Answer
The sketch of $c(t)=(t^2-4t,9-t^2)$ for $-4\leq t\leq 10$.
Work Step by Step
Notice that $x(t)=t^2-4t$ is neither odd nor even function but that $y(t)=9-t^2$ is an even function (Figure A).
The graph $x(t)=t^2-4t$ shows that
$x(t)<0$ for $00$ for $t<0$ or $t>4$.
For the interval $-4\leq t\leq 10$, the curve starts at $c(-4)=(32,-7)$, moves to the left of $y$-axis until it reaches $y$-axis at $t=0$, $c(0)=(0,9)$. Then it moves to the right of $y$-axis until it reaches $t=4$, $c(4)=(0,-7)$, and continues moving away from the $y$-axis until $t=10$ at $c(10)=(60,-91)$ (Figure B).
