NBT Mathematics: Functions and Graphs 4

Question 1

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?

Accepted Answer

x < 2 or x > 3

Answer

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 < 2). So the domain is x < 2 or x > 3.

Question 2

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?

Accepted Answer

a + d = 0

Answer

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.

Question 3

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

Accepted Answer

48

Answer

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.

Question 4

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

Accepted Answer

f(x) = cos(2x)

Answer

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.

Question 5

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

Accepted Answer

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

Answer

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.

(NBT) National Benchmark Test Practice Test

(NBT) National Benchmark Test Practice Test

NBT Mathematics: Functions and Graphs 4