The problem is divided into two parts. The first part asks to sketch the graph of the function $y = (1+x)(x-3)$. The second part asks to find the minimum value of the function. The second problem asks to write the four inequalities that define the unshaded region R on the XOY plane.

1. Problem Description

The problem is divided into two parts. The first part asks to sketch the graph of the function $y = (1+x)(x-3)$. The second part asks to find the minimum value of the function. The second problem asks to write the four inequalities that define the unshaded region R on the XOY plane.

2. Solution Steps

Part 1:

a. Sketch the graph of $y = (1+x)(x-3)$.

First, expand the expression:

$y = x^2 - 3x + x - 3 = x^2 - 2x - 3$.

This is a quadratic equation of the form $y = ax^2 + bx + c$, where $a=1$, $b=-2$, and $c=-3$. Since $a>0$, the parabola opens upwards.

To find the x-intercepts, set $y=0$:

$(1+x)(x-3) = 0$.

So, $x = -1$ or $x = 3$.

To find the y-intercept, set $x=0$:

$y = (1+0)(0-3) = -3$.

To find the vertex (minimum point), use the formula $x_v = -\frac{b}{2a}$:

$x_v = -\frac{-2}{2(1)} = 1$.

Then, $y_v = (1+1)(1-3) = 2(-2) = -4$.

The vertex is at $(1, -4)$.

The graph is a parabola opening upwards, intersecting the x-axis at $x=-1$ and $x=3$, the y-axis at $y=-3$, and has a vertex at $(1, -4)$.

b. Find the minimum value.

The minimum value of the function is the y-coordinate of the vertex, which is $-4$.

Part 2:

From the graph, we can identify the four boundary lines and the region R.

The lines are:

1. A vertical line at $x=-2$. The region R is to the right of the line, so $x > -2$. (Since the region is unshaded on the other side of x = -2, it means x >= -2)

2. A vertical line at $x=4$. The region R is to the left of the line, so $x < 4$. (Since the region is unshaded on the other side of x = 4, it means x <= 4)

3. A horizontal line at $y=2$. The region R is above the line, so $y > 2$. (Since the region is unshaded on the other side of y = 2, it means y >= 2)

4. An oblique line. It passes through (0,6) and (8,2). The slope of the line is $m = \frac{2-6}{8-0} = \frac{-4}{8} = -\frac{1}{2}$. The equation of the line is $y = -\frac{1}{2}x + 6$. The region R is below the line, so $y < -\frac{1}{2}x + 6$. (Since the region is unshaded on the other side of the line, it means $y <= -\frac{1}{2}x + 6$)

3. Final Answer

Part 1:

a. The sketch is a parabola with x-intercepts at -1 and 3, y-intercept at -3, and vertex at (1, -4).

b. The minimum value is -

4.

Part 2:

The four inequalities are:

$x \ge -2$

$x \le 4$

$y \ge 2$

$y \le -\frac{1}{2}x + 6$