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Quiz: Rational Functions

Test your understanding of rational functions, asymptotes, and holes with these review questions.

1. What is the vertical asymptote of \(f(x) = \frac{5}{x - 3}\)?

  1. \(x = 5\)
  2. \(x = 3\)
  3. \(x = -3\)
  4. \(y = 3\)
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The correct answer is B. A vertical asymptote occurs where the denominator equals zero (provided the numerator is nonzero there). Set \(x - 3 = 0\): \(x = 3\). This is a vertical line, not a horizontal one, so option D is incorrect notation. As \(x\) approaches \(3\) from either side, \(f(x)\) shoots off to \(\pm\infty\).

Concept Tested: Vertical Asymptote

2. What is the horizontal asymptote of \(f(x) = \frac{4x + 1}{2x - 5}\)?

  1. \(y = 0\)
  2. \(y = \frac{5}{2}\)
  3. \(y = 2\)
  4. \(y = 4\)
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The correct answer is C. When the numerator and denominator have the same degree, the horizontal asymptote is the ratio of leading coefficients: \(y = \frac{4}{2} = 2\). Option A would apply if the numerator had a lower degree. As \(x\) grows large in either direction, the lower-order terms become negligible, so \(f(x) \to \frac{4x}{2x} = 2\).

Concept Tested: Horizontal Asymptote

3. Which of the following describes a hole in a rational function's graph?

  1. A point where the function tends to infinity
  2. A horizontal line the graph approaches
  3. A single missing point caused by a common factor in numerator and denominator
  4. The \(y\)-intercept
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The correct answer is C. A hole (removable discontinuity) occurs when a factor appears in both the numerator and denominator. After cancellation, the function looks continuous everywhere except at that one excluded \(x\)-value, where there is a "missing dot." Option A describes a vertical asymptote, and option B describes a horizontal asymptote — these are different kinds of discontinuities.

Concept Tested: Holes in Rational Graphs

4. Find the hole (if any) in \(f(x) = \frac{x^2 - 9}{x - 3}\).

  1. No hole
  2. Hole at \((3, 6)\)
  3. Hole at \((-3, 0)\)
  4. Hole at \((3, 0)\)
Show Answer

The correct answer is B. Factor the numerator: \(\frac{(x - 3)(x + 3)}{x - 3}\). The factor \((x - 3)\) cancels, leaving \(f(x) = x + 3\) for \(x \neq 3\). At \(x = 3\), the simplified form gives \(y = 6\), so there is a hole at \((3, 6)\). The original function is undefined at \(x = 3\), but the graph otherwise looks like the line \(y = x + 3\).

Concept Tested: Holes in Rational Graphs

5. The graph of \(y = \frac{1}{x}\) is called a:

  1. Parabola
  2. Cubic
  3. Hyperbola
  4. Asymptote
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The correct answer is C. The reciprocal function \(y = \frac{1}{x}\) produces a curve with two separate branches — one in the first quadrant and one in the third quadrant — called a hyperbola. The two axes act as asymptotes that the graph approaches but never crosses. A parabola is the graph of a quadratic, and an asymptote is a line, not a curve.

Concept Tested: Hyperbola

6. For what value(s) of \(x\) is the function \(f(x) = \frac{2x + 5}{x + 4}\) undefined?

  1. \(x = -4\)
  2. \(x = 4\)
  3. \(x = -\frac{5}{2}\)
  4. \(x = 0\)
Show Answer

The correct answer is A. A rational function is undefined wherever its denominator equals zero. Set \(x + 4 = 0\), giving \(x = -4\). The function's domain excludes this single value. Option C is the \(x\)-intercept (where the numerator is zero), but that's a valid input — the function is defined there and equals \(0\).

Concept Tested: Rational Function

7. An oblique asymptote occurs in a rational function when:

  1. The degrees of numerator and denominator are equal
  2. The numerator's degree is exactly one more than the denominator's
  3. The numerator's degree is less than the denominator's
  4. The denominator has a repeated factor
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The correct answer is B. An oblique (slant) asymptote appears when the degree of the numerator is exactly one greater than the degree of the denominator. In this case, polynomial division produces a linear quotient, which becomes the oblique asymptote. Equal degrees give a horizontal asymptote, and smaller numerator degree gives the horizontal asymptote \(y = 0\).

Concept Tested: Oblique Asymptote

8. Find the \(y\)-intercept of \(f(x) = \frac{3x - 6}{x + 2}\).

  1. \((0, 3)\)
  2. \((0, 2)\)
  3. \((0, 6)\)
  4. \((0, -3)\)
Show Answer

The correct answer is D. The \(y\)-intercept is \(f(0) = \frac{3(0) - 6}{0 + 2} = \frac{-6}{2} = -3\), so the point is \((0, -3)\). Don't confuse this with the horizontal asymptote (\(y = 3\), which is the ratio of leading coefficients). A common mistake is to reverse the sign or to compute the \(x\)-intercept (which would be \(x = 2\), from setting the numerator to zero).

Concept Tested: Rational Function

9. What are the asymptotes of \(f(x) = \frac{x + 1}{x - 4}\)?

  1. Vertical: \(x = 4\); Horizontal: \(y = 1\)
  2. Vertical: \(x = -1\); Horizontal: \(y = 4\)
  3. Vertical: \(x = 4\); Horizontal: \(y = 0\)
  4. Vertical: \(x = 1\); Horizontal: \(y = 4\)
Show Answer

The correct answer is A. The vertical asymptote is found by setting the denominator to zero: \(x - 4 = 0\), so \(x = 4\). The horizontal asymptote is the ratio of leading coefficients (both have degree \(1\)): \(y = \frac{1}{1} = 1\). Together these lines form a cross-shape that the hyperbola branches avoid.

Concept Tested: Asymptotic Behavior

10. Which of the following is a rational function?

  1. \(f(x) = \sqrt{x + 1}\)
  2. \(f(x) = 2^x\)
  3. \(f(x) = \sin x\)
  4. \(f(x) = \frac{x^2 + 1}{x - 3}\)
Show Answer

The correct answer is D. A rational function is a ratio of two polynomials \(\frac{p(x)}{q(x)}\) where \(q(x)\) is not zero. Option D fits this definition exactly. Option A is a radical (square root) function, option B is an exponential function, and option C is a trigonometric function — none of these are rational functions, though they may have some similar features like domain restrictions.

Concept Tested: Rational Function