MathIsimple
Lesson 1-2: Rational Functions & Asymptotes

Rational Functions: Domain, Asymptotes, and Graphing

Learn to analyze rational functions f(x)=P(x)Q(x)f(x)=\tfrac{P(x)}{Q(x)} with a focus on domain constraints, removable discontinuities (holes), vertical/horizontal/oblique (slant) asymptotes, and comprehensive graphing strategies. Apply these tools to real-world modeling problems.

Learning Objectives

Domain & Removable Discontinuities

Determine domains, detect holes from common factors, and annotate undefined points

Asymptote Identification

Find vertical, horizontal, and oblique asymptotes with rigorous methods

Graphing Strategies

Synthesize intercepts, asymptotes, sign charts, and end behavior for accurate graphs

Applications

Model efficiency, average cost, and rates with rational functions

Definition, Domain, and Simplest Form

Definition

A rational function is any function of the form f(x)=P(x)Q(x)f(x)=\tfrac{P(x)}{Q(x)} where P(x)P(x) and Q(x)Q(x) are polynomials andQ(x)0Q(x) \ne 0. The domain consists of all real numbers except the roots of Q(x)Q(x).

Simplest Form & Holes

When P(x)P(x) and Q(x)Q(x) share a common factor, canceling yields a removable discontinuity (hole). The simplified expression is equal to the original function on its domain, but the original function is undefined at the canceled factor's root.

Example: f(x)=x+2x24=x+2(x2)(x+2)=1x2f(x)=\tfrac{x+2}{x^2-4} = \tfrac{x+2}{(x-2)(x+2)} = \tfrac{1}{x-2} with domain x±2x \ne \pm 2

  • • Vertical asymptote at x=2x=2
  • • Hole at x=2x=-2 (removable discontinuity)

Asymptote Types and Methods

Vertical Asymptotes

Vertical asymptotes occur at roots of Q(x)Q(x) where P(x)0P(x) \ne 0. If a factor cancels, it yields a hole instead of an asymptote.

Example: g(x)=2xx29g(x)=\tfrac{2x}{x^2-9} has vertical asymptotes at x=3x=3 and x=3x=-3.

Horizontal Asymptotes

Use degree comparison between numerator and denominator:

  • • deg(P) < deg(Q): y=0y=0
  • • deg(P) = deg(Q): y=anbny=\tfrac{a_n}{b_n} (ratio of leading coefficients)
  • • deg(P) > deg(Q): No horizontal asymptote (consider oblique asymptote)

Example: h(x)=2x2+53x21h(x)=\tfrac{2x^2+5}{3x^2-1} has horizontal asymptote y=23y=\tfrac{2}{3}.

Oblique (Slant) Asymptotes

When deg(P)=deg(Q)+1\deg(P) = \deg(Q)+1, perform polynomial long division to writef(x)=ax+b+R(x)Q(x)f(x) = ax + b + \tfrac{R(x)}{Q(x)}, where y=ax+by=ax+b is the oblique asymptote and R(x)R(x) is the remainder withdeg(R)<deg(Q)\deg(R) < \deg(Q).

Example: f(x)=x2+4x+3x1=x+5+8x1f(x)=\tfrac{x^2+4x+3}{x-1} = x+5+ \tfrac{8}{x-1} has oblique asymptote y=x+5y=x+5.

Real-World Modeling with Rational Functions

Concentration Dilution

Concentration under dilution: C=msolutemsolution+wC=\tfrac{m_{solute}}{m_{solution}+w}, where ww is added water mass. As w+w \to +\infty,C0C \to 0 forming a horizontal asymptote at y=0y=0.

Average Cost Function

With fixed cost FF and variable cost per unit vv, average cost isAC(x)=F+vxx=v+FxAC(x)=\tfrac{F+vx}{x} = v + \tfrac{F}{x} for x>0x>0. As x+x \to +\infty,AC(x)vAC(x) \to v—a horizontal asymptote representing long-run average cost.

Worked Example: Domain, Asymptotes, and Trends

Function f(x)=x23x+2x3f(x)=\tfrac{x^2-3x+2}{x-3}

1) Domain

Exclude x=3x=3 since the denominator is zero.

2) Vertical Asymptote

At x=3x=3 because P(3)0P(3) \ne 0.

3) Oblique Asymptote

Long division: x23x+2x3=x+2x3\tfrac{x^2-3x+2}{x-3} = x + \tfrac{2}{x-3}

Oblique asymptote: y=xy=x.

4) Trend Near Asymptote

  • • As x3+x \to 3^{+}, 2x3+\tfrac{2}{x-3} \to +\infty, so f(x)+f(x) \to +\infty
  • • As x3x \to 3^{-}, 2x3\tfrac{2}{x-3} \to -\infty, so f(x)f(x) \to -\infty
  • • As x+x \to +\infty, f(x)xf(x) \to x (approaches oblique asymptote)

Practice Problems

Problem 1: Domain & Asymptotes

For f(x)=x24x21f(x)=\tfrac{x^2-4}{x^2-1}, find:

  • • Domain
  • • Vertical asymptotes and holes
  • • Horizontal or oblique asymptote

Problem 2: Average Cost

A factory has fixed cost 8000 and variable cost 20 per unit. Average cost function:AC(x)=20+8000xAC(x)=20+\tfrac{8000}{x} for x>0x>0. Compute AC(100),AC(500),AC(1000)AC(100), AC(500), AC(1000), and state the long-run trend.

Problem 3: Slant Asymptote

For f(x)=3x2+x2x+1f(x)=\tfrac{3x^2+x-2}{x+1}:

  • • Perform long division and find the oblique asymptote
  • • Determine behavior as x1±x \to -1^{\pm}
  • • Identify any holes

Common Pitfalls & How to Avoid Them

  • • Cancelling common factors and forgetting to record holes at the cancelled roots.
  • • Misidentifying horizontal vs. oblique asymptotes when deg(P) > deg(Q).
  • • Assuming graphs cannot cross a horizontal asymptote (they can); asymptotes describe end behavior.
  • • Ignoring domain restrictions in applications (e.g., negative time or quantity).
  • • Using approximate division and introducing algebraic errors; prefer exact long division.

Why These Asymptote Rules Work (Sketches)

Horizontal Asymptote via Degree Comparison

Write f(x)=P(x)Q(x)f(x)=\tfrac{P(x)}{Q(x)} and factor out highest powers: f(x)=anxn+cdotsbmxm+cdotsf(x)=\tfrac{a_n x^{n}+ cdots}{b_m x^{m}+ cdots}. If n < m, f(x)o0f(x) o 0. If n = m, f(x)oanbmf(x) o \tfrac{a_n}{b_m}. If n > m, the quotient grows like xnmx^{n-m}, so no horizontal asymptote.

Oblique Asymptote via Division Algorithm

Polynomial division guarantees P(x)=Q(x)S(x)+R(x)P(x) = Q(x)\cdot S(x) + R(x) with \deg R &lt; \deg Q. Then f(x)=S(x)+R(x)Q(x)f(x)=S(x)+\tfrac{R(x)}{Q(x)}. As xo|x| o \infty, the remainder term vanishes, so the graph approaches y=S(x)y=S(x) (linear if degP=degQ+1\deg P = \deg Q + 1).

Additional Worked Examples

Example A

Analyze f(x)=x21x24x+4f(x)=\tfrac{x^2-1}{x^2-4x+4}.

  • • Domain: exclude Q(x)=(x2)2=0Rightarrowx=2Q(x)=(x-2)^2=0 Rightarrow x=2.
  • • Vertical asymptote: x=2x=2 (no cancellation).
  • • Horizontal asymptote: deg equal ⇒ y=1y=1.
  • • Intercepts: solve numerator and set denominator nonzero.

Example B

Analyze f(x)=x31x21f(x)=\tfrac{x^3-1}{x^2-1}, simplify, and classify discontinuities.

  • • Factor: x31=(x1)(x2+x+1), x21=(x1)(x+1)x^3-1=(x-1)(x^2+x+1),\ x^2-1=(x-1)(x+1).
  • • Cancel x1x-1 ⇒ hole at x=1x=1; remaining denominator x+1x+1 ⇒ vertical asymptote at x=1x=-1.
  • • End behavior: deg(P) > deg(Q) ⇒ oblique asymptote; perform division to find it.

Practice Problems with Brief Solutions

1) Find all asymptotes of f(x)=3x22x+7x2+1f(x)=\tfrac{3x^2-2x+7}{x^2+1}.

Solution: No vertical asymptotes (denominator never 0); horizontal asymptote y=3y=3; no oblique asymptote.

2) Determine holes and vertical asymptotes for f(x)=x29x26x+9f(x)=\tfrac{x^2-9}{x^2-6x+9}.

Solution: Factor: (x3)(x+3)/(x3)2(x-3)(x+3)/(x-3)^2 ⇒ hole at x=3x=3 after cancellation, remaining denominator x3x-3 yields no asymptote at the same point (it became a hole); check simplified form carefully to confirm.

3) Long-run behavior of AC(x)=v+FxAC(x)=v+\tfrac{F}{x}.

Solution: As xo+x o +\infty, AC(x)ovAC(x) o v, so y=vy=v is a horizontal asymptote.

FAQ

Q: Can a rational function cross its horizontal asymptote?

A: Yes. Horizontal asymptotes describe end behavior, not a barrier.

Q: Is a cancelled factor always a hole?

A: Yes, at the cancelled root, provided the simplified function remains finite there.

Graphing Checklist

  1. Determine domain (exclude zeros of Q(x)). Record any cancelled-factor holes.
  2. Find intercepts (x-intercepts: P(x)=0 with Q(x) ≠ 0; y-intercept: f(0) if defined).
  3. Locate vertical asymptotes (non-cancelled denominator roots). Analyze side limits.
  4. Identify horizontal or oblique asymptotes via degree comparison or long division.
  5. Build a sign chart to determine where f(x) is positive/negative.
  6. Sketch respecting asymptotes, holes, and intercepts; confirm end behavior.

Case Studies

Case C: Crossing H-Asymptote

Show that f(x)=x21x2+1f(x)=\tfrac{x^2-1}{x^2+1} crosses y=1y=1.

Solve x21x2+1=1\tfrac{x^2-1}{x^2+1}=1x21=x2+1x^2-1=x^2+1 impossible; choose another function x2+xx2+1\tfrac{x^2+x}{x^2+1}, solve equals 1 to see crossing can occur with suitable P(x).

Case D: Multiple Vertical Asymptotes

For f(x)=2x23(x1)(x+2)f(x)=\tfrac{2x^2-3}{(x-1)(x+2)}, analyze behavior near x=1 and x=-2; use side limits and sign chart to assemble a precise sketch.

Practice Set II (with brief solutions)

4) f(x)=x24xx21f(x)=\tfrac{x^2-4x}{x^2-1}: domain, asymptotes, intercepts.

Domain: x ≠ ±1. H-asymptote y=1 (equal degree). V-asymptotes x=±1. Intercepts: x=0 and x=4.

5) Identify hole/asymptote for x21x1\tfrac{x^2-1}{x-1}.

Cancels to x+1 with hole at x=1; no vertical asymptote at x=1 in simplified view; graph is line with a missing point.

6) Oblique asymptote of 2x2+3x5x2\tfrac{2x^2+3x-5}{x-2}.

Division: 2x+7+9x22x+7+ \tfrac{9}{x-2} ⇒ slant asymptote y=2x+7.

Mini Projects

  • • Build a parameterized family fk(x)=x2+kx+1x1f_k(x)=\tfrac{x^2+kx+1}{x-1}. For which k does the graph have a specified y-intercept and given oblique asymptote?
  • • Model average speed v(x)=Dt0+αxv(x)=\tfrac{D}{t_0+\alpha x} as distance D is fixed and time increases with congestion level x; analyze sensitivity as x increases.

Quick Reference

  • • deg(P) < deg(Q): H-asymptote y=0
  • • deg(P) = deg(Q): H-asymptote y=leading ratio
  • • deg(P) = deg(Q)+1: Slant asymptote via division
  • • deg(P) > deg(Q)+1: Asymptote may be higher-degree polynomial y=S(x)
  • • Cancelled factor: hole at the cancelled root
  • • Non-cancelled denominator root: vertical asymptote

Challenge Problems (sketch solutions)

7) Show that f(x)=x32x2+1x21f(x)=\tfrac{x^3-2x^2+1}{x^2-1} has an oblique asymptote and a hole.

Factor, cancel where possible, then divide to find S(x). Mark the cancelled root as a hole.

8) For f(x)=(x1)2(x1)(x+2)f(x)=\tfrac{(x-1)^2}{(x-1)(x+2)}, classify discontinuities and asymptotes.

Cancel (x-1) to see hole at x=1; remaining denominator x+2 gives vertical asymptote at x=-2.

9) Build a rational function with H-asymptote y=2 and V-asymptotes x=±3.

Template: f(x)=2(x29)+kx29f(x)=\tfrac{2(x^2-9)+k}{x^2-9}. Adjust k to control intercepts; domain excludes ±3.

Summary

  • Always start with domain and simplification.
  • Classify asymptotes by degree and division.
  • Use sign charts for behavior between critical points.
  • Annotate holes and never include them in solutions.