#### Answer

The sketch of $c(t)=(t^3-4t,t^2)$.

#### Work Step by Step

Step1. Use symmetric property of coordinate.
Notice that $x(t)=t^3-4t$ is an odd function and that $y(t)=t^2$ is an even function. This suggests that $c(t)$ is symmetric with respect to $y$-axis. Thus, we will plot the curve for $t\geq 0$ and reflect across the $y$-axis to obtain the part for $t\leq 0$.
Step 2. Analyze $x(t)$, $y(t)$ as functions of $t$.
The $y$-coordinate $y(t)=t^2$ increases to $\infty$ as $t\to \infty$. While the graph $x(t)=t^3-4t$ (Figure A) shows that
$x(t)<0$ for $00$ for $t>2$.
So, the curve starts at $c(0)=(0,0)$, moves to the left of $y$-axis, and returns to the $y$-axis at $t=2$. Afterwards, it moves to the right of $y$-axis.
Step 3. Plot points and join by an arc.
The points $c(0)$, $c(1)$, $c(2)$, $c(2.5)$ are tabulated below and plotted and joined by an arc to create the sketch for $t\geq 0$ as in Figure B.
$\left|
\begin{array}{ccc}
t & x(t)=t^3-4 t & y(t)=t^2 \\
0 & 0 & 0 \\
1 & -3 & 1 \\
2 & 0 & 4 \\
2.5 & 5.625 & 6.25 \\
\end{array}
\right|$
The completed sketch is obtained by reflecting across the $y$-axis as in Figure C.