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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