Use the rubric on the exam page. Award method credit for a correct model carried through with a minor arithmetic error, but check domain exclusions, sign conventions and units explicitly.
Shift right 2, stretch vertically by 3, then shift down 12. Vertex (2,−12); zeros 0 and 4; range [−12,∞) .
Midline 2 V, amplitude 3 V, range [−1,5] V , frequency 4 Hz, period 0.25 s. Factoring the angle gives 8π(t−1/16), so the shift is right by 0.0625 s. At zero, v=−1 V.
Factor to (x−2)(x+2)/((x−2)(x+3)). Exclude 2 and −3. The reduced rule is (x+2)/(x+3)=1−1/(x+3). Hole (2,4/5), vertical asymptote x=−3, horizontal asymptote y=1. The angle is 2π/3.
I=9/1500=0.006 A=6 mA, P=9×0.006=0.054 W, and E=0.054×20=1.08 J.
The pair is 2 kΩ and the total is 3 kΩ. Source current is 4 mA and pair voltage 8 V. Branch currents are 8/3≈2.667 mA and 8/6≈1.333 mA, summing to 4 mA.
Minimum resistance is 900 Ω. Imax=10/900≈11.11 mA, Pmax=100/900≈0.1111 W. A 0.1 W rating is below this result and 0.25 W is above it. Manufacturer temperature derating and operating conditions still matter.