Rational functions (extended) practice questions.

Determine the equation of the oblique asymptote of the graph of f(x) = (x² + 3x + 5)/(x − 1).

• y = x + 3
• y = x − 4
• y = x + 4
• y = x + 2

Answer: y = x + 4

Polynomial division gives x² + 3x + 5 = (x − 1)(x + 4) + 9, so f(x) = x + 4 + 9/(x − 1). As x → ±∞ the remainder term vanishes, giving the oblique asymptote y = x + 4.

State the equations of all asymptotes of the graph of f(x) = (2x + 1)/(x² − x − 6).

• x = 3, x = −2 and y = 0
• x = 3, x = −2 and y = 2
• x = −3, x = 2 and y = 0
• x = 3 and x = −2 only

Answer: x = 3, x = −2 and y = 0

The denominator factorises as (x − 3)(x + 2), giving vertical asymptotes x = 3 and x = −2. Since the degree of the numerator is less than that of the denominator, y = 0 is the horizontal asymptote.

The function f is defined by f(x) = (3x² − 12)/(x² − 1). Determine the equation of the horizontal asymptote of its graph.

• y = 0
• y = 3
• y = 12
• y = −12

Answer: y = 3

Numerator and denominator have equal degree, so the horizontal asymptote is the ratio of the leading coefficients: y = 3/1 = 3. Using the constant terms (12/1) is a common error.

Let f(x) = (x² + 1)/x, x ≠ 0. Determine the set of values that f(x) cannot take.

• y > 2 only
• y < −2 only
• −2 ≤ y ≤ 2
• −2 < y < 2

Answer: −2 < y < 2

Setting y = (x² + 1)/x gives x² − yx + 1 = 0, which has real solutions only when y² − 4 ≥ 0. Hence values with −2 < y < 2 are never attained, while y = ±2 are attained at x = ±1.

State the number of vertical asymptotes of the graph of f(x) = (2x − 3)/(x² + 4).

• 0
• 1
• 2
• 3

Answer: 0

Vertical asymptotes occur only where the denominator is zero, but x² + 4 > 0 for all real x. Hence the graph has no vertical asymptote.

The graph of f(x) = (x² + kx + 4)/(x − 2) has no vertical asymptote at x = 2. Determine the value of k.

• k = 4
• k = 2
• k = −4
• k = −2

Answer: k = −4

There is no vertical asymptote only if (x − 2) is also a factor of the numerator, so the numerator must vanish at x = 2: 4 + 2k + 4 = 0, giving k = −4. The graph then has a removable point (hole) instead.