The parametric equations provided in the image are:

$x = g(t) = t^2 + t, \quad y = h(t) = 2t - 1, \quad -1 \leq t \leq 1.$

Steps to sketch the parametric curve:

1. Determine key points: Evaluate $x(t)$ and $y(t)$ at several values of $t$ within the range $-1 \leq t \leq 1$. This will help locate points on the curve.

   • For $t = -1$: $x(-1) = (-1)^2 + (-1) = 1 - 1 = 0, \quad y(-1) = 2(-1) - 1 = -2 - 1 = -3.$ So, the point is $(0, -3)$.
   • For $t = 0$: $x(0) = 0^2 + 0 = 0, \quad y(0) = 2(0) - 1 = -1.$ So, the point is $(0, -1)$.
   • For $t = 1$: $x(1) = 1^2 + 1 = 1 + 1 = 2, \quad y(1) = 2(1) - 1 = 2 - 1 = 1.$ So, the point is $(2, 1)$.

2. Find the general shape of the curve:

   • As $t$ increases, $x(t) = t^2 + t$ behaves like a quadratic, and $y(t) = 2t - 1$ is a linear equation.
   • The $x$-values are non-negative because $t^2 \geq 0$, and the curve bends upward due to the $t^2$-term in the $x$-equation.

3. Sketch the curve:

   • The curve starts at $(0, -3)$ when $t = -1$, passes through $(0, -1)$ at $t = 0$, and ends at $(2, 1)$ when $t = 1$.
   • As $t$ varies from $-1$ to $1$, the curve traces a smooth path.

Would you like a more detailed graphical representation or have any other questions?

Follow-up Questions:

1. How can we eliminate $t$ to find a Cartesian equation for this parametric curve?
2. What effect does the quadratic nature of $x(t)$ have on the curve's shape?
3. How does the linearity of $y(t)$ influence the appearance of the curve?
4. Can you identify the slope of the curve at different points by differentiating the parametric equations?
5. What would happen if the range of $t$ was extended beyond $-1 \leq t \leq 1$?

Tip: When sketching parametric curves, evaluating points at key values of $t$ (like $t = -1, 0, 1$) helps to identify the general shape of the curve.