x(t) = t3y(t) = t4
x(t) = t2 + 1y(t) = 3t
We'll use the y-equation first since we'll find only one t-value that way.
When y = -6 we must have t = -2. Then
x = (-2)2 + 1 = 5.
The point (5, -6) occurs on the graph when t = -2.
x(t)= t2 + 1y(t) = 3t
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