You have shared a parametric equation problem. The equations are:

$x = 6 - t^2$ $y = \frac{t}{2}$

with the parameter $t$ in the domain $-2 \leq t \leq 4$.

Steps to graph the parametric equations:

1. Calculate key points by substituting different values of $t$ in the given range $[-2, 4]$.
2. Plot the points $(x(t), y(t))$.
3. Join the points smoothly to see the shape of the curve.

Let's calculate the values of $x$ and $y$ at key points of $t$:

For $t = -2$: $x = 6 - (-2)^2 = 6 - 4 = 2, \quad y = \frac{-2}{2} = -1$

For $t = -1$: $x = 6 - (-1)^2 = 6 - 1 = 5, \quad y = \frac{-1}{2} = -0.5$

For $t = 0$: $x = 6 - 0^2 = 6, \quad y = \frac{0}{2} = 0$

For $t = 1$: $x = 6 - 1^2 = 6 - 1 = 5, \quad y = \frac{1}{2} = 0.5$

For $t = 2$: $x = 6 - 2^2 = 6 - 4 = 2, \quad y = \frac{2}{2} = 1$

For $t = 3$: $x = 6 - 3^2 = 6 - 9 = -3, \quad y = \frac{3}{2} = 1.5$

For $t = 4$: $x = 6 - 4^2 = 6 - 16 = -10, \quad y = \frac{4}{2} = 2$

These points will give a curve when plotted on a coordinate plane.

Would you like a detailed graph of these parametric equations or have any further questions?

Here are 5 related questions you can explore:

1. What is the geometric shape formed by these parametric equations?
2. How would changing the domain of $t$ affect the graph?
3. Can you express $y$ as a function of $x$ by eliminating the parameter $t$?
4. How does the curve behave as $t \to \infty$ or $t \to -\infty$?
5. How do parametric equations differ from regular Cartesian equations?

Tip: When dealing with parametric equations, it’s helpful to choose symmetric values of $t$ (like negative and positive values) to better understand the curve's shape.