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Algebra and Functions Flashcards

6 cards from real Bluebook SAT Test practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 6 Algebra and Functions flashcards as text

  1. A function f satisfies f(2x) = 3f(x) − 2 for all real numbers x. If f(1) = 4, what is f(4)?

    Answer: 28

    Apply the functional equation step by step. f(2·1) = 3f(1) − 2 = 3(4) − 2 = 10, so f(2) = 10. Then f(2·2) = 3f(2) − 2 = 3(10) − 2 = 28, so f(4) = 28.

  2. The function f(x) = (x² − 9) / (x² − x − 6) has a removable discontinuity (hole) at x = a and a vertical asymptote at x = b. What is a − b?

    Answer: 5

    Factor numerator and denominator: x² − 9 = (x − 3)(x + 3) and x² − x − 6 = (x − 3)(x + 2). The common factor (x − 3) cancels, creating a hole at x = 3, so a = 3. The remaining denominator gives a vertical asymptote at x = −2, so b = −2. Therefore a − b = 3 − (−2) = 5.

  3. The parabola y = x² + 4x + c is tangent to the line y = 2x + 1 (meaning they intersect at exactly one point). What is the value of c?

    Answer: 2

    Set the equations equal: x² + 4x + c = 2x + 1, which gives x² + 2x + (c − 1) = 0. For exactly one intersection, the discriminant must equal zero: (2)² − 4(1)(c − 1) = 0 → 4 − 4c + 4 = 0 → 8 = 4c → c = 2. Verify: x² + 2x + 1 = (x + 1)² = 0 has exactly one solution at x = −1. ✓

  4. Let f(x) = 3x − 5. If f⁻¹(g(x)) = 2x + 1 for all x, what is g(7)?

    Answer: 40

    First find f⁻¹: if y = 3x − 5, then x = (y + 5)/3, so f⁻¹(y) = (y + 5)/3. Substituting into f⁻¹(g(x)) = 2x + 1: (g(x) + 5)/3 = 2x + 1 → g(x) + 5 = 6x + 3 → g(x) = 6x − 2. Therefore g(7) = 6(7) − 2 = 40.

  5. Consider f(x) = (3x² − 5x + 2) / (x² − 4). What is the sum of all x-values at which f(x) is either undefined or equal to zero?

    Answer: 5/3

    f(x) is undefined when x² − 4 = 0, giving x = 2 and x = −2. f(x) = 0 when the numerator equals zero: 3x² − 5x + 2 = 0 → (3x − 2)(x − 1) = 0 → x = 2/3 and x = 1. Note: neither 2/3 nor 1 make the denominator zero, so both are valid zeros. The sum is 2 + (−2) + 2/3 + 1 = 1 + 2/3 = 5/3.

  6. If 4ˣ + 4⁻ˣ = 7, what is the value of 8ˣ + 8⁻ˣ?

    Answer: 18

    Let u = 2ˣ + 2⁻ˣ. Then u² = 4ˣ + 2(2ˣ)(2⁻ˣ) + 4⁻ˣ = (4ˣ + 4⁻ˣ) + 2 = 7 + 2 = 9, so u = 3 (since u > 0). Now use the sum of cubes identity: 8ˣ + 8⁻ˣ = (2ˣ)³ + (2⁻ˣ)³ = (2ˣ + 2⁻ˣ)³ − 3(2ˣ)(2⁻ˣ)(2ˣ + 2⁻ˣ) = u³ − 3(1)(u) = 27 − 9 = 18.