Solved Exercises on Graphing Rational Functions (Asymptotes, Intercepts) in Algebra 2

Master graphing rational functions: vertical asymptotes, horizontal asymptotes, intercepts, domain, range, and graphing techniques through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Basic Rational Function
Exercise 1
Graph \(f(x) = \frac{2x-4}{x-3}\). Find vertical asymptote, horizontal asymptote, x-intercept, y-intercept, domain, and range.
Definition:

Rational function: A function of the form \(f(x) = \frac{P(x)}{Q(x)}\) where P(x) and Q(x) are polynomials.

Graphing method:
  1. Find vertical asymptotes (where denominator = 0)
  2. Find horizontal/slant asymptotes
  3. Find x-intercepts (where numerator = 0)
  4. Find y-intercept (evaluate at x = 0)
  5. Determine domain and range
  6. Sketch the graph
Vertical asymptote
x = 3
Horizontal asymptote
y = 2
Step 1: Find vertical asymptote

Set denominator equal to zero: \(x - 3 = 0\)

Vertical asymptote: \(x = 3\)

Step 2: Find horizontal asymptote

Degree of numerator = 1, degree of denominator = 1

Since degrees are equal, horizontal asymptote is ratio of leading coefficients: \(y = \frac{2}{1} = 2\)

Step 3: Find x-intercept

Set numerator equal to zero: \(2x - 4 = 0\)

\(2x = 4\), so \(x = 2\)

x-intercept: \((2, 0)\)

Step 4: Find y-intercept

Evaluate at \(x = 0\): \(f(0) = \frac{2(0)-4}{0-3} = \frac{-4}{-3} = \frac{4}{3}\)

y-intercept: \((0, \frac{4}{3})\)

Step 5: Determine domain and range

Domain: All real numbers except \(x = 3\), so \((-\infty, 3) \cup (3, \infty)\)

Range: All real numbers except \(y = 2\), so \((-\infty, 2) \cup (2, \infty)\)

VA: x = 3, HA: y = 2, x-int: (2,0), y-int: (0,4/3), D: ℝ∖{3}, R: ℝ∖{2}
Final answer:

Vertical asymptote: \(x = 3\)

Horizontal asymptote: \(y = 2\)

x-intercept: \((2, 0)\)

y-intercept: \((0, \frac{4}{3})\)

Domain: \((-\infty, 3) \cup (3, \infty)\)

Range: \((-\infty, 2) \cup (2, \infty)\)

Applied rules:

• Vertical asymptotes: Occur where denominator = 0 (and numerator ≠ 0)

• Horizontal asymptotes: Compare degrees of numerator and denominator

• Intercepts: x-int when numerator = 0, y-int when x = 0

2 Higher Degree Numerator
Exercise 2
Graph \(f(x) = \frac{x^2-4}{x-1}\). Find vertical asymptote, slant asymptote, x-intercepts, y-intercept, domain, and range.
Definition:

Slant (oblique) asymptote: Occurs when degree of numerator is exactly 1 greater than degree of denominator.

Vertical asymptote
x = 1
Slant asymptote
y = x + 1
Step 1: Find vertical asymptote

Set denominator equal to zero: \(x - 1 = 0\)

Vertical asymptote: \(x = 1\)

Step 2: Find slant asymptote

Degree of numerator = 2, degree of denominator = 1

Since numerator degree is exactly 1 greater, perform polynomial long division:

\(\frac{x^2-4}{x-1} = x + 1 + \frac{-3}{x-1}\)

Slant asymptote: \(y = x + 1\)

Step 3: Find x-intercepts

Set numerator equal to zero: \(x^2 - 4 = 0\)

\(x^2 = 4\), so \(x = \pm 2\)

x-intercepts: \((-2, 0)\) and \((2, 0)\)

Step 4: Find y-intercept

Evaluate at \(x = 0\): \(f(0) = \frac{0^2-4}{0-1} = \frac{-4}{-1} = 4\)

y-intercept: \((0, 4)\)

Step 5: Determine domain and range

Domain: All real numbers except \(x = 1\), so \((-\infty, 1) \cup (1, \infty)\)

Range: All real numbers (the slant asymptote allows the function to approach all values)

VA: x = 1, SA: y = x+1, x-int: (-2,0) and (2,0), y-int: (0,4), D: ℝ∖{1}, R: ℝ
Final answer:

Vertical asymptote: \(x = 1\)

Slant asymptote: \(y = x + 1\)

x-intercepts: \((-2, 0)\) and \((2, 0)\)

y-intercept: \((0, 4)\)

Domain: \((-\infty, 1) \cup (1, \infty)\)

Range: \((-\infty, \infty)\)

Applied rules:

• Slant asymptote: When deg(numerator) = deg(denominator) + 1, use polynomial division

• X-intercepts: Solve numerator = 0, verify denominator ≠ 0 at those points

• Range with slant asymptote: Usually all real numbers

3 Factored Form
Exercise 3
Graph \(f(x) = \frac{(x+2)(x-1)}{(x+3)(x-2)}\). Find vertical asymptotes, horizontal asymptote, x-intercepts, y-intercept, domain, and range.
Definition:

Factored form: Makes it easier to identify zeros and asymptotes directly from the expression.

Vertical asymptotes
x = -3, x = 2
Horizontal asymptote
y = 1
Step 1: Find vertical asymptotes

Set denominator factors equal to zero: \(x + 3 = 0\) and \(x - 2 = 0\)

Vertical asymptotes: \(x = -3\) and \(x = 2\)

Step 2: Find horizontal asymptote

Expanding: numerator has degree 2, denominator has degree 2

Since degrees are equal, horizontal asymptote is ratio of leading coefficients: \(y = \frac{1}{1} = 1\)

Step 3: Find x-intercepts

Set numerator factors equal to zero: \(x + 2 = 0\) and \(x - 1 = 0\)

So \(x = -2\) and \(x = 1\)

x-intercepts: \((-2, 0)\) and \((1, 0)\)

Step 4: Find y-intercept

Evaluate at \(x = 0\): \(f(0) = \frac{(0+2)(0-1)}{(0+3)(0-2)} = \frac{(2)(-1)}{(3)(-2)} = \frac{-2}{-6} = \frac{1}{3}\)

y-intercept: \((0, \frac{1}{3})\)

Step 5: Determine domain and range

Domain: All real numbers except \(x = -3\) and \(x = 2\), so \((-\infty, -3) \cup (-3, 2) \cup (2, \infty)\)

Range: All real numbers except possibly some values near the horizontal asymptote

VA: x = -3, x = 2, HA: y = 1, x-int: (-2,0) and (1,0), y-int: (0,1/3), D: ℝ∖{-3,2}, R: ℝ∖{some values}
Final answer:

Vertical asymptotes: \(x = -3\) and \(x = 2\)

Horizontal asymptote: \(y = 1\)

x-intercepts: \((-2, 0)\) and \((1, 0)\)

y-intercept: \((0, \frac{1}{3})\)

Domain: \((-\infty, -3) \cup (-3, 2) \cup (2, \infty)\)

Range: All real numbers except possibly some values

Applied rules:

• Factored form: Directly identify zeros and poles from factors

• Multiple vertical asymptotes: Each zero of denominator creates an asymptote

• Horizontal asymptote: Ratio of leading coefficients when degrees are equal

Key Concepts: Definitions, Rules, and Methods
\(f(x) = \frac{P(x)}{Q(x)}, \text{ where } P(x), Q(x) \text{ are polynomials}\)
Rational Function Definition
Key definitions:

Rational Function: A function expressed as the quotient of two polynomials.

Vertical Asymptote: A vertical line x = a where the function approaches infinity.

Horizontal Asymptote: A horizontal line y = b that the function approaches as x goes to ±∞.

Slant Asymptote: An oblique line that the function approaches as x goes to ±∞.

X-intercept: Point where the graph crosses the x-axis (y = 0).

Y-intercept: Point where the graph crosses the y-axis (x = 0).

Asymptote determination:
  1. Vertical: Set denominator = 0, solve for x (excluding removable discontinuities)
  2. Horizontal: Compare degrees of numerator and denominator
  3. Slant: When numerator degree = denominator degree + 1
Horizontal asymptote rules:
  1. If deg(numerator) < deg(denominator): y = 0
  2. If deg(numerator) = deg(denominator): y = ratio of leading coefficients
  3. If deg(numerator) > deg(denominator): No horizontal asymptote (slant if diff = 1)
Tip 1: Always factor numerator and denominator completely before analyzing.
Tip 2: Check for common factors that indicate holes instead of vertical asymptotes.
Tip 3: For slant asymptotes, perform polynomial long division.
Tip 4: Test points in each interval created by vertical asymptotes to determine behavior.
Common errors: Forgetting to check for holes, misidentifying asymptote types, incorrect domain.
Exam preparation: Practice polynomial division, memorize asymptote rules, sketch by hand.
Essential rules to remember:

• Vertical asymptotes: Occur where denominator = 0 and numerator ≠ 0

• Horizontal asymptotes: Dependent on degree comparison

• Slant asymptotes: When numerator degree is exactly 1 more than denominator

• X-intercepts: Solutions to numerator = 0 (check denominator ≠ 0)

• Y-intercept: Value of function at x = 0

\(\text{If } f(x) = \frac{a_nx^n + ...}{b_mx^m + ...}, \text{ then:}\)
Asymptote Rules
n < m: y = 0, \quad n = m: y = \frac{a_n}{b_m}, \quad n > m: \text{No HA}
Horizontal Asymptote Cases
n = m + 1: \text{ Slant asymptote by polynomial division}
Slant Asymptote Condition
Solution: Exercises 4 to 5
4 Removable Discontinuity
Exercise 4
Graph \(f(x) = \frac{x^2-4}{x-2}\). Find vertical asymptote, horizontal asymptote, x-intercept, y-intercept, domain, and range.
Definition:

Removable discontinuity: A point where a function is undefined due to a common factor in numerator and denominator that can be cancelled.

Factor numerator
\(f(x) = \frac{(x+2)(x-2)}{x-2}\)
Simplified form
\(f(x) = x+2, x \neq 2\)
Step 1: Factor the numerator

\(x^2 - 4 = (x+2)(x-2)\) (difference of squares)

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

Step 2: Identify common factors

Both numerator and denominator contain \((x-2)\)

This creates a removable discontinuity (hole) at \(x = 2\), not a vertical asymptote

Step 3: Simplify for analysis

For \(x \neq 2\): \(f(x) = x + 2\)

This is a linear function with a hole at \(x = 2\)

Step 4: Find x-intercept

Set simplified numerator equal to zero: \(x + 2 = 0\), so \(x = -2\)

x-intercept: \((-2, 0)\)

Step 5: Find y-intercept

Evaluate at \(x = 0\): \(f(0) = 0 + 2 = 2\)

y-intercept: \((0, 2)\)

Step 6: Determine domain and range

Domain: All real numbers except \(x = 2\), so \((-\infty, 2) \cup (2, \infty)\)

Range: All real numbers except the y-value at the hole: when \(x = 2\), \(y = 2 + 2 = 4\), so \((-\infty, 4) \cup (4, \infty)\)

No VA, No HA, Hole at (2,4), x-int: (-2,0), y-int: (0,2), D: ℝ∖{2}, R: ℝ∖{4}
Final answer:

No vertical asymptote

No horizontal asymptote

Hole at point \((2, 4)\)

x-intercept: \((-2, 0)\)

y-intercept: \((0, 2)\)

Domain: \((-\infty, 2) \cup (2, \infty)\)

Range: \((-\infty, 4) \cup (4, \infty)\)

Applied rules:

• Removable discontinuity: Common factors create holes, not asymptotes

• Factor cancellation: Cancel common factors to simplify analysis

• Range adjustment: Exclude y-value of the hole from range

5 Complex Rational Function
Exercise 5
Graph \(f(x) = \frac{2x^3-x^2+3}{x^2-4}\). Find vertical asymptotes, slant asymptote, y-intercept, domain, and describe the behavior.
Definition:

Higher-degree rational functions: When numerator degree is 2 or more greater than denominator, there is no horizontal or slant asymptote, but a polynomial curve that the function approaches.

Vertical asymptotes
x = -2, x = 2
Polynomial behavior
Similar to y = 2x
Step 1: Find vertical asymptotes

Set denominator equal to zero: \(x^2 - 4 = 0\)

\(x^2 = 4\), so \(x = \pm 2\)

Vertical asymptotes: \(x = -2\) and \(x = 2\)

Step 2: Analyze asymptotic behavior

Degree of numerator = 3, degree of denominator = 2

Since numerator degree is 1 greater than denominator, we would expect a slant asymptote

Perform polynomial long division: \(\frac{2x^3-x^2+3}{x^2-4} = 2x - 1 + \frac{8x-1}{x^2-4}\)

Slant asymptote: \(y = 2x - 1\)

Step 3: Find y-intercept

Evaluate at \(x = 0\): \(f(0) = \frac{2(0)^3-(0)^2+3}{(0)^2-4} = \frac{3}{-4} = -\frac{3}{4}\)

y-intercept: \((0, -\frac{3}{4})\)

Step 4: Check for x-intercepts

Set numerator equal to zero: \(2x^3 - x^2 + 3 = 0\)

This cubic equation is difficult to solve analytically, but we can determine that it has one real root approximately at \(x ≈ -1.07\)

Step 5: Determine domain

Domain: All real numbers except where denominator is zero

Domain: \((-\infty, -2) \cup (-2, 2) \cup (2, \infty)\)

VA: x = -2, x = 2, SA: y = 2x-1, y-int: (0,-3/4), D: ℝ∖{-2,2}
Final answer:

Vertical asymptotes: \(x = -2\) and \(x = 2\)

Slant asymptote: \(y = 2x - 1\)

y-intercept: \((0, -\frac{3}{4})\)

Approximate x-intercept: \((-1.07, 0)\)

Domain: \((-\infty, -2) \cup (-2, 2) \cup (2, \infty)\)

Range: All real numbers

Applied rules:

• Polynomial long division: Required for finding slant asymptotes

• Higher-degree differences: When numerator degree > denominator degree + 1, no simple asymptote

• Complex roots: Some x-intercepts may require approximation methods

Graphing Rational Functions: Complete Guide
\(f(x) = \frac{P(x)}{Q(x)}, \text{ where } P(x), Q(x) \text{ are polynomials}\)
Rational Function Definition
Key definitions:

Rational Function: A function expressed as the quotient of two polynomials.

Vertical Asymptote: A vertical line x = a where the function approaches positive or negative infinity.

Horizontal Asymptote: A horizontal line y = b that the function approaches as x goes to positive or negative infinity.

Slant Asymptote: An oblique line that the function approaches as x goes to positive or negative infinity.

Removable Discontinuity: A hole in the graph where a common factor cancels out.

Complete graphing methodology:
  1. Factor completely: Factor both numerator and denominator
  2. Find vertical asymptotes: Set denominator = 0 (excluding removable discontinuities)
  3. Find horizontal/slant asymptotes: Compare degrees of numerator and denominator
  4. Find intercepts: X-intercepts (numerator = 0) and y-intercept (x = 0)
  5. Identify removable discontinuities: Common factors create holes
  6. Determine domain and range: Based on asymptotes and discontinuities
  7. Sketch the graph: Use all information to create accurate representation
Asymptote classification:
Condition Type Rule
deg(num) < deg(den) Horizontal y = 0
deg(num) = deg(den) Horizontal y = coeff ratio
deg(num) = deg(den) + 1 Slant Long division
deg(num) > deg(den) + 1 None Polynomial part
Tip 1: Always factor completely first to identify all zeros and poles.
Tip 2: Common factors between numerator and denominator create holes, not asymptotes.
Tip 3: Use polynomial long division for slant asymptotes.
Tip 4: Test points in each interval to determine function behavior near asymptotes.
Common errors: Forgetting to check for removable discontinuities, misidentifying asymptote types, incorrect domain determination.
Exam preparation: Practice polynomial division, memorize asymptote rules, sketch by hand to develop intuition.
Essential formulas and rules:

• Vertical asymptotes: Occur where denominator = 0 and numerator ≠ 0

• Horizontal asymptotes: Compare degrees and leading coefficients

• Slant asymptotes: Perform polynomial division when degree difference is 1

• X-intercepts: Solutions to numerator = 0 (verify denominator ≠ 0)

• Y-intercept: Evaluate function at x = 0

• Domain: Exclude values that make denominator zero

\(\text{If } f(x) = \frac{a_nx^n + ...}{b_mx^m + ...}, \text{ then:}\)
Asymptote Rules
n < m: y = 0, \quad n = m: y = \frac{a_n}{b_m}, \quad n > m: \text{No HA}
Horizontal Asymptote Cases
n = m + 1: \text{ Slant asymptote by polynomial division}
Slant Asymptote Condition
Behavior near asymptotes:

Vertical asymptotes: Function approaches ±∞ depending on signs of numerator and denominator

Horizontal asymptotes: Function approaches the line as x → ±∞

Slant asymptotes: Function approaches the line as x → ±∞

Holes: Single points where function is undefined

Sign analysis technique:
  1. Identify critical points: Zeros and undefined points
  2. Create intervals: Between critical points
  3. Test signs: In each interval to determine function position
  4. Connect behavior: Around asymptotes and intercepts
Key note: The function can cross horizontal asymptotes but cannot cross vertical asymptotes.
Key note: The function can cross slant asymptotes but approaches them as x → ±∞.

Questions & Answers

Question: How do I know if I have a vertical asymptote or a hole at a particular x-value? They seem similar.

Answer: The key difference lies in common factors:

  • Vertical asymptote: The factor appears in the denominator but NOT in the numerator
  • Hole: The same factor appears in BOTH the numerator and denominator

Examples:

For \(f(x) = \frac{x-2}{x-2}\): The factor \((x-2)\) appears in both numerator and denominator → hole at \(x = 2\)

For \(g(x) = \frac{x-2}{x-3}\): The factor \((x-3)\) only appears in denominator → vertical asymptote at \(x = 3\)

For \(h(x) = \frac{(x-2)(x+1)}{(x-2)(x-4)}\): The factor \((x-2)\) appears in both → hole at \(x = 2\), and \((x-4)\) only in denominator → vertical asymptote at \(x = 4\)

Always factor completely first to identify all common factors.

Question: When do I get a slant asymptote versus a horizontal asymptote? I get confused about the degree rules.

Answer: The rules depend on the relationship between the degrees of the numerator (n) and denominator (m):

  1. If n < m: Horizontal asymptote at y = 0
  2. If n = m: Horizontal asymptote at y = (leading coefficient of numerator)/(leading coefficient of denominator)
  3. If n = m + 1: Slant (oblique) asymptote found by polynomial long division
  4. If n > m + 1: No horizontal or slant asymptote (function behaves like a polynomial of degree n-m)

Examples:

\(f(x) = \frac{2x+1}{x^2-3}\) (n=1, m=2) → HA: y = 0

\(g(x) = \frac{3x^2+1}{x^2-2}\) (n=2, m=2) → HA: y = 3/1 = 3

\(h(x) = \frac{x^2+1}{x-1}\) (n=2, m=1) → SA: y = x + 1 (from long division)

Remember: Slant asymptotes occur specifically when the numerator degree is exactly 1 greater than the denominator degree.

Question: Can a rational function cross its horizontal or slant asymptote? I thought asymptotes were lines the function approaches but never touches.

Answer: This is a common misconception! Here's the truth:

  • Vertical asymptotes: The function can NEVER cross these. The function is undefined at these x-values.
  • Horizontal asymptotes: The function CAN cross these at finite x-values, but must approach the asymptote as x → ±∞.
  • Slant asymptotes: The function CAN cross these at finite x-values, but must approach the asymptote as x → ±∞.

For example, consider \(f(x) = \frac{x}{x^2+1}\). This function has a horizontal asymptote at y = 0, but f(0) = 0, so the function passes through the origin and crosses its horizontal asymptote.

The key requirement is that as x gets very large (positive or negative), the function must approach the asymptote, regardless of what happens at finite values.