Answer
The answer is $c(t)=(t+2,3t+2)$.
Work Step by Step
At $t=0$, we have $(x(0),y(0))=(2,2)$.
We may take $x(t)=t+2$. So, $y(t)=3(t+2)-4=3t+2$.
Thus, the parametrization of $y=3 x-4$ is $c(t)=(t+2,3t+2)$ for $-\infty
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.1 Parametric Equations - Exercises - Page 603: 39
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