Log inSign up
AP Precalculus · Question of the Day

AP Precalculus Question of the Day

A fresh daily question to build accuracy, reinforce recall, and turn practice into a steady habit.
Monday, September 21, 2026

In Charleston, South Carolina, a simplified tide model uses h(t)=4.0sin⁡ ⁣(2π12.4(t−1.5))+5.8h(t)=4.0\sin\!\left(\frac{2\pi}{12.4}(t-1.5)\right)+5.8, where tt is hours after midnight and h(t)h(t) is tide height in feet. Here, 4.0 is the amplitude, 12.4 hours is the period, t−1.5t-1.5 represents the phase shift aligning a peak near 1:30 AM, and 5.8 is the vertical shift (mean level). Based on the data presented, if the phase shift was increased by 2 hours, how would the model change?

Keep practicing AP Precalculus

Question of the Day

Answer today's AP Precalculus question, reveal the full explanation, then keep the streak going with a new question every day.

In Charleston, South Carolina, a simplified tide model uses h(t)=4.0sin⁡ ⁣(2π12.4(t−1.5))+5.8h(t)=4.0\sin\!\left(\frac{2\pi}{12.4}(t-1.5)\right)+5.8, where tt is hours after midnight and h(t)h(t) is tide height in feet. Here, 4.0 is the amplitude, 12.4 hours is the period, t−1.5t-1.5 represents the phase shift aligning a peak near 1:30 AM, and 5.8 is the vertical shift (mean level). Based on the data presented, if the phase shift was increased by 2 hours, how would the model change?

  1. Replace (t−1.5)(t-1.5) with (t−3.5)(t-3.5). (correct answer)
  2. Replace (t−1.5)(t-1.5) with (t+0.5)(t+0.5).
  3. Replace 12.412.4 with 14.414.4.
  4. Replace +5.8+5.8 with +7.8+7.8.

Explanation: This question tests AP level understanding of sinusoidal functions and data modeling, specifically identifying or adjusting parameters like amplitude, period, phase shift, and vertical shift. Sinusoidal functions model periodic behavior where amplitude indicates the peak value, period the cycle length, phase shift the horizontal displacement, and vertical shift the baseline adjustment. In this scenario, the Charleston tide model has phase shift (t-1.5), placing a peak at t = 1.5 hours (1:30 AM), and increasing the phase shift by 2 hours means the peak should occur 2 hours later at t = 3.5 hours (3:30 AM). Choice A is correct because it accurately applies the phase shift change: replacing (t-1.5) with (t-3.5) moves the peak from 1:30 AM to 3:30 AM, demonstrating that increasing phase shift delays the wave. Choice B is incorrect because it decreases the phase shift by 2 hours (from 1.5 to -0.5), moving the peak earlier rather than later, a common error when students confuse the direction of phase shifts. To help students: Emphasize that in the form sin(t-h), increasing h moves the graph RIGHT (delays the wave). Practice with concrete examples showing how (t-2) peaks later than (t-1), and always verify by substituting the peak time.