NBT Mathematics: Functions and Graphs Flashcards

A rational function is given by f(x) = (x² − x − 6) / (x² − 4). Which statement correctly identifies ALL features of its graph?

Answer: Vertical asymptote at x = 2 only; hole at x = −2; horizontal asymptote y = 1

Factor the numerator: x² − x − 6 = (x − 3)(x + 2). Factor the denominator: x² − 4 = (x − 2)(x + 2). The common factor (x + 2) cancels, producing a hole at x = −2 (not an asymptote). The remaining simplified function is (x − 3)/(x − 2), giving a vertical asymptote at x = 2. Because the original numerator and denominator have equal degree, the horizontal asymptote is y = 1 (ratio of leading coefficients).

Let f(x) = log₂(x + 4) and g(x) = 2ˣ − 4. What is (f ∘ g)(x) and its domain?

Answer: x, for all real numbers

(f ∘ g)(x) = f(g(x)) = log₂((2ˣ − 4) + 4) = log₂(2ˣ) = x. The argument of the logarithm simplifies to 2ˣ, which is strictly positive for every real x, so no domain restriction is introduced. f and g are inverses of each other, and their composition is the identity function on all real numbers.

The function f(x) = 3ˣ is used to define g(x) = f(x + 2) − f(x). Which of the following is equivalent to g(x)?

Answer: 8 · 3ˣ

g(x) = 3^(x+2) − 3ˣ. Using the exponent law: 3^(x+2) = 3ˣ · 3² = 9 · 3ˣ. Therefore g(x) = 9 · 3ˣ − 3ˣ = (9 − 1) · 3ˣ = 8 · 3ˣ.

The graph of f(x) = −2(x − 3)² + 8 is first reflected about the x-axis, then shifted 5 units upward to form g(x). What are the x-intercepts of g(x)?

Answer: x = 3 ± √(3/2)

Reflecting f(x) about the x-axis gives −f(x) = 2(x − 3)² − 8. Shifting up 5 units: g(x) = 2(x − 3)² − 8 + 5 = 2(x − 3)² − 3. Setting g(x) = 0: 2(x − 3)² = 3, so (x − 3)² = 3/2, giving x = 3 ± √(3/2).

The function f(x) = x² − 6x + 5 is restricted to the domain [1, 5]. What is the range of f on this interval, and on which sub-interval is f strictly decreasing?

Answer: Range [−4, 0]; decreasing on [1, 3]

Completing the square: f(x) = (x − 3)² − 4. The vertex is (3, −4), the minimum. Evaluate at the endpoints: f(1) = (1−3)² − 4 = 4 − 4 = 0 and f(5) = (5−3)² − 4 = 4 − 4 = 0. The minimum value is −4 (at x = 3) and the maximum is 0 (at both endpoints). Range = [−4, 0]. Since the parabola opens upward with axis of symmetry x = 3, f is strictly decreasing on [1, 3] and strictly increasing on [3, 5].

A function is defined piecewise: f(x) = x² − 4 for x < 1, and f(x) = 2x + k for x ≥ 1. For what value of k is f continuous at x = 1, and on which interval is the resulting function strictly decreasing?

Answer: k = −5; strictly decreasing on (−∞, 0)

For continuity at x = 1, the left-hand limit must equal f(1): lim(x→1⁻)(x² − 4) = 1 − 4 = −3, and f(1) = 2(1) + k = 2 + k. Setting equal: 2 + k = −3, so k = −5. With k = −5: for x 0 (always increasing). The function is strictly decreasing only on (−∞, 0).