Bluebook SAT Test Algebra and Functions 4

Question 1

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

Accepted Answer

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

Answer

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.

Question 2

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

Accepted Answer

3

Answer

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.

Question 3

The system of equations below has no solution. What is the value of c?

3x − 6y = 12
cx − 4y = 7

Accepted Answer

2

Answer

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.

Question 4

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

Accepted Answer

4

Answer

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.

Question 5

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

Accepted Answer

135

Answer

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.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Algebra and Functions 4