Algebra and Functions Flashcards

The function f is defined by f(x) = (x² − 9) / (x² − x − 6). Which of the following correctly describes the graph of f?

Answer: A hole at x = 3 and a vertical asymptote at x = −2

Factor both: numerator = (x−3)(x+3), denominator = (x−3)(x+2). The (x−3) factor cancels, producing a hole at x = 3. The remaining denominator factor (x+2) gives a vertical asymptote at x = −2. So there is a hole at x = 3 and a vertical asymptote at x = −2.

If f(x) = 2x + k and f(f(x)) = 4x + 9, what is the value of k?

Answer: 3

Compute f(f(x)): f(f(x)) = 2(2x + k) + k = 4x + 2k + k = 4x + 3k. Setting 4x + 3k = 4x + 9 gives 3k = 9, so k = 3.

The system of equations below has no solution. What is the value of c? 3x − 6y = 12 cx − 4y = 7

Answer: 2

A system has no solution when the lines are parallel — same slope, different y-intercepts. Rewrite the first equation: y = (1/2)x − 2. Rewrite the second: y = (c/4)x − 7/4. For parallel lines, c/4 = 1/2, so c = 2. Check that the y-intercepts differ (−2 ≠ −7/4), confirming no solution.

For all x > 0, which expression is equivalent to (x^(1/2) + x^(−1/2))² − (x^(1/2) − x^(−1/2))²?

Answer: 4

Use the difference of squares identity: (A+B)² − (A−B)² = 4AB, where A = x^(1/2) and B = x^(−1/2). So the expression equals 4 · x^(1/2) · x^(−1/2) = 4 · x^0 = 4.

A function g satisfies g(2x) = 3·g(x) for all real x, and g(1) = 5. What is g(8)?

Answer: 135

Apply the rule repeatedly: g(2·1) = 3·g(1), so g(2) = 15. Then g(2·2) = 3·g(2), so g(4) = 45. Then g(2·4) = 3·g(4), so g(8) = 135.

The graph of y = f(x) passes through (2, 5). The graph of y = f(3x − 6) + 4 passes through (a, 9). What is the value of a?

Answer: 8/3

We need f(3a − 6) + 4 = 9, so f(3a − 6) = 5. Since f(2) = 5, we set 3a − 6 = 2, giving 3a = 8, so a = 8/3.