Mathfolis

Rational Functions and Holes

Unit 1 · Polynomial and Rational Functions

What AP Precalc asks here

A hole (removable discontinuity) at x=cx = c requires that both numerator and denominator vanish AND share the factor (x−c)(x - c) with the numerator's multiplicity high enough to fully cancel the denominator's. After cancellation, the simplified function r~\tilde r is defined and continuous at cc — but the original rr is not. The hole is at (c,r~(c))(c, \tilde r(c)).

Hole vs vertical asymptote (factor (x−c)(x - c) shared)

Numerator multiplicity ≥\geq denominator multiplicity
Hole at (c,r~(c))\text{Hole at } (c, \tilde r(c))
Numerator multiplicity << denominator multiplicity
Vertical asymptote at x=c\text{Vertical asymptote at } x = c

Hole y-coordinate

Compute via the simplified function
y=r~(c), where r~ is r with the shared factor cancelledy = \tilde r(c) \text{, where } \tilde r \text{ is } r \text{ with the shared factor cancelled}
Caution: The simplified function r~\tilde r is not literally equal to rr. They differ at x=cx = c: r~(c)\tilde r(c) is defined; r(c)r(c) is not.
Type 1

Locate a hole and its y-coordinate

Cancel the shared factor in r(x)r(x) and substitute x=cx = c into the simplified r~\tilde r to find the y-coordinate of the hole.

Example 1
Identify the hole of r(x)=(x−3)(x+1)(x−3)(x−5)r(x) = \dfrac{(x - 3)(x + 1)}{(x - 3)(x - 5)}. (A) (3,−2)(3, -2) (B) (3,0)(3, 0) (C) (−1,0)(-1, 0) (D) (5,0)(5, 0)

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Type 2

Classify a shared factor as hole or vertical asymptote (VA)

Compare the multiplicity of the shared factor in the numerator and the denominator. If the numerator's is at least as high, the factor fully cancels in the denominator → hole. Otherwise, the denominator still vanishes → VA.

Example 2
For r(x)=x−2(x−2)2(x+3)r(x) = \dfrac{x - 2}{(x - 2)^2 (x + 3)}, classify x=2x = 2. (A) Hole at (2,0)(2, 0) (B) Hole at (2,−1/5)(2, -1/5) (C) Vertical asymptote at x=2x = 2 (D) Removable, with no discontinuity

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Type 3

Construct a rational with a specified hole

Start from a simplified r~(x)\tilde r(x) that passes through (c,k)(c, k), then introduce a shared factor (x−c)(x - c) in numerator and denominator to create a hole at x=cx = c.

Example 3
Which rational function has a hole at (1,5)(1, 5)? (A) (x−1)(x+4)x−1\dfrac{(x - 1)(x + 4)}{x - 1} (B) x+4x−1\dfrac{x + 4}{x - 1} (C) (x−1)(x−5)x−1\dfrac{(x - 1)(x - 5)}{x - 1} (D) x−1x+4\dfrac{x - 1}{x + 4}

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