Question 3 • Precalculus
At t = −1, the equations give (x, y) = (−5, 6). At t = 0, they give (−4, 5), and at t = 2, they give (−2, 9). Since y = t² + 5 decreases until t = 0 and then increases, the curve passes down to (−4, 5) and then back up, matching the graph that starts at (−5, 6): choice B.
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