Topic 1.7 - Rational Functions and End Behavior

Case I: n = d

  • If the leading terms have the same degree, the horizontal asymptote is: y=aby = \frac{a}{b}

Case II: n < d

  • If the denominator dominates, the horizontal asymptote is: y=0y = 0

Case III: n > d

  • End behavior: f(x)abxndf(x) \sim \frac{a}{b} x^{n-d} (dominant term behaves like this polynomial)

  • Note: If the degree difference is 1 (n = d + 1), there is a slant (oblique) asymptote.

  • Slant asymptote direction: parallel to the line y=abxy = \frac{a}{b} x

Rational Function basics

  • Rational Function: y=f(x)g(x)y = \frac{f(x)}{g(x)} where f(x)f(x) and g(x)g(x) are polynomials and g(x)0g(x) \neq 0
  • Examples:
    • y=2x+3y = \frac{2}{x+3}
    • y=x33x+13x+4y = \frac{x^{3} - 3x + 1}{3x + 4}
    • y=2x3+4x6x27x+11y = \frac{2x^{3} + 4x - 6}{x^{2} - 7x + 11}

Example 1: Determine horizontal/slant end behavior

  • a) f(x)=3x2+4x75x23f(x) = \frac{3x^{2} + 4x - 7}{5x^{2} - 3}
    • Degrees equal ⇒ horizontal asymptote: y=35y = \frac{3}{5}
  • b) y=2x5x2+3x+2y = \frac{2x - 5}{x^{2} + 3x + 2}
    • Denominator degree > numerator ⇒ horizontal asymptote: y=0y = 0
  • c) g(x)=2x245x+9g(x) = \frac{2x^{2} - 4}{5x + 9}
    • Numerator degree 2, denominator degree 1 ⇒ slant asymptote
    • Slope of slant: ab=25\frac{a}{b} = \frac{2}{5}
    • Exact line (from long division): y=25x1825y = \frac{2}{5}x - \frac{18}{25}
  • d) y=4x+58x1y = \frac{4x + 5}{8x - 1}
    • Degrees equal ⇒ horizontal asymptote: y=48=12y = \frac{4}{8} = \frac{1}{2}
  • e) k(x)=3x2+3x7k(x) = \frac{3}{x^{2} + 3x - 7}
    • Numerator degree 0, denominator degree 2 ⇒ horizontal asymptote: y=0y = 0
  • f) p(x)=42x+1p(x) = \frac{-4}{2x + 1}
    • Horizontal asymptote: y=0y = 0

Example 2: End behavior limits

  • a) f(x)=2x4+4x16x4x3+4f(x) = \frac{2x^{4} + 4x - 1}{6x^{4} - x^{3} + 4}
    • Leading terms ratio: 26=13\frac{2}{6} = \frac{1}{3}
    • Thus: lim<em>xf(x)=lim</em>xf(x)=13\lim<em>{x \to \infty} f(x) = \lim</em>{x \to -\infty} f(x) = \frac{1}{3}
  • b) g(x)=5x48x+92x4+x1g(x) = \frac{5x^{4} - 8x + 9}{2x^{4} + x - 1}
    • Leading terms ratio: 52\frac{5}{2}
    • Thus: lim<em>xg(x)=lim</em>xg(x)=52\lim<em>{x \to \infty} g(x) = \lim</em>{x \to -\infty} g(x) = \frac{5}{2}
  • c) h(x)=3x2x2+xx2+4x+4h(x) = \frac{-3x^{2} - x^{2} + x}{x^{2} + 4x + 4}
    • Leading terms ratio: 31=3\frac{-3}{1} = -3
    • Thus: limx±h(x)=3\lim_{x \to \pm \infty} h(x) = -3
  • d) (content not provided in the transcript)
  • e) (content not provided in the transcript)

Example 3: Slant asymptote parallel to y = (1/2) x

  • A slant asymptote exists when the numerator degree is exactly one more than the denominator degree, and its slope is the leading ratio ab\frac{a}{b}
  • I. f(x)=x2+32x2+x+6f(x) = \frac{x^{2} + 3}{2x^{2} + x + 6} → Degrees equal ⇒ no slant
  • II. g(x)=x3+4x+12x5+x2+2g(x) = \frac{x^{3} + 4x + 1}{2x^{5} + x^{2} + 2} → Denominator degree 5 > numerator ⇒ no slant
  • III. h(x)=x2+3x+52x+4h(x) = \frac{x^{2} + 3x + 5}{2x + 4} → Degrees 2 and 1; n = d + 1 ⇒ slant; slope ab=12\frac{a}{b} = \frac{1}{2}
  • IV. k(x)=x2+x+52x2+x1k(x) = \frac{x^{2} + x + 5}{2x^{2} + x - 1} → Degrees equal ⇒ no slant
  • Answer: III only (option C).

Slant asymptotes recap

  • If the numerator degree is exactly 1 greater than the denominator degree, the slant asymptote is parallel to the line y=abxy = \frac{a}{b} x, where aa and bb are the leading coefficients of the numerator and denominator, respectively.