NBT Algebra and Functions Flashcards

Given f(x) = (2x + 1)/(x − 3) and g(x) = x² − 4, find the value(s) of x for which (f ∘ g)(x) is undefined.

Answer: x = ±2 and x = ±√7

f(g(x)) = f(x² − 4) = (2(x² − 4) + 1) / ((x² − 4) − 3) = (2x² − 7) / (x² − 7). This is undefined when g(x) puts us outside the domain of f, i.e. when x² − 4 = 3 → x² = 7 → x = ±√7, AND when g(x) itself is undefined — but g(x) = x² − 4 is a polynomial, so it's defined everywhere. However, f is also undefined when its own input causes division by zero: x² − 4 = 3 gives x = ±√7. There is no restriction from g itself. Wait — we also need to consider whether g(x) sends values outside f's domain. f is undefined when x − 3 = 0, i.e. input = 3, so g(x) = 3 → x² − 4 = 3 → x = ±√7. Additionally, g is undefined nowhere, but f(g(x)) requires the output of g to be a valid input for f. Thus (f ∘ g)(x) is undefined only when x² − 4 = 3, giving x = ±√7. Combined with no domain restriction on g, the answer should be x = ±√7. Re-checking the options — option D includes ±2 which would only apply if g were in a denominator. The correct set is x = ±√7 only, which is option C partially. Let me restate: (f ∘ g)(x) is undefined when g(x) = 3, i.e. x² − 4 = 3, x = ±√7. The answer is x = ±√7.

The function h(x) = (x² − 9) / (x² − x − 6) is simplified and its graph is drawn. Which statement correctly describes the graph?

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

Factor: numerator = (x − 3)(x + 3), denominator = (x − 3)(x + 2). The common factor (x − 3) cancels, leaving (x + 3)/(x + 2). The cancelled factor creates a hole at x = 3 (not an asymptote), while x = −2 still makes the denominator zero → vertical asymptote at x = −2. Since degrees of numerator and denominator are equal after cancellation, the horizontal asymptote is y = 1 (leading coefficients both 1).

If f(x) = log₂(x − 1) + 3, which transformation maps y = log₂(x) onto f(x), and what is the domain of f(x)?

Answer: Shift right 1 unit and up 3 units; domain: x > 1

Replacing x with (x − 1) shifts the graph 1 unit to the RIGHT (horizontal shift), and adding 3 outside the function shifts it 3 units UP. The basic log₂(x) has domain x > 0; after the horizontal shift right by 1, the argument (x − 1) must be positive: x − 1 > 0 → x > 1. So domain is x > 1.

A function is defined as f(x) = { 2x + k, x < 1 { x² + 3, x ≥ 1. For f to be continuous at x = 1, what must k equal, and what is f⁻¹(4) if it exists in the region x < 1?

Answer: k = 2, and f⁻¹(4) = 0.5

For continuity at x = 1, the left-hand limit must equal the right-hand value: lim(x→1⁻) f(x) = 2(1) + k = 2 + k must equal f(1) = 1² + 3 = 4. So 2 + k = 4 → k = 2. Now f(x) = 2x + 2 for x < 1. To find f⁻¹(4) in this region, solve 2x + 2 = 4 → x = 1, but x = 1 is NOT in the region x < 1 (it's excluded). However, x = 1 is the boundary — strictly f⁻¹(4) in the open region x < 1 approaches but never reaches 1, so it doesn't exist there. But at x = 1 the piece x² + 3 applies. Re-checking: the inverse in region x < 1 for value 4: 2x + 2 = 4 → x = 1, which is excluded, so f⁻¹(4) = 0.5 is wrong. Let me try f⁻¹(3): 2x+2=3 → x=0.5 < 1 ✓. The question asks f⁻¹(4) for x < 1 — x=1 is excluded, so it doesn't exist there. The best answer that correctly identifies k=2 and handles the inverse question is B (acknowledging the boundary issue), but technically f⁻¹(4) in x<1 doesn't exist since x=1 is excluded. Answer B is the closest correct option for k=2.

Given that the roots of 3x² − 7x + k = 0 are α and β, and α² + β² = 25/9, find the value of k.

Answer: k = 2

By Vieta's formulas: α + β = 7/3 and αβ = k/3. Use the identity α² + β² = (α + β)² − 2αβ = (7/3)² − 2(k/3) = 49/9 − 2k/3. Set equal to 25/9: 49/9 − 2k/3 = 25/9 → 2k/3 = 49/9 − 25/9 = 24/9 = 8/3 → k = (8/3)(3/2) = 4. Wait: 49/9 − 25/9 = 24/9 = 8/3, so 2k/3 = 8/3 → k = 4. The correct answer is k = 4.

The graph of y = a·bˣ passes through the points (2, 12) and (5, 96). What are the values of a and b?

Answer: a = 3, b = 2

From the two points: a·b² = 12 and a·b⁵ = 96. Dividing the second by the first: b³ = 96/12 = 8 → b = 2. Substituting back: a·4 = 12 → a = 3. Verify: at x = 5, 3·2⁵ = 3·32 = 96 ✓. So a = 3 and b = 2.