Questions: Rational Functions and Asymptotes Review

Question 1

The function f(x) = (x-3)(x+1) / [(x-3)(x-7)] has what behavior at x = 3?

• A. A vertical asymptote, because x = 3 makes the denominator zero
• B. A hole (removable discontinuity), because (x-3) cancels from both numerator and denominator
• C. An x-intercept, because x = 3 makes the numerator zero
• D. No feature — the function equals zero there

Answer: (B)

When a factor appears in both numerator and denominator, it cancels — the simplified function has a finite limit at x = 3, but the original denominator is still zero there, so the point is excluded from the domain. This produces a hole, not a vertical asymptote. At x = 7, (x-7) remains in the denominator after cancellation, so the function grows without bound: that is a genuine vertical asymptote. The key distinction is cancellation: cancel → hole, no cancel → vertical asymptote.

Question 2

A student determines that f(x) = (5x² + 2)/(2x² - 1) has horizontal asymptote y = 5/2. She then argues the graph can never equal 5/2 for any finite x. Is she right?

• A. Yes — horizontal asymptotes are barriers the graph approaches but never crosses
• B. No — horizontal asymptotes describe limiting behavior as x → ±∞ and place no restriction on the graph's interior values
• C. Yes — because f(x) = 5/2 would require the denominator to be infinite
• D. Only correct if the function has no vertical asymptotes

Answer: (B)

Horizontal asymptotes describe end behavior — what f(x) approaches as x → ±∞ — but say nothing about the graph's values at finite x. A graph can cross, touch, or oscillate across its horizontal asymptote in the interior of the domain. The student has confused horizontal asymptotes (limiting behavior) with vertical asymptotes (true barriers where the function is undefined and never crossed).

Question 3

If the degree of the numerator polynomial is exactly 1 greater than the degree of the denominator, the rational function has no horizontal asymptote.

• T. True
• F. False

Answer: True

When deg(numerator) = deg(denominator) + 1, polynomial long division produces a linear quotient — this is an oblique (slant) asymptote, not a horizontal one. A horizontal asymptote requires deg(numerator) ≤ deg(denominator). If the degree difference is greater than 1, the function grows without bound and has neither type. Only equal degrees produce a nonzero horizontal asymptote; numerator degree less than denominator produces y = 0.

Question 4

A hole in a rational function's graph and a vertical asymptote both occur at x-values excluded from the domain, so they are the same type of discontinuity.

• T. True
• F. False

Answer: False

Both occur where the denominator is zero, but they are fundamentally different. A hole (removable discontinuity) occurs when a factor cancels from both numerator and denominator — the function approaches a finite limit there and could theoretically be 'repaired' by defining the value at that one point. A vertical asymptote occurs when a denominator factor does not cancel — the function grows without bound and cannot be made continuous there.

Question 5

Explain why a graph CAN cross its horizontal asymptote in the middle of its domain, even though the asymptote represents the function's long-run behavior.

Model answer: A horizontal asymptote is a statement about limits at infinity — it describes what f(x) approaches as x grows large. It places no restriction on f(x) at finite x values. The function may equal the asymptotic value y = L at some finite x₀ (i.e., f(x₀) = L) while still satisfying lim(x→∞) f(x) = L. Only vertical asymptotes are true barriers, because the function is undefined at those x-values.

Many students treat all asymptotes as uncrossable walls. Vertical asymptotes are genuinely uncrossable because the function doesn't exist there. Horizontal asymptotes describe tail behavior; the function is perfectly free to pass through that y-value finitely many times in the interior before its ends settle near the asymptote.