The problem consists of two parts. Part a requires sketching the graph of the function $y = (1+x)(x-3)$. Part b asks to find the minimum value of the function. Then, we need to write four inequalities that define the unshaded region R on the XOY plane.

1. Problem Description

The problem consists of two parts.

Part a requires sketching the graph of the function $y = (1+x)(x-3)$.

Part b asks to find the minimum value of the function.

Then, we need to write four inequalities that define the unshaded region R on the XOY plane.

2. Solution Steps

Part a: Sketching the graph

The function is a quadratic function. We can expand it to find $y = x^2 - 2x - 3$.

The roots of the equation $y=0$ are $x=-1$ and $x=3$. The vertex of the parabola lies at the average of the roots, which is $x = \frac{-1+3}{2} = 1$.

The y-coordinate of the vertex is $y = (1+1)(1-3) = 2(-2) = -4$.

Therefore, the vertex of the parabola is at $(1,-4)$.

The parabola opens upwards since the coefficient of $x^2$ is positive.

Part b: Finding the minimum value

The minimum value of the quadratic function is the y-coordinate of the vertex. From part a, we found the vertex to be $(1,-4)$. Therefore, the minimum value is $-4$.

For the inequalities, we observe that the unshaded region is bounded by four lines.

Line 1: A horizontal line at $y=2$. Since the region is above this line, the inequality is $y > 2$.

Line 2: A vertical line at $x=-2$. Since the region is to the right of this line, the inequality is $x > -2$.

Line 3: A vertical line at $x=4$. Since the region is to the left of this line, the inequality is $x < 4$.

Line 4: An oblique line. The line passes through approximately $(0,6)$ and $(8,0)$. The slope is $\frac{0-6}{8-0} = -\frac{3}{4}$. The y-intercept is

6. Therefore, the equation of the line is $y = -\frac{3}{4}x + 6$. Since the region is below this line, the inequality is $y < -\frac{3}{4}x + 6$.

3. Final Answer

Part a: The graph is a parabola opening upwards with roots at $x=-1$ and $x=3$, and vertex at $(1, -4)$.

Part b: The minimum value is $-4$.

The four inequalities that define the unshaded region R are:

$y > 2$

$x > -2$

$x < 4$

$y < -\frac{3}{4}x + 6$