The question asks to determine the range of the function $y = \frac{1}{4}x^2$ for two different domains:

1. $-4 \leq x \leq 2$
2. $2 \leq x \leq 6$

Step 1: Recall the function properties

The function $y = \frac{1}{4}x^2$ is a parabola that opens upwards, and its vertex is at $(0, 0)$.

Step 2: Analyze each domain

(1) For $-4 \leq x \leq 2$:

• Compute the values of $y$ at the endpoints:
  • At $x = -4$, $y = \frac{1}{4}(-4)^2 = \frac{1}{4}(16) = 4$.
  • At $x = 2$, $y = \frac{1}{4}(2)^2 = \frac{1}{4}(4) = 1$.
• The function $y = \frac{1}{4}x^2$ is symmetric about $x = 0$, so the minimum $y$-value in this range occurs at $x = 0$, which gives $y = 0$.
• The range is: $0 \leq y \leq 4$

(2) For $2 \leq x \leq 6$:

• Compute the values of $y$ at the endpoints:
  • At $x = 2$, $y = \frac{1}{4}(2)^2 = 1$.
  • At $x = 6$, $y = \frac{1}{4}(6)^2 = \frac{1}{4}(36) = 9$.
• The function increases in this interval, so the minimum $y$-value is at $x = 2$ and the maximum is at $x = 6$.
• The range is: $1 \leq y \leq 9$

Final Answer:

1. For $-4 \leq x \leq 2$, the range is $0 \leq y \leq 4$.
2. For $2 \leq x \leq 6$, the range is $1 \leq y \leq 9$.

Do you have any questions or need further clarification?

Here are 5 related questions you might explore:

1. How does the symmetry of the function affect the range calculation?
2. What would the range be if the domain was $-6 \leq x \leq 6$?
3. How do you determine the vertex of a parabola like $y = \frac{1}{4}x^2$?
4. What happens to the range if the coefficient of $x^2$ changes?
5. How does the range differ for a function like $y = -\frac{1}{4}x^2$?

Tip: Always substitute the endpoints of the domain into the function to determine the range efficiently.