From the image, we have the parametric equations:

$x = 4t, \quad y = t + 6$

Part (a): Sketching the curve

To determine the direction of the curve as $t$ increases:

1. Solve for a relationship between $x$ and $y$ by eliminating $t$.

   • From $x = 4t$, solve for $t$: $t = \frac{x}{4}$.
   • Substitute $t = \frac{x}{4}$ into $y = t + 6$: $y = \frac{x}{4} + 6$ This is the equation of a straight line with slope $\frac{1}{4}$ and $y$-intercept $6$.

2. The curve's direction is determined by increasing $t$. As $t$ increases:

   • $x = 4t$: $x$ increases.
   • $y = t + 6$: $y$ increases as well.

Thus, the direction of the curve is from bottom-left to top-right, matching the lower-right graph in the image.

Part (b): Find the rectangular-coordinate equation

From above, the rectangular equation is:

$y = \frac{x}{4} + 6$

This is the required rectangular-coordinate equation.

Let me know if you'd like a detailed breakdown of the solution!

Questions for further exploration:

1. What are the steps for solving a parametric equation when the parameter appears in nonlinear terms?
2. How can you confirm the direction of a parametric curve numerically?
3. What is the geometric interpretation of eliminating a parameter from equations?
4. How does the slope in the rectangular equation relate to the parametric equations?
5. Can we parameterize this line using different equations?

Tip:

When sketching parametric curves, always calculate the direction by considering changes in $x$ and $y$ as $t$ increases!