Bluebook SAT Test Interpreting Nonlinear Functions 5

Question 1

A ball is launched from a raised platform, and its height in feet above the ground is modeled by h(t) = −16t² + 96t + 112, where t is the number of seconds after launch. The ball strikes the ground exactly 7 seconds after launch. At what time, in seconds, does the ball reach its maximum height?

Accepted Answer

t = 3

Answer

The maximum of a downward-opening parabola occurs at the axis of symmetry, which lies midway between the two roots. Setting h(t) = 0: −16t² + 96t + 112 = 0 simplifies to t² − 6t − 7 = 0, giving (t − 7)(t + 1) = 0, so t = 7 or t = −1. The axis of symmetry is at t = (−1 + 7)/2 = 3. The common mistake is to take half of 7 (the landing time), but that ignores the second root at t = −1, which arises because the ball was launched from an elevated platform, not from ground level.

Question 2

The table below shows selected values of a function f.

x:    1,  2,  3,  4,  5
f(x): 6, 14, 26, 42, 62

Assuming f is a quadratic function, what is the value of f(6)?

Accepted Answer

86

Answer

The first differences are 8, 12, 16, 20 — not constant, so f is not linear. The second differences are all 4 (constant), confirming f is quadratic. Using the system of equations with f(1)=6, f(2)=14, f(3)=26 yields f(x) = 2x² + 2x + 2. Therefore f(6) = 2(36) + 2(6) + 2 = 72 + 12 + 2 = 86. The trap answer 82 comes from naively extending the last observed difference of 20 linearly (62 + 20 = 82), which ignores that the differences themselves are growing by 4 each step — so the next difference is 24, giving 62 + 24 = 86.

Question 3

The number of active users on a platform t months after launch is modeled by N(t) = 1,200 · 2^(t/4). Which of the following correctly describes the relationship between N(t) and N(t + 4) for any value of t?

Accepted Answer

N(t + 4) = 2 · N(t)

Answer

Computing the ratio: N(t + 4) / N(t) = [1200 · 2^((t+4)/4)] / [1200 · 2^(t/4)] = 2^((t+4)/4 − t/4) = 2^(4/4) = 2^1 = 2. So N(t + 4) = 2 · N(t) for any t — the user count doubles every 4 months regardless of the current value of t. Choice A is wrong because exponential growth is multiplicative, not additive. Choice B would imply quadrupling every 4 months. Choice D describes a different kind of growth entirely.

Question 4

Let g(x) = x² − 8x + 15. If g(2) = g(n) for some integer n where n ≠ 2, what is the value of n?

Accepted Answer

6

Answer

Rewriting in vertex form: g(x) = (x − 4)² − 1, so the axis of symmetry is x = 4. Because a parabola is symmetric about its axis, g(a) = g(b) whenever a and b are equidistant from the axis. The point x = 2 is 2 units to the left of x = 4, so the symmetric point is x = 4 + 2 = 6. Verification: g(2) = 4 − 16 + 15 = 3 and g(6) = 36 − 48 + 15 = 3. ✓ Choice A (n = 4) is the x-coordinate of the vertex, where g(4) = −1 ≠ 3. Choice D is wrong because the symmetry of any non-linear quadratic guarantees such a point exists.

Question 5

A company's annual revenue R(t), in thousands of dollars, is modeled by R(t) = −2t² + 20t + 50, where t is the number of years since founding. Between which pair of consecutive years does the revenue increase by the greatest amount?

Accepted Answer

Years 0 and 1

Answer

The year-over-year increase is R(t + 1) − R(t). Expanding: R(t+1) − R(t) = −2(t+1)² + 20(t+1) + 50 − (−2t² + 20t + 50) = −4t + 18. This expression is a decreasing linear function of t, meaning the annual increase is greatest when t is smallest — at t = 0 (years 0 to 1), where the gain is 18 thousand dollars. Many students intuitively pick years 4–5 (the year before the revenue peak at t = 5), but that interval has only a 2-thousand-dollar increase. For a concave-down parabola, the steepest gains occur at the beginning of the increasing phase, not near the peak.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Interpreting Nonlinear Functions 5