The answer is $c(t)=(t+3,t^2+6t+9)$.
Work Step by Step
At $t=0$, we have $(x(0),y(0))=(3,9)$. We may take $x(t)=t+3$. So, $y(t)=(t+3)^2=t^2+6t+9$. Thus, the parametrization of $y=x^2$ is $c(t)=(t+3,t^2+6t+9)$ for $-\infty
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