(x+2) = 2x^2+4x \). Subtract: \( -3x-3 \).
\( -3x \div x = -3 \). Multiply: \( -3(x+2) = -3x-6 \). Subtract: \( 3 \).
Result: \( f(x) = 2x – 3 + \dfrac{3}{x+2} \).
Oblique asymptote: \( y = 2x – 3 \).
Q11. A rational function has a vertical asymptote at \( x=4 \), a horizontal asymptote at \( y=0 \), and no holes. Write a possible equation for such a function.
Since \( y=0 \) is the horizontal asymptote, degree of numerator must be less than degree of denominator.
Since vertical asymptote is at \( x=4 \), denominator must have factor \( (x-4) \).
A simple answer: \( f(x) = \dfrac{1}{x-4} \)
Another valid answer: \( f(x) = \dfrac{x+1}{(x-4)(x+2)} \) or any similar form.
Since vertical asymptote is at \( x=4 \), denominator must have factor \( (x-4) \).
A simple answer: \( f(x) = \dfrac{1}{x-4} \)
Another valid answer: \( f(x) = \dfrac{x+1}{(x-4)(x+2)} \) or any similar form.
End Behavior of Rational Functions
The end behavior of a rational function describes what happens to \( f(x) \) as \( x \to \infty \) and as \( x \to -\infty \). The horizontal (or oblique) asymptote tells us the end behavior.
Example 22
Describe the end behavior of \( f(x) = \dfrac{3x^2-2x+1}{6x^2+5} \).
Solution:
Both degrees are 2, so horizontal asymptote: \( y = \dfrac{3}{6} = \dfrac{1}{2} \).
As \( x \to \infty \), \( f(x) \to \dfrac{1}{2} \) from above or below.
As \( x \to -\infty \), \( f(x) \to \dfrac{1}{2} \) from above or below.
We write: \( f(x) \to \frac{1}{2} \) as \( x \to \pm\infty \).
Both degrees are 2, so horizontal asymptote: \( y = \dfrac{3}{6} = \dfrac{1}{2} \).
As \( x \to \infty \), \( f(x) \to \dfrac{1}{2} \) from above or below.
As \( x \to -\infty \), \( f(x) \to \dfrac{1}{2} \) from above or below.
We write: \( f(x) \to \frac{1}{2} \) as \( x \to \pm\infty \).
Solving Equations Involving Rational Expressions
To solve equations with rational expressions, we multiply both sides by the LCD to eliminate all denominators, then solve the resulting polynomial equation. But remember: we must check all solutions against the restricted values!
Example 23
Solve: \( \dfrac{3}{x-2} = \dfrac{5}{x+1} \).
Solution:
LCD \( = (x-2)(x+1) \). Multiply both sides:
Solution: \( x = \dfrac{13}{2} \)
LCD \( = (x-2)(x+1) \). Multiply both sides:
\[
3(x+1) = 5(x-2)
\]
\[
3x+3 = 5x-10
\]
\[
13 = 2x \implies x = \frac{13}{2}
\]
Check: \( x = \frac{13}{2} \) does not make any denominator zero. ✓Solution: \( x = \dfrac{13}{2} \)
Example 24
Solve: \( \dfrac{x}{x-3} + \dfrac{1}{x+3} = \dfrac{18}{x^2-9} \).
Solution:
LCD \( = (x-3)(x+3) = x^2-9 \). Multiply both sides by LCD:
\( x = -7 \) is valid. ✓
Solution: \( x = -7 \)
LCD \( = (x-3)(x+3) = x^2-9 \). Multiply both sides by LCD:
\[
x(x+3) + 1(x-3) = 18
\]
\[
x^2+3x+x-3 = 18
\]
\[
x^2+4x-21 = 0
\]
\[
(x+7)(x-3) = 0 \implies x = -7 \text{ or } x = 3
\]
Check: \( x = 3 \) makes the original denominators zero → extraneous solution, reject!\( x = -7 \) is valid. ✓
Solution: \( x = -7 \)
Key Exam Note: Always check your solutions by substituting back into the ORIGINAL equation. Solutions that make any denominator zero are called extraneous solutions and must be rejected. This is a very common exam trap!
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Exam-Style Questions — Solving Rational Equations
Q12. Solve \( \dfrac{2x}{x-1} = \dfrac{x+3}{x-1} + 1 \).
LCD \( = x-1 \). Multiply both sides by \( x-1 \):
\( 2x = (x+3) + (x-1) \)
\( 2x = 2x + 2 \)
\( 0 = 2 \) — Contradiction!
No solution. (Note: \( x \neq 1 \) is a restriction, but the equation has no valid solution regardless.)
\( 2x = (x+3) + (x-1) \)
\( 2x = 2x + 2 \)
\( 0 = 2 \) — Contradiction!
No solution. (Note: \( x \neq 1 \) is a restriction, but the equation has no valid solution regardless.)
Q13. Solve \( \dfrac{x+2}{x-4} – \dfrac{3}{x+1} = \dfrac{4x-5}{x^2-3x-4} \).
Factor: \( x^2-3x-4 = (x-4)(x+1) \). LCD \( = (x-4)(x+1) \).
Multiply both sides by LCD:
\( (x+2)(x+1) – 3(x-4) = 4x-5 \)
\( x^2+x+2x+2 – 3x+12 = 4x-5 \)
\( x^2 + 14 = 4x – 5 \)
\( x^2 – 4x + 19 = 0 \)
Discriminant: \( \Delta = 16-76 = -60 < 0 \)
No real solution.
Multiply both sides by LCD:
\( (x+2)(x+1) – 3(x-4) = 4x-5 \)
\( x^2+x+2x+2 – 3x+12 = 4x-5 \)
\( x^2 + 14 = 4x – 5 \)
\( x^2 – 4x + 19 = 0 \)
Discriminant: \( \Delta = 16-76 = -60 < 0 \)
No real solution.
Q14. Solve \( \dfrac{1}{x-2} + \dfrac{1}{x+2} = \dfrac{4}{x^2-4} \).
LCD \( = (x-2)(x+2) \). Multiply both sides:
\( (x+2) + (x-2) = 4 \)
\( 2x = 4 \)
\( x = 2 \)
But \( x=2 \) makes the original denominator zero → extraneous!
No solution.
\( (x+2) + (x-2) = 4 \)
\( 2x = 4 \)
\( x = 2 \)
But \( x=2 \) makes the original denominator zero → extraneous!
No solution.
Solving Rational Inequalities
Solving rational inequalities is similar to solving rational equations, but with one important difference: when you multiply or divide by a negative number, you must flip the inequality sign. For rational inequalities, the safest method is the sign analysis (test point) method.
Steps for Solving Rational Inequalities
1. Move everything to one side so the inequality is in the form \( \frac{P(x)}{Q(x)} > 0 \) (or \( \geq 0 \), \( < 0 \), \( \leq 0 \)).
2. Factor numerator and denominator completely.
3. Find all critical values (zeros of numerator and denominator).
4. Use a sign chart to test intervals.
5. Select intervals that satisfy the inequality.
6. Check endpoints: include zeros of numerator for \( \geq \) or \( \leq \); never include zeros of denominator.
2. Factor numerator and denominator completely.
3. Find all critical values (zeros of numerator and denominator).
4. Use a sign chart to test intervals.
5. Select intervals that satisfy the inequality.
6. Check endpoints: include zeros of numerator for \( \geq \) or \( \leq \); never include zeros of denominator.
Example 25
Solve: \( \dfrac{x-1}{x+3} > 0 \).
Solution:
Critical values: Numerator zero at \( x=1 \). Denominator zero at \( x=-3 \).
Intervals: \( (-\infty,-3) \), \( (-3,1) \), \( (1,\infty) \).
Test each interval:
– For \( x=-4 \): \( \frac{-4-1}{-4+3} = \frac{-5}{-1} = 5 > 0 \) ✓
– For \( x=0 \): \( \frac{0-1}{0+3} = \frac{-1}{3} < 0 \) ✗
– For \( x=2 \): \( \frac{2-1}{2+3} = \frac{1}{5} > 0 \) ✓
The inequality is strict (\(>\)), so we don’t include \( x=1 \), and we never include \( x=-3 \).
Solution: \( x \in (-\infty,-3) \cup (1,\infty) \)
Critical values: Numerator zero at \( x=1 \). Denominator zero at \( x=-3 \).
Intervals: \( (-\infty,-3) \), \( (-3,1) \), \( (1,\infty) \).
Test each interval:
– For \( x=-4 \): \( \frac{-4-1}{-4+3} = \frac{-5}{-1} = 5 > 0 \) ✓
– For \( x=0 \): \( \frac{0-1}{0+3} = \frac{-1}{3} < 0 \) ✗
– For \( x=2 \): \( \frac{2-1}{2+3} = \frac{1}{5} > 0 \) ✓
The inequality is strict (\(>\)), so we don’t include \( x=1 \), and we never include \( x=-3 \).
Solution: \( x \in (-\infty,-3) \cup (1,\infty) \)
Example 26
Solve: \( \dfrac{x^2-4}{x^2-1} \leq 0 \).
Solution:
Factor: \( \dfrac{(x-2)(x+2)}{(x-1)(x+1)} \leq 0 \)
Critical values: \( x=-2, -1, 1, 2 \)
Intervals: \( (-\infty,-2) \), \( (-2,-1) \), \( (-1,1) \), \( (1,2) \), \( (2,\infty) \)
Test each:
– \( x=-3 \): \( \frac{(+)(-)}{(-)(-)} = \frac{-}{+} < 0 \) ✓
– \( x=-1.5 \): \( \frac{(-)(+)}{(-)(+)} = \frac{-}{+} < 0 \) ✓
– \( x=0 \): \( \frac{(-)(+)}{(-)(+)} = \frac{-}{+} < 0 \) ✓
– \( x=1.5 \): \( \frac{(-)(+)}{(+)(+)} = \frac{-}{+} < 0 \) ✓
– \( x=3 \): \( \frac{(+)(+)}{(+)(+)} > 0 \) ✗
Include \( x=-2 \) and \( x=2 \) (zeros of numerator, since \( \leq \)).
Never include \( x=-1 \) and \( x=1 \) (zeros of denominator).
Solution: \( x \in (-\infty,-2] \cup (-1,1) \cup [2,\infty) \)
Factor: \( \dfrac{(x-2)(x+2)}{(x-1)(x+1)} \leq 0 \)
Critical values: \( x=-2, -1, 1, 2 \)
Intervals: \( (-\infty,-2) \), \( (-2,-1) \), \( (-1,1) \), \( (1,2) \), \( (2,\infty) \)
Test each:
– \( x=-3 \): \( \frac{(+)(-)}{(-)(-)} = \frac{-}{+} < 0 \) ✓
– \( x=-1.5 \): \( \frac{(-)(+)}{(-)(+)} = \frac{-}{+} < 0 \) ✓
– \( x=0 \): \( \frac{(-)(+)}{(-)(+)} = \frac{-}{+} < 0 \) ✓
– \( x=1.5 \): \( \frac{(-)(+)}{(+)(+)} = \frac{-}{+} < 0 \) ✓
– \( x=3 \): \( \frac{(+)(+)}{(+)(+)} > 0 \) ✗
Include \( x=-2 \) and \( x=2 \) (zeros of numerator, since \( \leq \)).
Never include \( x=-1 \) and \( x=1 \) (zeros of denominator).
Solution: \( x \in (-\infty,-2] \cup (-1,1) \cup [2,\infty) \)
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Exam-Style Questions — Rational Inequalities
Q15. Solve \( \dfrac{x+1}{x-2} \geq 0 \).
Critical values: \( x=-1 \) (numerator), \( x=2 \) (denominator).
Intervals: \( (-\infty,-1) \), \( (-1,2) \), \( (2,\infty) \).
Test: \( x=-2 \): \( + \) ✓; \( x=0 \): \( – \) ✗; \( x=3 \): \( + \) ✓.
Include \( x=-1 \) (numerator zero, \( \geq \)). Exclude \( x=2 \) (denominator zero).
Solution: \( [-1, 2) \)
Intervals: \( (-\infty,-1) \), \( (-1,2) \), \( (2,\infty) \).
Test: \( x=-2 \): \( + \) ✓; \( x=0 \): \( – \) ✗; \( x=3 \): \( + \) ✓.
Include \( x=-1 \) (numerator zero, \( \geq \)). Exclude \( x=2 \) (denominator zero).
Solution: \( [-1, 2) \)
Q16. Solve \( \dfrac{x}{x^2-9} < 0 \).
Factor: \( \dfrac{x}{(x-3)(x+3)} < 0 \)
Critical values: \( x=-3, 0, 3 \)
Intervals: \( (-\infty,-3) \), \( (-3,0) \), \( (0,3) \), \( (3,\infty) \)
Test: \( x=-4 \): \( \frac{-}{(-)(-)} = \frac{-}{+} < 0 \) ✓
\( x=-1 \): \( \frac{-}{(-)(+)} = \frac{-}{-} > 0 \) ✗
\( x=1 \): \( \frac{+}{(+)(-)} = \frac{+}{-} < 0 \) ✓
\( x=4 \): \( \frac{+}{(+)(+)} > 0 \) ✗
Strict inequality: exclude all critical values.
Solution: \( (-\infty,-3) \cup (0,3) \)
Critical values: \( x=-3, 0, 3 \)
Intervals: \( (-\infty,-3) \), \( (-3,0) \), \( (0,3) \), \( (3,\infty) \)
Test: \( x=-4 \): \( \frac{-}{(-)(-)} = \frac{-}{+} < 0 \) ✓
\( x=-1 \): \( \frac{-}{(-)(+)} = \frac{-}{-} > 0 \) ✗
\( x=1 \): \( \frac{+}{(+)(-)} = \frac{+}{-} < 0 \) ✓
\( x=4 \): \( \frac{+}{(+)(+)} > 0 \) ✗
Strict inequality: exclude all critical values.
Solution: \( (-\infty,-3) \cup (0,3) \)
Q17. Solve \( \dfrac{2x-1}{x+4} \leq 3 \).
Move everything to one side: \( \frac{2x-1}{x+4} – 3 \leq 0 \)
Combine: \( \frac{2x-1-3(x+4)}{x+4} \leq 0 \)
\( \frac{2x-1-3x-12}{x+4} \leq 0 \)
\( \frac{-x-13}{x+4} \leq 0 \), which is \( \frac{x+13}{x+4} \geq 0 \) (multiplying by \(-1\) flips sign).
Critical values: \( x=-13, x=-4 \).
Test: \( x=-14 \): \( \frac{-}{-} > 0 \) ✓; \( x=-10 \): \( \frac{+}{-} < 0 \) ✗; \( x=0 \): \( \frac{+}{+} > 0 \) ✓.
Include \( x=-13 \). Exclude \( x=-4 \).
Solution: \( (-\infty,-13] \cup (-4,\infty) \)
Combine: \( \frac{2x-1-3(x+4)}{x+4} \leq 0 \)
\( \frac{2x-1-3x-12}{x+4} \leq 0 \)
\( \frac{-x-13}{x+4} \leq 0 \), which is \( \frac{x+13}{x+4} \geq 0 \) (multiplying by \(-1\) flips sign).
Critical values: \( x=-13, x=-4 \).
Test: \( x=-14 \): \( \frac{-}{-} > 0 \) ✓; \( x=-10 \): \( \frac{+}{-} < 0 \) ✗; \( x=0 \): \( \frac{+}{+} > 0 \) ✓.
Include \( x=-13 \). Exclude \( x=-4 \).
Solution: \( (-\infty,-13] \cup (-4,\infty) \)
Quick Revision Notes — Rational Expressions and Rational Functions
Key Definitions
Rational Expression: A fraction \( \frac{P(x)}{Q(x)} \) where both \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) \neq 0 \).
Rational Function: A function \( f(x) = \frac{P(x)}{Q(x)} \) where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) \neq 0 \).
Restricted Values: Values of \( x \) that make the denominator zero. These are excluded from the domain.
Domain: All real numbers except the restricted values.
Vertical Asymptote: A vertical line \( x = a \) where the function approaches \( \pm\infty \). Occurs when a denominator factor does NOT cancel.
Horizontal Asymptote: A horizontal line \( y = L \) that the graph approaches as \( x \to \pm\infty \).
Oblique (Slant) Asymptote: A non-horizontal, non-vertical line that the graph approaches. Occurs when degree of numerator = degree of denominator + 1.
Hole (Removable Discontinuity): A point where the function is undefined but the limit exists. Occurs when a factor cancels from both numerator and denominator.
Extraneous Solution: A solution obtained algebraically that does not satisfy the original equation (usually because it makes a denominator zero).
Rational Function: A function \( f(x) = \frac{P(x)}{Q(x)} \) where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) \neq 0 \).
Restricted Values: Values of \( x \) that make the denominator zero. These are excluded from the domain.
Domain: All real numbers except the restricted values.
Vertical Asymptote: A vertical line \( x = a \) where the function approaches \( \pm\infty \). Occurs when a denominator factor does NOT cancel.
Horizontal Asymptote: A horizontal line \( y = L \) that the graph approaches as \( x \to \pm\infty \).
Oblique (Slant) Asymptote: A non-horizontal, non-vertical line that the graph approaches. Occurs when degree of numerator = degree of denominator + 1.
Hole (Removable Discontinuity): A point where the function is undefined but the limit exists. Occurs when a factor cancels from both numerator and denominator.
Extraneous Solution: A solution obtained algebraically that does not satisfy the original equation (usually because it makes a denominator zero).
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Key Formulas and Rules
Operations on Rational Expressions:
Multiplication: \( \frac{A}{B} \times \frac{C}{D} = \frac{AC}{BD} \)
Division: \( \frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C} = \frac{AD}{BC} \)
Addition (same denominator): \( \frac{A}{C} + \frac{B}{C} = \frac{A+B}{C} \)
Subtraction (same denominator): \( \frac{A}{C} – \frac{B}{C} = \frac{A-B}{C} \)
Addition/Subtraction (different denominators): Find LCD, rewrite, then add/subtract numerators.
Multiplication: \( \frac{A}{B} \times \frac{C}{D} = \frac{AC}{BD} \)
Division: \( \frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C} = \frac{AD}{BC} \)
Addition (same denominator): \( \frac{A}{C} + \frac{B}{C} = \frac{A+B}{C} \)
Subtraction (same denominator): \( \frac{A}{C} – \frac{B}{C} = \frac{A-B}{C} \)
Addition/Subtraction (different denominators): Find LCD, rewrite, then add/subtract numerators.
Horizontal Asymptote Rules:
Let \( n = \) degree of numerator, \( m = \) degree of denominator.
• If \( n < m \): HA at \( y = 0 \)
• If \( n = m \): HA at \( y = \frac{a_n}{b_m} \) (ratio of leading coefficients)
• If \( n > m \): No horizontal asymptote
• If \( n = m + 1 \): Oblique asymptote (find by long division)
Let \( n = \) degree of numerator, \( m = \) degree of denominator.
• If \( n < m \): HA at \( y = 0 \)
• If \( n = m \): HA at \( y = \frac{a_n}{b_m} \) (ratio of leading coefficients)
• If \( n > m \): No horizontal asymptote
• If \( n = m + 1 \): Oblique asymptote (find by long division)
Complex Fractions — Shortcut Method:
Multiply the numerator and denominator of the main fraction by the LCD of ALL small fractions inside.
Multiply the numerator and denominator of the main fraction by the LCD of ALL small fractions inside.
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Solving Rational Equations — Steps
1. Identify all restricted values (values making any denominator zero).
2. Find the LCD of all fractions.
3. Multiply both sides of the equation by the LCD.
4. Simplify and solve the resulting polynomial equation.
5. CHECK all solutions against restricted values. Reject extraneous solutions.
2. Find the LCD of all fractions.
3. Multiply both sides of the equation by the LCD.
4. Simplify and solve the resulting polynomial equation.
5. CHECK all solutions against restricted values. Reject extraneous solutions.
Solving Rational Inequalities — Steps
1. Move all terms to one side.
2. Factor numerator and denominator completely.
3. Find all critical values (zeros of numerator AND denominator).
4. Draw a sign chart and test one point in each interval.
5. Select intervals satisfying the inequality.
6. For \( \geq \) or \( \leq \): include zeros of numerator (but NEVER zeros of denominator).
2. Factor numerator and denominator completely.
3. Find all critical values (zeros of numerator AND denominator).
4. Draw a sign chart and test one point in each interval.
5. Select intervals satisfying the inequality.
6. For \( \geq \) or \( \leq \): include zeros of numerator (but NEVER zeros of denominator).
Common Mistakes to Avoid
Mistake 1: Canceling terms instead of factors
Wrong: \( \frac{x+3}{x-2} = \frac{3}{-2} \)
Right: Cannot cancel — \( x \) is not a factor here.
Mistake 2: Forgetting restrictions after canceling
If \( \frac{(x-2)(x+1)}{(x-2)(x-3)} = \frac{x+1}{x-3} \), you must still note \( x \neq 2 \).
Mistake 3: Not checking for extraneous solutions
Always substitute back into the ORIGINAL equation.
Mistake 4: Confusing holes with vertical asymptotes
Canceled factor → Hole. Non-canceled denominator factor → Vertical asymptote.
Mistake 5: Wrong horizontal asymptote
Remember to compare DEGREES first, then use the appropriate rule. Don’t just look at coefficients.
Mistake 6: Including denominator zeros in inequality solutions
Even with \( \leq \) or \( \geq \), never include values that make the denominator zero.
Mistake 7: Forgetting to flip the inequality sign
When multiplying or dividing an inequality by a negative number, flip the sign.
Wrong: \( \frac{x+3}{x-2} = \frac{3}{-2} \)
Right: Cannot cancel — \( x \) is not a factor here.
Mistake 2: Forgetting restrictions after canceling
If \( \frac{(x-2)(x+1)}{(x-2)(x-3)} = \frac{x+1}{x-3} \), you must still note \( x \neq 2 \).
Mistake 3: Not checking for extraneous solutions
Always substitute back into the ORIGINAL equation.
Mistake 4: Confusing holes with vertical asymptotes
Canceled factor → Hole. Non-canceled denominator factor → Vertical asymptote.
Mistake 5: Wrong horizontal asymptote
Remember to compare DEGREES first, then use the appropriate rule. Don’t just look at coefficients.
Mistake 6: Including denominator zeros in inequality solutions
Even with \( \leq \) or \( \geq \), never include values that make the denominator zero.
Mistake 7: Forgetting to flip the inequality sign
When multiplying or dividing an inequality by a negative number, flip the sign.
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Quick Examples for Recap
Simplify: \( \dfrac{x^2-25}{x^2+10x+25} = \dfrac{(x-5)(x+5)}{(x+5)(x+5)} = \dfrac{x-5}{x+5}, \; x \neq -5 \)
Multiply: \( \dfrac{x}{x+1} \times \dfrac{x+1}{x-2} = \dfrac{x}{x-2}, \; x \neq -1, 2 \)
Divide: \( \dfrac{x^2-9}{x} \div \dfrac{x-3}{x^2} = \dfrac{(x-3)(x+3)}{x} \times \dfrac{x^2}{x-3} = x(x+3), \; x \neq 0, 3 \)
Add: \( \dfrac{1}{x} + \dfrac{1}{x+1} = \dfrac{x+1+x}{x(x+1)} = \dfrac{2x+1}{x(x+1)} \)
Complex fraction: \( \dfrac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}} = \dfrac{b-a}{b+a} \) (multiply top and bottom by \( ab \))
Asymptotes of \( f(x)=\dfrac{2x}{x^2-9} \): VA: \( x=3, x=-3 \); HA: \( y=0 \)
Solve \( \dfrac{3}{x}=\dfrac{6}{x+2} \): \( 3(x+2)=6x \Rightarrow 3x+6=6x \Rightarrow x=2 \). Check: valid. ✓
Challenge Exam Questions — Rational Expressions and Rational Functions
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Multiple Choice Questions
Q1. What is the simplified form of \( \dfrac{x^3-27}{x^2-9} \)?
A) \( \dfrac{x^2+3x+9}{x+3} \)
B) \( \dfrac{x^2-3x+9}{x+3} \)
C) \( x-3 \)
D) \( \dfrac{x^2+3x+9}{x-3} \)
A) \( \dfrac{x^2+3x+9}{x+3} \)
B) \( \dfrac{x^2-3x+9}{x+3} \)
C) \( x-3 \)
D) \( \dfrac{x^2+3x+9}{x-3} \)
Answer: A
Factor numerator using difference of cubes: \( x^3-27 = (x-3)(x^2+3x+9) \)
Factor denominator: \( x^2-9 = (x-3)(x+3) \)
Cancel \( (x-3) \):
\[ \frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)} = \frac{x^2+3x+9}{x+3}, \; x \neq 3, -3 \]
Factor numerator using difference of cubes: \( x^3-27 = (x-3)(x^2+3x+9) \)
Factor denominator: \( x^2-9 = (x-3)(x+3) \)
Cancel \( (x-3) \):
\[ \frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)} = \frac{x^2+3x+9}{x+3}, \; x \neq 3, -3 \]
Q2. The rational function \( f(x) = \dfrac{2x^2+5x-3}{x^2-x-6} \) has:
A) A vertical asymptote at \( x=3 \) and a hole at \( x=-2 \)
B) A vertical asymptote at \( x=-2 \) and a hole at \( x=3 \)
C) Vertical asymptotes at \( x=3 \) and \( x=-2 \)
D) A hole at both \( x=3 \) and \( x=-2 \)
A) A vertical asymptote at \( x=3 \) and a hole at \( x=-2 \)
B) A vertical asymptote at \( x=-2 \) and a hole at \( x=3 \)
C) Vertical asymptotes at \( x=3 \) and \( x=-2 \)
D) A hole at both \( x=3 \) and \( x=-2 \)
Answer: A
Factor numerator: \( 2x^2+5x-3 = (2x-1)(x+3) \)
Factor denominator: \( x^2-x-6 = (x-3)(x+2) \)
No common factors → no holes. Both denominator factors remain.
Vertical asymptotes at \( x=3 \) and \( x=-2 \).
Correction: The correct answer is actually C. Since no factors cancel, there are vertical asymptotes at both \( x=3 \) and \( x=-2 \), and no holes. I apologize for the initial labeling — answer C is correct.
Factor numerator: \( 2x^2+5x-3 = (2x-1)(x+3) \)
Factor denominator: \( x^2-x-6 = (x-3)(x+2) \)
No common factors → no holes. Both denominator factors remain.
Vertical asymptotes at \( x=3 \) and \( x=-2 \).
Correction: The correct answer is actually C. Since no factors cancel, there are vertical asymptotes at both \( x=3 \) and \( x=-2 \), and no holes. I apologize for the initial labeling — answer C is correct.
Q3. What is the horizontal asymptote of \( f(x) = \dfrac{5x^3-2x+1}{3x^3+x^2-7} \)?
A) \( y = 0 \)
B) \( y = \dfrac{5}{3} \)
C) \( y = \dfrac{3}{5} \)
D) There is no horizontal asymptote
A) \( y = 0 \)
B) \( y = \dfrac{5}{3} \)
C) \( y = \dfrac{3}{5} \)
D) There is no horizontal asymptote
Answer: B
Both numerator and denominator have degree 3. The horizontal asymptote is the ratio of leading coefficients:
\[ y = \frac{5}{3} \]
Both numerator and denominator have degree 3. The horizontal asymptote is the ratio of leading coefficients:
\[ y = \frac{5}{3} \]
Q4. Which of the following is equivalent to \( \dfrac{\dfrac{2}{x}-\dfrac{3}{y}}{\dfrac{4}{x}+\dfrac{1}{y}} \)?
A) \( \dfrac{2y-3x}{4y+x} \)
B) \( \dfrac{2x-3y}{4x+y} \)
C) \( \dfrac{2y+3x}{4y-x} \)
D) \( \dfrac{2-3}{4+1} = -\dfrac{1}{5} \)
A) \( \dfrac{2y-3x}{4y+x} \)
B) \( \dfrac{2x-3y}{4x+y} \)
C) \( \dfrac{2y+3x}{4y-x} \)
D) \( \dfrac{2-3}{4+1} = -\dfrac{1}{5} \)
Answer: A
LCD of small fractions \( = xy \). Multiply top and bottom by \( xy \):
\[ \frac{xy\left(\frac{2}{x}-\frac{3}{y}\right)}{xy\left(\frac{4}{x}+\frac{1}{y}\right)} = \frac{2y-3x}{4y+x} \]
Option D is a common mistake (canceling terms instead of using LCD).
LCD of small fractions \( = xy \). Multiply top and bottom by \( xy \):
\[ \frac{xy\left(\frac{2}{x}-\frac{3}{y}\right)}{xy\left(\frac{4}{x}+\frac{1}{y}\right)} = \frac{2y-3x}{4y+x} \]
Option D is a common mistake (canceling terms instead of using LCD).
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Fill in the Blank
Q5. The domain of \( f(x) = \dfrac{x+7}{x^2-5x-14} \) is all real numbers except \( x = \) _______ and \( x = \) _______.
Factor denominator: \( x^2-5x-14 = (x-7)(x+2) \)
Set each factor to zero: \( x=7 \) and \( x=-2 \).
Answer: \( 7 \) and \( -2 \)
Set each factor to zero: \( x=7 \) and \( x=-2 \).
Answer: \( 7 \) and \( -2 \)
Q6. The oblique asymptote of \( f(x) = \dfrac{x^2-4x+7}{x-1} \) is \( y = \) _______.
Divide \( x^2-4x+7 \) by \( x-1 \):
\( x^2 \div x = x \). \( x(x-1) = x^2-x \). Subtract: \( -3x+7 \).
\( -3x \div x = -3 \). \( -3(x-1) = -3x+3 \). Subtract: \( 4 \).
Result: \( x – 3 + \frac{4}{x-1} \).
Answer: \( y = x – 3 \)
\( x^2 \div x = x \). \( x(x-1) = x^2-x \). Subtract: \( -3x+7 \).
\( -3x \div x = -3 \). \( -3(x-1) = -3x+3 \). Subtract: \( 4 \).
Result: \( x – 3 + \frac{4}{x-1} \).
Answer: \( y = x – 3 \)
Q7. If \( f(x) = \dfrac{(x+1)(x-4)}{(x+1)(x+5)} \), then the hole is at the point \( ( \)_____, _____\( ) \).
\( (x+1) \) cancels → hole at \( x=-1 \).
Simplified: \( f(x) = \frac{x-4}{x+5} \).
\( y \)-coordinate: \( \frac{-1-4}{-1+5} = \frac{-5}{4} \).
Answer: \( (-1, -\frac{5}{4}) \)
Simplified: \( f(x) = \frac{x-4}{x+5} \).
\( y \)-coordinate: \( \frac{-1-4}{-1+5} = \frac{-5}{4} \).
Answer: \( (-1, -\frac{5}{4}) \)
Q8. The solution to \( \dfrac{x^2-16}{x-4} = 0 \) is \( x = \) _______.
Simplify: \( \frac{(x-4)(x+4)}{x-4} = x+4 \), \( x \neq 4 \).
Set equal to zero: \( x+4=0 \Rightarrow x=-4 \).
Check: \( x=-4 \) does not violate \( x \neq 4 \). ✓
Answer: \( -4 \)
Set equal to zero: \( x+4=0 \Rightarrow x=-4 \).
Check: \( x=-4 \) does not violate \( x \neq 4 \). ✓
Answer: \( -4 \)
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Short Answer Questions
Q9. Simplify completely: \( \dfrac{x^3-8}{x^2-4} \times \dfrac{x+2}{x^2+2x+4} \).
Factor each part:
\( x^3-8 = (x-2)(x^2+2x+4) \)
\( x^2-4 = (x-2)(x+2) \)
\[ \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} \times \frac{x+2}{x^2+2x+4} \]
Cancel everything: \( (x-2) \), \( (x+2) \), \( (x^2+2x+4) \) all cancel.
Answer: \( 1 \), with restrictions \( x \neq 2, -2 \).
\( x^3-8 = (x-2)(x^2+2x+4) \)
\( x^2-4 = (x-2)(x+2) \)
\[ \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} \times \frac{x+2}{x^2+2x+4} \]
Cancel everything: \( (x-2) \), \( (x+2) \), \( (x^2+2x+4) \) all cancel.
Answer: \( 1 \), with restrictions \( x \neq 2, -2 \).
Q10. For \( f(x) = \dfrac{3x^2-x-4}{x^2+x-6} \), find all asymptotes and holes.
Factor numerator: \( 3x^2-x-4 = (3x-4)(x+1) \)
Factor denominator: \( x^2+x-6 = (x-2)(x+3) \)
No common factors → no holes.
Vertical asymptotes: \( x=2 \) and \( x=-3 \)
Horizontal asymptote: Both degree 2. \( y = \frac{3}{1} = 3 \)
No oblique asymptote (degrees are equal, not differing by 1).
Factor denominator: \( x^2+x-6 = (x-2)(x+3) \)
No common factors → no holes.
Vertical asymptotes: \( x=2 \) and \( x=-3 \)
Horizontal asymptote: Both degree 2. \( y = \frac{3}{1} = 3 \)
No oblique asymptote (degrees are equal, not differing by 1).
Q11. Simplify the complex fraction: \( \dfrac{\dfrac{2}{x^2}+\dfrac{3}{x}}{\dfrac{1}{x}-\dfrac{4}{x^2}} \).
LCD of small fractions \( = x^2 \). Multiply top and bottom by \( x^2 \):
\[ \frac{x^2\left(\frac{2}{x^2}+\frac{3}{x}\right)}{x^2\left(\frac{1}{x}-\frac{4}{x^2}\right)} = \frac{2+3x}{x-4} \]
Answer: \( \dfrac{3x+2}{x-4} \), \( x \neq 0, 4 \).
\[ \frac{x^2\left(\frac{2}{x^2}+\frac{3}{x}\right)}{x^2\left(\frac{1}{x}-\frac{4}{x^2}\right)} = \frac{2+3x}{x-4} \]
Answer: \( \dfrac{3x+2}{x-4} \), \( x \neq 0, 4 \).
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Step-by-Step Calculation Questions
Q12. Perform the operation and simplify: \( \dfrac{3x-6}{x^2+4x+3} \div \dfrac{x^2-4}{x^2+6x+9} \).
Step 1: Rewrite as multiplication by reciprocal.
\[ \frac{3x-6}{x^2+4x+3} \times \frac{x^2+6x+9}{x^2-4} \]
Step 2: Factor everything.
\( 3x-6 = 3(x-2) \)
\( x^2+4x+3 = (x+1)(x+3) \)
\( x^2+6x+9 = (x+3)(x+3) \)
\( x^2-4 = (x-2)(x+2) \)
Step 3: Cancel common factors.
\[ \frac{3(x-2)}{(x+1)(x+3)} \times \frac{(x+3)(x+3)}{(x-2)(x+2)} \]
Cancel \( (x-2) \) and one \( (x+3) \):
\[ = \frac{3(x+3)}{(x+1)(x+2)} = \frac{3x+9}{x^2+3x+2} \]
Restrictions: \( x \neq -3, -1, 2, -2 \).
\[ \frac{3x-6}{x^2+4x+3} \times \frac{x^2+6x+9}{x^2-4} \]
Step 2: Factor everything.
\( 3x-6 = 3(x-2) \)
\( x^2+4x+3 = (x+1)(x+3) \)
\( x^2+6x+9 = (x+3)(x+3) \)
\( x^2-4 = (x-2)(x+2) \)
Step 3: Cancel common factors.
\[ \frac{3(x-2)}{(x+1)(x+3)} \times \frac{(x+3)(x+3)}{(x-2)(x+2)} \]
Cancel \( (x-2) \) and one \( (x+3) \):
\[ = \frac{3(x+3)}{(x+1)(x+2)} = \frac{3x+9}{x^2+3x+2} \]
Restrictions: \( x \neq -3, -1, 2, -2 \).
Q13. Add and simplify: \( \dfrac{2x+1}{x^2+x-6} + \dfrac{x-3}{x^2-x-12} \).
Factor denominators:
\( x^2+x-6 = (x+3)(x-2) \)
\( x^2-x-12 = (x-4)(x+3) \)
LCD \( = (x+3)(x-2)(x-4) \)
\[ \frac{(2x+1)(x-4)}{(x+3)(x-2)(x-4)} + \frac{(x-3)(x-2)}{(x+3)(x-2)(x-4)} \]
Expand numerators:
\( (2x+1)(x-4) = 2x^2-8x+x-4 = 2x^2-7x-4 \)
\( (x-3)(x-2) = x^2-5x+6 \)
Add: \( 2x^2-7x-4+x^2-5x+6 = 3x^2-12x+2 \)
\[ \frac{3x^2-12x+2}{(x+3)(x-2)(x-4)} \]
Check if numerator factors: discriminant \( = 144-24 = 120 \), not a perfect square. Does not factor nicely.
Answer: \( \dfrac{3x^2-12x+2}{(x+3)(x-2)(x-4)} \)
\( x^2+x-6 = (x+3)(x-2) \)
\( x^2-x-12 = (x-4)(x+3) \)
LCD \( = (x+3)(x-2)(x-4) \)
\[ \frac{(2x+1)(x-4)}{(x+3)(x-2)(x-4)} + \frac{(x-3)(x-2)}{(x+3)(x-2)(x-4)} \]
Expand numerators:
\( (2x+1)(x-4) = 2x^2-8x+x-4 = 2x^2-7x-4 \)
\( (x-3)(x-2) = x^2-5x+6 \)
Add: \( 2x^2-7x-4+x^2-5x+6 = 3x^2-12x+2 \)
\[ \frac{3x^2-12x+2}{(x+3)(x-2)(x-4)} \]
Check if numerator factors: discriminant \( = 144-24 = 120 \), not a perfect square. Does not factor nicely.
Answer: \( \dfrac{3x^2-12x+2}{(x+3)(x-2)(x-4)} \)
Q14. Solve: \( \dfrac{x-1}{x+2} + \dfrac{x+3}{x-1} = \dfrac{2x^2+3}{x^2+x-2} \).
Factor: \( x^2+x-2 = (x+2)(x-1) \). LCD \( = (x+2)(x-1) \).
Multiply both sides by LCD:
\( (x-1)(x-1) + (x+3)(x+2) = 2x^2+3 \)
\( (x-1)^2 + (x+3)(x+2) = 2x^2+3 \)
\( x^2-2x+1 + x^2+5x+6 = 2x^2+3 \)
\( 2x^2+3x+7 = 2x^2+3 \)
\( 3x+7 = 3 \)
\( 3x = -4 \)
\( x = -\dfrac{4}{3} \)
Check: \( x = -\frac{4}{3} \) does not make any denominator zero. ✓
Solution: \( x = -\dfrac{4}{3} \)
Multiply both sides by LCD:
\( (x-1)(x-1) + (x+3)(x+2) = 2x^2+3 \)
\( (x-1)^2 + (x+3)(x+2) = 2x^2+3 \)
\( x^2-2x+1 + x^2+5x+6 = 2x^2+3 \)
\( 2x^2+3x+7 = 2x^2+3 \)
\( 3x+7 = 3 \)
\( 3x = -4 \)
\( x = -\dfrac{4}{3} \)
Check: \( x = -\frac{4}{3} \) does not make any denominator zero. ✓
Solution: \( x = -\dfrac{4}{3} \)
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Q15. Solve the inequality: \( \dfrac{x^2-2x-8}{x^2-x-6} > 0 \).
Factor numerator: \( x^2-2x-8 = (x-4)(x+2) \)
Factor denominator: \( x^2-x-6 = (x-3)(x+2) \)
Simplified: \( \dfrac{(x-4)(x+2)}{(x-3)(x+2)} = \dfrac{x-4}{x-3} \), \( x \neq -2 \)
Critical values: \( x=-2 \) (hole), \( x=3 \) (VA), \( x=4 \) (zero of numerator).
Intervals: \( (-\infty,-2) \), \( (-2,3) \), \( (3,4) \), \( (4,\infty) \)
Test using simplified form \( \frac{x-4}{x-3} \):
\( x=-3 \): \( \frac{-7}{-6} > 0 \) ✓
\( x=0 \): \( \frac{-4}{-3} > 0 \) ✓
\( x=3.5 \): \( \frac{-0.5}{0.5} < 0 \) ✗
\( x=5 \): \( \frac{1}{2} > 0 \) ✓
Strict inequality: exclude \( x=4 \), exclude hole at \( x=-2 \), exclude VA at \( x=3 \).
Solution: \( (-\infty,-2) \cup (-2,3) \cup (4,\infty) \)
Factor denominator: \( x^2-x-6 = (x-3)(x+2) \)
Simplified: \( \dfrac{(x-4)(x+2)}{(x-3)(x+2)} = \dfrac{x-4}{x-3} \), \( x \neq -2 \)
Critical values: \( x=-2 \) (hole), \( x=3 \) (VA), \( x=4 \) (zero of numerator).
Intervals: \( (-\infty,-2) \), \( (-2,3) \), \( (3,4) \), \( (4,\infty) \)
Test using simplified form \( \frac{x-4}{x-3} \):
\( x=-3 \): \( \frac{-7}{-6} > 0 \) ✓
\( x=0 \): \( \frac{-4}{-3} > 0 \) ✓
\( x=3.5 \): \( \frac{-0.5}{0.5} < 0 \) ✗
\( x=5 \): \( \frac{1}{2} > 0 \) ✓
Strict inequality: exclude \( x=4 \), exclude hole at \( x=-2 \), exclude VA at \( x=3 \).
Solution: \( (-\infty,-2) \cup (-2,3) \cup (4,\infty) \)
Q16. Fully analyze the rational function \( f(x) = \dfrac{2x^2-8}{x^2-2x-8} \): domain, intercepts, asymptotes, and holes.
Factor:
Numerator: \( 2x^2-8 = 2(x-2)(x+2) \)
Denominator: \( x^2-2x-8 = (x-4)(x+2) \)
Simplify:
\( f(x) = \dfrac{2(x-2)(x+2)}{(x-4)(x+2)} = \dfrac{2(x-2)}{x-4} \), \( x \neq -2, 4 \)
Domain: \( (-\infty,-2) \cup (-2,4) \cup (4,\infty) \)
Hole: at \( x=-2 \). \( y \)-coordinate: \( \frac{2(-2-2)}{-2-4} = \frac{2(-4)}{-6} = \frac{8}{6} = \frac{4}{3} \).
Hole at \( \left(-2, \frac{4}{3}\right) \).
Vertical Asymptote: \( x = 4 \)
Horizontal Asymptote: Both degree 2. \( y = \frac{2}{1} = 2 \).
\(x\)-intercept: Set \( 2(x-2)=0 \Rightarrow x=2 \). Point: \( (2, 0) \).
\(y\)-intercept: \( f(0) = \dfrac{2(0-2)}{0-4} = \dfrac{-4}{-4} = 1 \). Point: \( (0, 1) \).
Numerator: \( 2x^2-8 = 2(x-2)(x+2) \)
Denominator: \( x^2-2x-8 = (x-4)(x+2) \)
Simplify:
\( f(x) = \dfrac{2(x-2)(x+2)}{(x-4)(x+2)} = \dfrac{2(x-2)}{x-4} \), \( x \neq -2, 4 \)
Domain: \( (-\infty,-2) \cup (-2,4) \cup (4,\infty) \)
Hole: at \( x=-2 \). \( y \)-coordinate: \( \frac{2(-2-2)}{-2-4} = \frac{2(-4)}{-6} = \frac{8}{6} = \frac{4}{3} \).
Hole at \( \left(-2, \frac{4}{3}\right) \).
Vertical Asymptote: \( x = 4 \)
Horizontal Asymptote: Both degree 2. \( y = \frac{2}{1} = 2 \).
\(x\)-intercept: Set \( 2(x-2)=0 \Rightarrow x=2 \). Point: \( (2, 0) \).
\(y\)-intercept: \( f(0) = \dfrac{2(0-2)}{0-4} = \dfrac{-4}{-4} = 1 \). Point: \( (0, 1) \).
Q17. A rational function satisfies ALL of the following conditions:
— Vertical asymptote at \( x = -3 \)
— Horizontal asymptote at \( y = 2 \)
— Hole at \( x = 1 \)
— \( y \)-intercept at \( (0, -2) \)
Find a possible equation for this function.
— Vertical asymptote at \( x = -3 \)
— Horizontal asymptote at \( y = 2 \)
— Hole at \( x = 1 \)
— \( y \)-intercept at \( (0, -2) \)
Find a possible equation for this function.
Let’s build the function step by step.
Vertical asymptote at \( x=-3 \): Denominator has factor \( (x+3) \).
Hole at \( x=1 \): Both numerator and denominator have factor \( (x-1) \).
Horizontal asymptote at \( y=2 \): Degrees must be equal, and ratio of leading coefficients = 2.
So far: \( f(x) = \dfrac{2(x-1)(\cdots)}{(x+3)(x-1)(\cdots)} \)
We need equal degrees. If numerator and denominator both have degree 2:
\( f(x) = \dfrac{2(x-1)(x-a)}{(x+3)(x-1)} = \dfrac{2(x-a)}{x+3} \), \( x \neq 1 \)
\(y\)-intercept at \( (0,-2) \): \( f(0) = \dfrac{2(0-a)}{0+3} = \dfrac{-2a}{3} = -2 \)
\( -2a = -6 \Rightarrow a = 3 \)
But if \( a=3 \), then the numerator becomes \( 2(x-3) \), which has a zero at \( x=3 \). The denominator has a factor \( (x+3) \) — no issue there. But wait, we originally said numerator has factor \( (x-1)(x-3) \) and denominator has \( (x+3)(x-1) \). After canceling \( (x-1) \):
\( f(x) = \dfrac{2(x-3)}{x+3} \), \( x \neq 1 \)
Verify: VA at \( x=-3 \) ✓, HA at \( y=2 \) ✓, Hole at \( x=1 \) (y-coordinate: \( \frac{2(1-3)}{1+3} = \frac{-4}{4} = -1 \), so hole at \( (1,-1) \)) ✓, y-intercept: \( f(0) = \frac{-6}{3} = -2 \) ✓
Answer: \( f(x) = \dfrac{2(x-3)}{x+3}, \; x \neq 1 \)
Or equivalently: \( f(x) = \dfrac{2x^2-8x+6}{x^2-2x-3}, \; x \neq 1 \)
Vertical asymptote at \( x=-3 \): Denominator has factor \( (x+3) \).
Hole at \( x=1 \): Both numerator and denominator have factor \( (x-1) \).
Horizontal asymptote at \( y=2 \): Degrees must be equal, and ratio of leading coefficients = 2.
So far: \( f(x) = \dfrac{2(x-1)(\cdots)}{(x+3)(x-1)(\cdots)} \)
We need equal degrees. If numerator and denominator both have degree 2:
\( f(x) = \dfrac{2(x-1)(x-a)}{(x+3)(x-1)} = \dfrac{2(x-a)}{x+3} \), \( x \neq 1 \)
\(y\)-intercept at \( (0,-2) \): \( f(0) = \dfrac{2(0-a)}{0+3} = \dfrac{-2a}{3} = -2 \)
\( -2a = -6 \Rightarrow a = 3 \)
But if \( a=3 \), then the numerator becomes \( 2(x-3) \), which has a zero at \( x=3 \). The denominator has a factor \( (x+3) \) — no issue there. But wait, we originally said numerator has factor \( (x-1)(x-3) \) and denominator has \( (x+3)(x-1) \). After canceling \( (x-1) \):
\( f(x) = \dfrac{2(x-3)}{x+3} \), \( x \neq 1 \)
Verify: VA at \( x=-3 \) ✓, HA at \( y=2 \) ✓, Hole at \( x=1 \) (y-coordinate: \( \frac{2(1-3)}{1+3} = \frac{-4}{4} = -1 \), so hole at \( (1,-1) \)) ✓, y-intercept: \( f(0) = \frac{-6}{3} = -2 \) ✓
Answer: \( f(x) = \dfrac{2(x-3)}{x+3}, \; x \neq 1 \)
Or equivalently: \( f(x) = \dfrac{2x^2-8x+6}{x^2-2x-3}, \; x \neq 1 \)
Q18. Solve: \( \dfrac{1}{x-1} – \dfrac{2}{x+1} = \dfrac{4}{x^2-1} \).
LCD \( = (x-1)(x+1) = x^2-1 \). Multiply both sides:
\( (x+1) – 2(x-1) = 4 \)
\( x+1-2x+2 = 4 \)
\( -x+3 = 4 \)
\( -x = 1 \)
\( x = -1 \)
Check: \( x=-1 \) makes the denominators \( x+1 \) and \( x^2-1 \) equal to zero.
Extraneous solution! No solution.
\( (x+1) – 2(x-1) = 4 \)
\( x+1-2x+2 = 4 \)
\( -x+3 = 4 \)
\( -x = 1 \)
\( x = -1 \)
Check: \( x=-1 \) makes the denominators \( x+1 \) and \( x^2-1 \) equal to zero.
Extraneous solution! No solution.
Q19. Simplify: \( \dfrac{\dfrac{x}{x-1}-1}{\dfrac{x}{x-1}+1} \div \dfrac{x-1}{x+1} \).
Part 1: Simplify the complex fraction.
LCD of \( \frac{x}{x-1} \) and \( 1 \) is \( x-1 \). Multiply top and bottom by \( x-1 \):
\[ \frac{x-(x-1)}{x+(x-1)} = \frac{x-x+1}{2x-1} = \frac{1}{2x-1} \]
Part 2: Divide by \( \frac{x-1}{x+1} \).
\[ \frac{1}{2x-1} \div \frac{x-1}{x+1} = \frac{1}{2x-1} \times \frac{x+1}{x-1} = \frac{x+1}{(2x-1)(x-1)} \]
Restrictions: \( x \neq 1, \; x \neq \frac{1}{2}, \; x \neq -1 \) (from original expression’s \( x-1 \) factors and the denominator).
Answer: \( \dfrac{x+1}{(2x-1)(x-1)} \)
LCD of \( \frac{x}{x-1} \) and \( 1 \) is \( x-1 \). Multiply top and bottom by \( x-1 \):
\[ \frac{x-(x-1)}{x+(x-1)} = \frac{x-x+1}{2x-1} = \frac{1}{2x-1} \]
Part 2: Divide by \( \frac{x-1}{x+1} \).
\[ \frac{1}{2x-1} \div \frac{x-1}{x+1} = \frac{1}{2x-1} \times \frac{x+1}{x-1} = \frac{x+1}{(2x-1)(x-1)} \]
Restrictions: \( x \neq 1, \; x \neq \frac{1}{2}, \; x \neq -1 \) (from original expression’s \( x-1 \) factors and the denominator).
Answer: \( \dfrac{x+1}{(2x-1)(x-1)} \)
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Q20. The cost \( C \) (in birr per item) of producing \( x \) items is given by \( C(x) = \dfrac{5000+3x}{x} \). Find: (a) the horizontal asymptote and explain its meaning, (b) the cost per item when 1000 items are produced, (c) how many items must be produced so the cost per item is less than 4 birr.
(a) Simplify: \( C(x) = \frac{5000}{x} + 3 \).
As \( x \to \infty \), \( \frac{5000}{x} \to 0 \), so \( C(x) \to 3 \).
Horizontal asymptote: \( y = 3 \).
Meaning: As production increases indefinitely, the cost per item approaches 3 birr (the variable cost per item). Fixed costs are spread over more items.
(b) \( C(1000) = \frac{5000+3000}{1000} = \frac{8000}{1000} = 8 \) birr per item.
(c) Solve \( \frac{5000+3x}{x} < 4 \), \( x > 0 \).
\( \frac{5000}{x} + 3 < 4 \)
\( \frac{5000}{x} < 1 \)
\( 5000 < x \) (since \( x > 0 \), multiplying by \( x \) doesn’t flip the sign)
More than 5000 items must be produced.
As \( x \to \infty \), \( \frac{5000}{x} \to 0 \), so \( C(x) \to 3 \).
Horizontal asymptote: \( y = 3 \).
Meaning: As production increases indefinitely, the cost per item approaches 3 birr (the variable cost per item). Fixed costs are spread over more items.
(b) \( C(1000) = \frac{5000+3000}{1000} = \frac{8000}{1000} = 8 \) birr per item.
(c) Solve \( \frac{5000+3x}{x} < 4 \), \( x > 0 \).
\( \frac{5000}{x} + 3 < 4 \)
\( \frac{5000}{x} < 1 \)
\( 5000 < x \) (since \( x > 0 \), multiplying by \( x \) doesn’t flip the sign)
More than 5000 items must be produced.