The answer is $c(t)=(t-1,3t-7)$.
Work Step by Step
At $t=3$, we have $(x(3),y(3))=(2,2)$. We may take $x(t)=t-1$. So, $y(t)=3(t-1)-4=3t-7$. Thus, the parametrization of $y=3 x-4$ is $c(t)=(t-1,3t-7)$ for $-\infty
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