#### 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

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Published by
W. H. Freeman

ISBN 10:
1464125260

ISBN 13:
978-1-46412-526-3

The answer is $c(t)=(t+2,3t+2)$.

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