The x y-coordinate plane is given. There is a curve, two lines, and a region on the graph.
The curve labeled y = x2 enters the window in the second quadrant, goes down and right becoming less steep, passes through the point (−2, 4) crossing the first line, changes direction at the origin, goes up and right becoming more steep, passes through the point (2, 4) crossing the second line, and exits the window in the first quadrant.
The first line labeled 3y = 2x + 16 enters the window in the second quadrant, goes up and right, passes through the point (−2, 4) crossing the curve, crosses the y-axis at approximately y = 5.3, passes through the point (1, 6) crossing the second line, and exits the window in the first quadrant.
The second line labeled y = 8 − 2x enters the window in the second quadrant, goes down and right, crosses the y-axis at y = 8, passes through the point (1, 6) crossing the first line, passes through the point (2, 4) crossing the curve, crosses the x-axis at x = 4, and exits the window in the fourth quadrant.
The region is above the curve and below the two lines.

From your description, we are dealing with a graphical setup involving three elements: a **parabolic curve**, two **linear equations**, and a **region** formed between these curves. Let’s break this down:

### Key Equations and Points:
1. **Parabola**: \( y = x^2 \)
   - This is a standard parabola opening upwards.
   - It passes through key points like \( (-2, 4) \), \( (0, 0) \), and \( (2, 4) \).

2. **First Line**: \( 3y = 2x + 16 \) or equivalently \( y = \frac{2}{3}x + \frac{16}{3} \)
   - This line has a positive slope and intersects the curve at \( (-2, 4) \).
   - It crosses the y-axis at approximately \( y \approx 5.3 \) and intersects the second line at \( (1, 6) \).

3. **Second Line**: \( y = 8 - 2x \)
   - This line has a negative slope and intersects both the first line and the parabola.
   - It crosses the y-axis at \( y = 8 \), intersects the first line at \( (1, 6) \), the parabola at \( (2, 4) \), and crosses the x-axis at \( x = 4 \).

### Region of Interest:
- The **region** is described as the area **above the parabola** \( y = x^2 \) and **below the two lines**. This forms a bounded region between the curves and the lines in the coordinate plane.

### Goal:
If you are interested in the **area of this region**, we need to compute the area between these curves. To do that, we need to:
1. Find the points of intersection between the curves and lines.
2. Set up an integral for the area between the parabola and the two lines.

### Intersection Points:
From the description, we know the following key points:
- \( (-2, 4) \): Intersection of the parabola and the first line.
- \( (1, 6) \): Intersection of the two lines.
- \( (2, 4) \): Intersection of the parabola and the second line.

### Finding the Area:
To compute the area of the region, we need to evaluate the following:
- Set up an integral from \( x = -2 \) to \( x = 1 \) for the area between the parabola \( y = x^2 \) and the first line \( y = \frac{2}{3}x + \frac{16}{3} \).
- Set up another integral from \( x = 1 \) to \( x = 2 \) for the area between the parabola \( y = x^2 \) and the second line \( y = 8 - 2x \).

We can now compute this area if you're interested in continuing with that step. Would you like to proceed with the calculations?

### 5 Relative Questions:
1. How do you find the points of intersection between a line and a parabola algebraically?
2. What is the significance of the slope in determining the shape and direction of a line?
3. Can you generalize the method for finding the area between any two curves?
4. How does changing the equation of the parabola affect the bounded region?
5. What are some other applications of finding areas between curves in real-world problems?

### Tip:
When solving problems involving areas between curves, always sketch the graph first to identify the points of intersection and visualize the region you're calculating.

The x y-coordinate plane is given. There is a curve, two lines, and a region on the graph. The curve labeled y = x2 enters the window in the second quadrant, goes down and right becoming less steep, passes through the point (−2, 4) crossing the first line, changes direction at the origin, goes up and right becoming more steep, passes through the point (2, 4) crossing the second line, and exits the window in the first quadrant. The first line labeled 3y = 2x + 16 enters the window in the second quadrant, goes up and right, passes through the point (−2, 4) crossing the curve, crosses the y-axis at approximately y = 5.3, passes through the point (1, 6) crossing the second line, and exits the window in the first quadrant. The second line labeled y = 8 − 2x enters the window in the second quadrant, goes down and right, crosses the y-axis at y = 8, passes through the point (1, 6) crossing the first line, passes through the point (2, 4) crossing the curve, crosses the x-axis at x = 4, and exits the window in the fourth quadrant. The region is above the curve and below the two lines.

Solve a problem involving a parabola y = x^2 and two lines, y = (2/3)x + 16/3 and y = 8 - 2x. This setup includes finding points of intersection and computing the area between curves using integration, suitable for students in Grades 10-12.

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