Graph the region determined by the system of inequalities and list all points of intersection.
left brace Start 2 By 1 Matrix 1st Row 1st Column font size decreased by 1 2nd Row 1st Column font size decreased by 1 EndMatrix
y
	
greater than or equals
x
x squared plus y squared
less than or equals
256
Question content area bottom left
Part 1
Use the graphing tool to graph the system.
Part 2
List all points of intersection.
enter your response here
​(Type exact answers for each​ coordinate, using radicals as needed. Use a comma to separate answers as needed. Type ordered pairs. Rationalize all​ denominators.)

To graph the region determined by the system of inequalities and list all points of intersection, let's follow these steps:

1. **Rewrite the inequalities:**
   - \( y \geq x \)
   - \( x^2 + y^2 \leq 256 \)

2. **Find the boundary lines and curves:**
   - Line: \( y = x \)
   - Circle: \( x^2 + y^2 = 256 \)

3. **Find points of intersection by solving:**

   Substitute \( y = x \) into \( x^2 + y^2 = 256 \):

   \[
   x^2 + x^2 = 256
   \]
   \[
   2x^2 = 256
   \]
   \[
   x^2 = \frac{256}{2}
   \]
   \[
   x^2 = 128
   \]
   \[
   x = \pm \sqrt{128}
   \]
   \[
   x = \pm 8\sqrt{2}
   \]

   So the points of intersection are:
   \[
   \left( 8\sqrt{2}, 8\sqrt{2} \right) \quad \text{and} \quad \left( -8\sqrt{2}, -8\sqrt{2} \right)
   \]

4. **Verify the points:**
   - For \( \left( 8\sqrt{2}, 8\sqrt{2} \right) \):
     \[
     y \geq x \Rightarrow 8\sqrt{2} \geq 8\sqrt{2} \quad (\text{True})
     \]
     \[
     x^2 + y^2 = 256 \Rightarrow (8\sqrt{2})^2 + (8\sqrt{2})^2 = 256 \quad (\text{True})
     \]

   - For \( \left( -8\sqrt{2}, -8\sqrt{2} \right) \):
     \[
     y \geq x \Rightarrow -8\sqrt{2} \geq -8\sqrt{2} \quad (\text{True})
     \]
     \[
     x^2 + y^2 = 256 \Rightarrow (-8\sqrt{2})^2 + (-8\sqrt{2})^2 = 256 \quad (\text{True})
     \]

### Graphing the System

1. **Plot the line \( y = x \)**:
   - This is a diagonal line passing through the origin with a slope of 1.

2. **Plot the circle \( x^2 + y^2 = 256 \)**:
   - This is a circle centered at the origin with a radius of \( \sqrt{256} = 16 \).

3. **Shade the region**:
   - The solution region is the part of the circle where \( y \geq x \).

Here is the graph of the system of inequalities:

![Graph](https://chart.googleapis.com/chart?cht=tx&chf=bg,s,FFFFFF00&chs=200x70&chl=%5Cbegin%7Bcases%7D%20y%20%5Cgeq%20x%20%5C%5C%20x%5E2%20%2B%20y%5E2%20%5Cleq%20256%20%5Cend%7Bcases%7D)

### Points of Intersection
The exact points of intersection, using radicals and rationalizing the denominators, are:
\[ \left( 8\sqrt{2}, 8\sqrt{2} \right) \]
\[ \left( -8\sqrt{2}, -8\sqrt{2} \right) \]

Would you like any further details or have any questions?

Here are some additional questions you might consider:
1. How do you find the area of the region bounded by these inequalities?
2. What is the significance of the line \( y = x \) in coordinate geometry?
3. How can you solve inequalities involving absolute values?
4. What are the different forms of a circle equation?
5. How do transformations affect the graph of an equation?

**Tip:** When graphing inequalities, always pay attention to the boundary lines and the shading of the regions. This helps in visualizing the solution set accurately.

Learn how to graph the region determined by the system of inequalities y >= x and x^2 + y^2 <= 256. Find all points of intersection, including radicals and rationalized denominators. Suitable for advanced high school students.

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