The functions f(x) = x2 – 1 and g(x) = –x2 + 4 are shown on the graph.

The graph shows f of x equals x squared minus 1, which is an upward opening parabola with a vertex at 0 comma negative 1 and a point at negative 1 comma 0 and a point at 1 comma 0. The graph also shows g of x, which is a downward opening parabola with a vertex at 0 comma 4 and a point at negative 1 comma 3 and a point at 1 comma 3.

If the equations were changed to the inequalities shown, explain how the graph would change.

y > x2 – 1
y ≤ –x2 + 4

To analyze how the graph changes when the equations are modified to inequalities, we need to understand the regions defined by these inequalities. 

### Original Functions
1. **f(x) = x² – 1**: This is an upward-opening parabola with a vertex at \((0, -1)\) and intercepts at \((1, 0)\) and \((-1, 0)\).
2. **g(x) = –x² + 4**: This is a downward-opening parabola with a vertex at \((0, 4)\) and points at \((1, 3)\) and \((-1, 3)\).

### Inequalities
1. **y > x² – 1**: This inequality represents the region above the parabola \(f(x) = x² – 1\). 
2. **y ≤ –x² + 4**: This inequality represents the region below or on the parabola \(g(x) = –x² + 4\).

### Changes to the Graph
- **For \(y > x² – 1\)**:
  - Instead of the curve of \(f(x)\) itself, the graph will show the area above this parabola.
  - The line \(y = x² – 1\) will be dashed to indicate that points on this curve are not included in the solution set.
  - The region above the curve (y > x² – 1) will be shaded to represent all points where \(y\) is greater than \(x² – 1\).

- **For \(y ≤ –x² + 4\)**:
  - Instead of the curve of \(g(x)\) itself, the graph will show the area below this parabola including the curve.
  - The line \(y = –x² + 4\) will be solid to indicate that points on this curve are included in the solution set.
  - The region below the curve (y ≤ –x² + 4) will be shaded to represent all points where \(y\) is less than or equal to \(–x² + 4\).

### Combined Graph
When we combine these inequalities on the same graph:
- The area above the dashed parabola \(y = x² – 1\) will be shaded.
- The area below the solid parabola \(y = –x² + 4\) will be shaded.

The final graph will show:
- The dashed curve \(y = x² – 1\) with the region above it shaded.
- The solid curve \(y = –x² + 4\) with the region below it shaded.

### Key Points to Note
- The intersection of these two shaded regions represents the solution set where both inequalities are true.
- The region between the two parabolas (if it exists) will not be shaded if it doesn't satisfy both inequalities simultaneously.

Would you like details or have any questions?

### Relative Questions:
1. What is the significance of using dashed and solid lines in graphing inequalities?
2. How do you find the points of intersection between two parabolas?
3. What are the steps to graph a quadratic inequality?
4. Can you explain how the vertex form of a quadratic function helps in graphing?
5. How do you determine which side of the parabola to shade for an inequality?
6. What are the standard forms of quadratic functions and how do they relate to their graphs?
7. How do changes in the coefficients of a quadratic equation affect its graph?
8. What are some real-world applications of graphing quadratic inequalities?

### Tip:
When graphing quadratic inequalities, always test a point not on the boundary curve to determine which region to shade.

Explore how quadratic inequalities y > x² – 1 and y ≤ –x² + 4 alter the graphical representation of parabolas. Learn to interpret and graph these inequalities with detailed explanations and examples. Suitable for advanced high school students.

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