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The problem asks to define the region \( R \) in Figure 3 using inequalities. To approach this, let's break down the given information and derive the necessary equations.

### Information Provided:
1. **Curve \( C \)**: The equation of \( C \) is \( y = f(x) \), where \( f(x) \) is a quadratic function in \( x \).  
   - \( C \) cuts the \( x \)-axis at \( x = -2 \) and \( x = 6 \), implying that these are the roots of the quadratic equation.
   
2. **Line \( l \)**: Line \( l \) cuts the \( y \)-axis at \( y = 8 \) and intersects the curve \( C \) at the point \( (4, -36) \).

3. **Objective**: Use inequalities to define the shaded region \( R \) bounded by the curve \( C \) and the line \( l \).

### Step-by-Step Solution:

1. **Equation of the Curve \( C \)**:
   Since the curve is a quadratic, we can express it as:
   \[
   f(x) = a(x + 2)(x - 6)
   \]
   because the roots are \( x = -2 \) and \( x = 6 \).

   To find the constant \( a \), use the fact that the curve passes through the point \( (4, -36) \). Substituting \( x = 4 \) and \( y = -36 \) into the equation:
   \[
   -36 = a(4 + 2)(4 - 6)
   \]
   \[
   -36 = a(6)(-2)
   \]
   \[
   -36 = -12a
   \]
   \[
   a = 3
   \]
   So, the equation of the curve \( C \) is:
   \[
   y = 3(x + 2)(x - 6)
   \]
   Expanding this:
   \[
   y = 3(x^2 - 4x - 12)
   \]
   \[
   y = 3x^2 - 12x - 36
   \]

2. **Equation of Line \( l \)**:
   The line cuts the \( y \)-axis at \( 8 \), so the equation is of the form:
   \[
   y = mx + 8
   \]
   To find the slope \( m \), use the point \( (4, -36) \) where the line intersects the curve:
   \[
   -36 = m(4) + 8
   \]
   \[
   -36 = 4m + 8
   \]
   \[
   4m = -44
   \]
   \[
   m = -11
   \]
   Thus, the equation of the line \( l \) is:
   \[
   y = -11x + 8
   \]

3. **Inequalities for the Region \( R \)**:
   The shaded region \( R \) lies between the curve \( C \) and the line \( l \). 
   - The curve is below the line, so the inequality for the region is:
   \[
   3x^2 - 12x - 36 \leq -11x + 8
   \]
   Simplifying this inequality:
   \[
   3x^2 - 12x - 36 \leq -11x + 8
   \]
   \[
   3x^2 - x - 44 \leq 0
   \]
   This inequality defines the region between the curve and the line.

4. **Domain for the Region**:
   The region \( R \) is bounded horizontally by the points where the curve cuts the \( x \)-axis, which are \( x = -2 \) and \( x = 6 \). Therefore, the domain for \( R \) is:
   \[
   -2 \leq x \leq 4
   \]
   since the region is shaded between \( x = -2 \) and \( x = 4 \), where the line intersects the curve.

### Final Inequalities:
Thus, the region \( R \) is defined by the system of inequalities:
\[
3x^2 - x - 44 \leq 0 \quad \text{and} \quad -2 \leq x \leq 4
\]

Would you like further details or clarifications on any part of the solution?

### Related Questions:
1. How would the inequality change if the line had a different slope?
2. What would be the effect on the region \( R \) if the curve had different roots?
3. How can you find the area of the region \( R \)?
4. What happens if the curve and line do not intersect?
5. How does the quadratic equation change if the curve is flipped upside down?

### Tip:
When solving inequalities involving quadratics, remember to check the signs of the expression over the intervals determined by the roots to ensure correct inequality direction.

Figure 3 shows a sketch of a curve C and a straight line l. Given that C has equation y = f(x) where f(x) is a quadratic expression in x, C cuts the x-axis at -2 and 6, l cuts the y-axis at 8 and intersects C at the point (4, -36), use inequalities to define the region R shown shaded in Figure 3.

Learn how to define the shaded region bounded by a quadratic curve and a straight line using inequalities. This solution involves finding the equations of both the curve and the line, and applying the relevant inequalities to define the region in the coordinate plane. Suitable for students in Grades 10-12.

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