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Identify vertical and horizontal asymptotes

A vertical asymptote is a line x=ax=a that the graph approaches as function values increase or decrease without bound near an excluded or undefined input; a horizontal asymptote is a line y=by=b that describes the function’s end behavior as xx approaches positive or negative infinity. For rational functions, these can be identified from simplified factors and degree or leading-coefficient comparisons, while distinguishing vertical asymptotes from removable holes; formal advanced asymptotic analysis is not included.

Detailed Explanation: Identify vertical and horizontal asymptotes

To identify asymptotes of a rational function:

  1. Factor the numerator and denominator.
  2. Find vertical asymptotes:
    • Set the remaining denominator equal to zero.
    • A factor that cancels creates a hole, not a vertical asymptote.
  3. Find the horizontal asymptote by comparing the degrees and leading coefficients:
    • Same degree: divide the leading coefficients.
    • Numerator degree smaller: y=0y=0.
    • Numerator degree larger: there is no horizontal asymptote.

Consider the function

f(x)=x2−x−2x2+x−6.f(x)=\frac{x^2-x-2}{x^2+x-6}.

Step 1: Factor

Factor the numerator and denominator:

f(x)=(x−2)(x+1)(x−2)(x+3).f(x)=\frac{(x-2)(x+1)}{(x-2)(x+3)}.

The original denominator is zero when x=2x=2 or x=−3x=-3, so these values are excluded from the domain.

Step 2: Identify a hole and a vertical asymptote

Cancel the common factor (x−2)(x-2):

f(x)=x+1x+3,x≠2.f(x)=\frac{x+1}{x+3}, \qquad x\ne 2.

Because (x−2)(x-2) canceled, x=2x=2 creates a removable hole, not a vertical asymptote.

The remaining denominator is x+3x+3. Set it equal to zero:

x+3=0⇒x=−3.x+3=0 \quad\Rightarrow\quad x=-3.

Therefore, the vertical asymptote is

x=−3.\boxed{x=-3}.

As xx gets close to −3-3, the function values increase or decrease without bound.

Step 3: Identify the horizontal asymptote

Use the simplified function:

x+1x+3.\frac{x+1}{x+3}.

The numerator and denominator have the same degree. Compare their leading coefficients:

11=1.\frac{1}{1}=1.

Therefore, the horizontal asymptote is

y=1.\boxed{y=1}.

So the function has:

  • Vertical asymptote: x=−3\boxed{x=-3}
  • Horizontal asymptote: y=1\boxed{y=1}
  • Hole: at x=2x=2; its corresponding yy-value is 2+12+3=35\frac{2+1}{2+3}=\frac35, so the hole is (2,35)\left(2,\frac35\right).

Learn by doing: Identify vertical and horizontal asymptotes

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    Rational Functions and Asymptotes - Calculate Double Vertical Asymptotes (Expanded)

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