Reciprocal/rational (linear/linear); exponential and log functions practice questions.

State the equations of the asymptotes of the graph of f(x) = (2x + 1)/(x − 3).

• x = −3 and y = 2
• x = 2 and y = 3
• x = 3 and y = 2
• x = 3 and y = −1/3

Answer: x = 3 and y = 2

The vertical asymptote occurs where the denominator is zero, x = 3, and the horizontal asymptote is the ratio of leading coefficients, y = 2/1 = 2.

Which of the following functions has range y > 4?

• y = e^(x+4)
• y = eˣ + 4
• y = ln(x − 4)
• y = eˣ − 4

Answer: y = eˣ + 4

Since eˣ > 0 for all x, adding 4 shifts the range up to y > 4. In contrast, e^(x+4) is a horizontal shift and keeps range y > 0, while ln(x − 4) has range all real numbers.

Let f(x) = (x + 2)/(x − 1). Determine f⁻¹(x).

• f⁻¹(x) = (x − 1)/(x + 2)
• f⁻¹(x) = (x − 2)/(x + 1)
• f⁻¹(x) = (2 − x)/(x + 1)
• f⁻¹(x) = (x + 2)/(x − 1)

Answer: f⁻¹(x) = (x + 2)/(x − 1)

Setting y = (x + 2)/(x − 1) and solving for x gives x = (y + 2)/(y − 1), so the function is self-inverse: f⁻¹ = f. Taking the reciprocal instead of solving algebraically is the common trap.

Which single transformation description maps the graph of y = eˣ onto the graph of y = eˣ⁻² + 3?

• Translation 2 units right and 3 units up
• Translation 2 units left and 3 units up
• Translation 3 units right and 2 units up
• Translation 2 units right and 3 units down

Answer: Translation 2 units right and 3 units up

Replacing x by x − 2 shifts the graph 2 units to the right (opposite to the sign), and adding 3 outside shifts it 3 units up.

State the domain of the function f(x) = ln(x + 5).

• x > 5
• x ≥ −5
• x > −5
• x ∈ ℝ, x ≠ −5

Answer: x > −5

The logarithm is defined only for a positive argument: x + 5 > 0, so x > −5. The inequality is strict because ln 0 is undefined.

The graphs of y = eˣ and y = ln x are reflections of each other in which line?

• the x-axis
• the y-axis
• y = −x
• y = x

Answer: y = x

eˣ and ln x are inverse functions, and the graph of any inverse is the reflection of the original in the line y = x.