Unit 3 · Chapter 3.7

3.7Rational Functions

Analyze rational functions by finding domain, vertical asymptotes, holes, horizontal and oblique asymptotes, and intercepts — then sketch complete graphs using sign analysis.

Rational functions model real-world phenomena like average cost, concentration over time, and population growth rates. Understanding asymptotes and holes is essential for calculus limits and for interpreting the behavior of any ratio of polynomials.

Essential Question: How do the degrees of the numerator and denominator of a rational function determine its asymptotes, holes, and overall behavior?

Lesson Overview

A rational function has the form f(x) = p(x)/q(x) where p and q are polynomials and q(x) ≠ 0. The domain excludes all x-values where q(x) = 0. Depending on whether the numerator also equals zero at those points, the graph has either a vertical asymptote (a boundary the curve approaches but never crosses) or a hole (a single missing point, drawn as an open circle). The relationship between the degrees of p and q governs the horizontal asymptote — the long-run behavior of the function. When deg(p) = deg(q) + 1, there is an oblique (slant) asymptote found by polynomial division. Mastering these features lets you sketch rational functions accurately and interpret real-world models involving rates, concentrations, and averages.

Key Features to Find

  • Domain: set q(x) ≠ 0
  • Vertical asymptotes (VA): q(a) = 0, p(a) ≠ 0 after canceling
  • Holes: both p(a) = 0 and q(a) = 0 (common factor cancels)
  • Horizontal asymptote (HA): compare degrees of p and q
  • Oblique asymptote: deg(p) = deg(q) + 1; use long division
  • x-intercepts: zeros of p(x) not canceled as holes
  • y-intercept: f(0) if 0 is in the domain

Sketching Steps

  1. Factor p(x) and q(x) fully
  2. Cancel common factors → find holes
  3. Find domain (exclude VA and hole x-values)
  4. Determine VA (uncanceled zeros of q)
  5. Determine HA or oblique asymptote
  6. Find x-intercepts and y-intercept
  7. Test signs in each interval between VAs and intercepts
  8. Sketch, marking asymptotes as dashed lines and holes as open circles

f(x) = (x+2) / ((x−1)(x+3)) — VA: x=1, x=−3 · HA: y=0

xy-4-3-2-11234-11x=−3x=1y=0 (HA)(−2, 0)Vertical Asymptotef(x) curvex-intercept

Hole vs. Vertical Asymptote

f(x) = (x²−4)/(x−2) = x+2, x≠2hole (2,4)24Open circle = HOLEg(x) = 1/(x−2)x=2 (VA)Dashed line = VA

Horizontal Asymptote Rules — Degree Comparison

Degree ConditionHA RuleExampleAsymptotedeg(p) < deg(q)y = 03/(x²+1)y = 0numerator degree lower → y=0deg(p) = deg(q)y = a/b2x²/(3x²−1)y = 2/3same degree → ratio of leading coeffsdeg(p) > deg(q)No HAx³/(x²+1)Obliquedeg(p)=deg(q)+1 → polynomial division

Worked Examples

Example 1

Find the domain, vertical asymptotes, horizontal asymptote, and intercepts of f(x) = (x + 2) / ((x − 1)(x + 3)).

Step 1 — Factor: numerator = (x + 2); denominator = (x − 1)(x + 3). No common factors, so no holes.

Step 2 — Domain: set denominator ≠ 0 → x ≠ 1 and x ≠ −3. Domain: all reals except x = 1 and x = −3.

Step 3 — Vertical asymptotes: zeros of denominator not canceled → VA at x = 1 and x = −3.

Step 4 — Horizontal asymptote: deg(p) = 1, deg(q) = 2. Since deg(p) < deg(q), HA is y = 0.

Step 5 — x-intercept: set numerator = 0 → x + 2 = 0 → x = −2. Check: −2 is in domain. x-intercept: (−2, 0).

Step 6 — y-intercept: f(0) = (0 + 2) / ((0 − 1)(0 + 3)) = 2 / (−3) = −2/3. y-intercept: (0, −2/3).

Answer:Domain: x ≠ 1, x ≠ −3 · VA: x = 1, x = −3 · HA: y = 0 · x-int: (−2, 0) · y-int: (0, −2/3)
Example 2

Find any holes and vertical asymptotes of f(x) = (x² − 4) / (x² − x − 2).

Step 1 — Factor numerator: x² − 4 = (x + 2)(x − 2).

Step 2 — Factor denominator: x² − x − 2 = (x − 2)(x + 1).

Step 3 — Cancel common factor (x − 2): f(x) = (x + 2)/(x + 1) for x ≠ 2.

Step 4 — Hole at x = 2: substitute into simplified form → y = (2 + 2)/(2 + 1) = 4/3. Hole: (2, 4/3).

Step 5 — Vertical asymptote: remaining zero of denominator → x + 1 = 0 → x = −1. VA: x = −1.

Answer:Hole at (2, 4/3) · VA at x = −1
Example 3

Determine the horizontal asymptote (if any) of each: (a) f(x) = 3/(x² + 1), (b) f(x) = (2x² − 1)/(x² + 3), (c) f(x) = (x³ + 1)/(x² − 1).

(a) deg(p) = 0, deg(q) = 2. Since deg(p) < deg(q): HA is y = 0.

(b) deg(p) = 2, deg(q) = 2. Equal degrees: HA = leading coeff of p / leading coeff of q = 2/1 = 2. HA: y = 2.

(c) deg(p) = 3, deg(q) = 2. Since deg(p) > deg(q): no horizontal asymptote.

Note: for (c), deg(p) = deg(q) + 1, so there is an oblique (slant) asymptote instead.

Answer:(a) y = 0 (b) y = 2 (c) No HA (oblique asymptote exists)
Example 4

Find the oblique (slant) asymptote of f(x) = (x² + 2x − 3) / (x − 1).

Step 1 — Check degrees: deg(p) = 2, deg(q) = 1. Since deg(p) = deg(q) + 1, an oblique asymptote exists.

Step 2 — Perform polynomial long division of x² + 2x − 3 by (x − 1):

x² + 2x − 3 ÷ (x − 1):

x² ÷ x = x. Multiply: x(x − 1) = x² − x. Subtract: (x² + 2x − 3) − (x² − x) = 3x − 3.

3x ÷ x = 3. Multiply: 3(x − 1) = 3x − 3. Subtract: (3x − 3) − (3x − 3) = 0.

Step 3 — Quotient is x + 3, remainder is 0. So f(x) = x + 3 exactly (with a hole at x = 1).

Step 4 — Oblique asymptote: y = x + 3.

Answer:Oblique asymptote: y = x + 3
Example 5

Sketch a complete graph of f(x) = 2x / (x² − 9). Label all asymptotes and intercepts.

Step 1 — Factor denominator: x² − 9 = (x − 3)(x + 3). Numerator 2x shares no factors. No holes.

Step 2 — Domain: x ≠ 3 and x ≠ −3.

Step 3 — Vertical asymptotes: x = 3 and x = −3.

Step 4 — Horizontal asymptote: deg(p) = 1, deg(q) = 2. deg(p) < deg(q) → HA: y = 0.

Step 5 — x-intercept: 2x = 0 → x = 0. Check domain: 0 is fine. x-intercept: (0, 0).

Step 6 — y-intercept: f(0) = 0/(−9) = 0. y-intercept: (0, 0). (Same as x-intercept.)

Step 7 — Sign analysis: test x = −4 (left of −3): f(−4) = −8/(16−9) = −8/7 < 0 (below x-axis).

Test x = −1 (between −3 and 0): f(−1) = −2/(1−9) = −2/(−8) = 1/4 > 0.

Test x = 1 (between 0 and 3): f(1) = 2/(1−9) = 2/(−8) = −1/4 < 0.

Test x = 4 (right of 3): f(4) = 8/(16−9) = 8/7 > 0.

Step 8 — Sketch: draw dashed VAs at x = ±3, dashed HA at y = 0, plot (0,0), and draw four branches.

Answer:VA: x = ±3 · HA: y = 0 · Intercept: (0, 0) · Four branches approaching asymptotes

Guided Practice

Guided Problem 1

Find the domain and all vertical asymptotes of f(x) = (x − 5) / (x² + x − 6).

Hint: Factor the denominator first. Check whether any factor also appears in the numerator (which would create a hole instead of a VA).

Guided Problem 2

Identify any holes and vertical asymptotes of g(x) = (x² − 9) / (x² + 2x − 3).

Hint: Factor both numerator and denominator completely. Cancel any common factors — each canceled factor gives a hole. Remaining zeros of the denominator are VAs.

Guided Problem 3

Determine the horizontal asymptote of h(x) = (5x³ − 2x) / (3x³ + x² − 1).

Hint: Compare the degrees of numerator and denominator. If they are equal, the HA is the ratio of the leading coefficients.

Guided Problem 4

Find the oblique asymptote of f(x) = (x² − 3x + 5) / (x − 2).

Hint: Since deg(numerator) = deg(denominator) + 1, perform polynomial long division. The quotient (ignoring the remainder) is the oblique asymptote equation.

Guided Problem 5

Find all intercepts of f(x) = (x² − 4x + 3) / (x² − 9).

Hint: Factor both. Cancel any common factors to find holes. Then set the simplified numerator = 0 for x-intercepts, and evaluate f(0) for the y-intercept.

Key Vocabulary

Rational function

A function of the form f(x) = p(x)/q(x) where p and q are polynomials and q(x) ≠ 0. The domain excludes all x-values where q(x) = 0.

Example: f(x) = (x + 1)/(x² − 4) is rational; domain excludes x = ±2

Vertical asymptote (VA)

A vertical line x = a where the function grows without bound. Occurs when q(a) = 0 and p(a) ≠ 0 after fully canceling common factors.

Example: f(x) = 1/(x − 3) has VA at x = 3

Hole (removable discontinuity)

A single missing point on the graph where both p(a) = 0 and q(a) = 0. The common factor (x − a) cancels; plot an open circle at (a, simplified value).

Example: f(x) = (x²−4)/(x−2): hole at (2, 4) after canceling (x−2)

Horizontal asymptote (HA)

A horizontal line y = L that the graph approaches as x → ±∞. Determined by comparing degrees: y=0 if deg(p) < deg(q); y = a/b if degrees equal; none if deg(p) > deg(q).

Example: f(x) = 3x/(2x+1) has HA y = 3/2 (equal degrees)

Oblique (slant) asymptote

A non-horizontal, non-vertical line the graph approaches as x → ±∞. Exists when deg(p) = deg(q) + 1. Found by polynomial long division; the quotient (without remainder) is the asymptote.

Example: f(x) = (x²+1)/(x−1): divide to get y = x+1 as oblique asymptote

Workbook Quiz

Interactive Practice — 5 Questions

1

What is the domain of f(x) = (x − 1) / (x² − x − 6)?

2

For f(x) = (x² − 1) / ((x − 1)(x + 2)), what feature occurs at x = 1?

3

What is the horizontal asymptote of f(x) = (5x³ − 2) / (3x³ + x)?

4

Which function has an oblique (slant) asymptote?

5

Which rational function has NO x-intercepts?

Independent Practice

Independent Practice

1

Find the domain of f(x) = (x + 4) / (x² − 16).

2

Identify all vertical asymptotes and holes of g(x) = (x² − 5x + 6) / (x² − 4).

3

State the horizontal asymptote of h(x) = (7x⁴ − 3) / (2x⁴ + x).

4

Find all intercepts of f(x) = (x² − 3x) / (x² + x − 2).

5

Find the oblique asymptote of f(x) = (x² − 2x + 5) / (x + 1).

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Common Mistakes

Labeling every zero of the denominator as a vertical asymptote.

First cancel common factors with the numerator. If (x − a) cancels, x = a is a HOLE, not a VA. Only uncanceled zeros of the denominator are VAs.

Concluding there is no horizontal asymptote whenever deg(p) = deg(q).

When degrees are equal, the HA is y = (leading coeff of p) / (leading coeff of q). Only when deg(p) {">"} deg(q) is there no HA.

Forgetting to plot the hole as an open circle on the graph.

A hole is a removable discontinuity — the function is undefined at that point. Always mark it with an open circle at the correct (x, y) coordinate.

Using the original (unfactored) function to find the y-coordinate of a hole.

Substitute the hole x-value into the SIMPLIFIED (canceled) form of the function to find the y-coordinate of the hole.

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Math Tips

✂️

Always factor and cancel before analyzing — it reveals holes and simplifies finding asymptotes and intercepts.

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The degree comparison shortcut for HA: "bottom heavy → y=0, equal → ratio, top heavy → oblique."

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To find the y-coordinate of a hole, substitute the hole x-value into the simplified (reduced) rational function, not the original.

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Sign analysis between vertical asymptotes and intercepts tells you whether the curve is above or below the x-axis in each region — essential for accurate sketching.

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A rational function can cross its horizontal asymptote at a finite x-value — the HA only describes end behavior, not the entire graph.