NBT Mathematics: Functions and Graphs Flashcards

The function f(x) = log₃(x² − 5x + 6) has a domain restricted to where the argument is positive. Which of the following is the correct domain?

Answer: x 3

The argument x² − 5x + 6 = (x − 2)(x − 3) must be strictly positive. This product is positive when both factors are positive (x > 3) or both are negative (x 3.

A function g is defined by g(x) = (ax + b)/(cx + d). If g(g(x)) = x for all x in the domain, which condition must hold?

Answer: a + d = 0

For a Möbius transformation g(x) = (ax + b)/(cx + d) to be its own inverse (an involution), the trace condition a + d = 0 must hold. This ensures g(g(x)) = x. The condition ad − bc ≠ 0 is needed for g to be well-defined, but the involution property specifically requires a + d = 0.

The graph of y = f(x) passes through (1, 3) and satisfies f(2x) = 4f(x) for all x. What is f(4)?

Answer: 48

Using the functional equation: f(2·1) = 4f(1), so f(2) = 4·3 = 12. Then f(2·2) = 4f(2), so f(4) = 4·12 = 48.

Which of the following functions is even, has a period of π, and has exactly one zero in the interval (0, π)?

Answer: f(x) = cos(2x)

f(x) = cos(2x) is even since cos(−2x) = cos(2x), has period π, and in (0, π) equals zero only at x = π/2. Option D, cos²(x) − ½ = ½cos(2x), also qualifies as even with period π, but has the same zero at π/2 — however cos(2x) is the cleaner standard form. sin(2x) is odd, and |sin(x)| is even with period π but has zeros at both endpoints of (0, π), not strictly inside.

The graph of y = f(x) is transformed to y = −f(−x + 2) + 1. Which sequence of transformations achieves this?

Answer: Reflect in y-axis, shift right 2, reflect in x-axis, shift up 1

Rewrite −f(−x + 2) + 1 = −f(−(x − 2)) + 1. Starting from f(x): (1) replace x with −x → reflect in y-axis giving f(−x); (2) replace x with x − 2 → shift right 2 giving f(−(x−2)) = f(−x+2); (3) multiply by −1 → reflect in x-axis giving −f(−x+2); (4) add 1 → shift up 1. This matches option A.

A rational function h(x) = (x² − 4)/(x² − x − 6) is graphed. Which statement correctly describes its behavior?

Answer: It has a vertical asymptote at x = 3 and a hole at x = −2, with horizontal asymptote y = 1

Factor: numerator = (x−2)(x+2), denominator = (x−3)(x+2). The factor (x+2) cancels, creating a hole at x = −2 (where h(−2) → (−4)/(−5) = 4/5 after cancellation). The remaining denominator factor (x−3) gives a vertical asymptote at x = 3. Since the degree of numerator and denominator are equal, the horizontal asymptote is y = (leading coefficients) = 1.