The problem has two parts. Part a: Sketch the graph of the function $y = (1+x)(x-3)$. Part b: Find the minimum value of the function. The second problem asks to define the unshaded region R on the XOY plane with four inequalities.

1. Problem Description

The problem has two parts.

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

Part b: Find the minimum value of the function.

The second problem asks to define the unshaded region R on the XOY plane with four inequalities.

2. Solution Steps

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

First, expand the expression:

$y = (x+1)(x-3) = x^2 - 3x + x - 3 = x^2 - 2x - 3$.

The graph is a parabola.

The roots of the equation $y = 0$ are $x = -1$ and $x = 3$.

The x-coordinate of the vertex is the midpoint of the roots:

$x_v = \frac{-1 + 3}{2} = \frac{2}{2} = 1$.

The y-coordinate of the vertex is:

$y_v = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4$.

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

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

To sketch the graph: Draw a parabola passing through $(-1, 0)$ and $(3, 0)$ with the vertex at $(1, -4)$.

Part b: Find the minimum value.

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

Second problem.

The unshaded region R is defined by the following four inequalities:

1. $x \geq -2$

2. $x \leq 4$

3. $y \geq 2$

4. The line passes through $(0,6)$ and $(8,2)$. The slope is $\frac{2-6}{8-0} = \frac{-4}{8} = -\frac{1}{2}$.

The equation of the line is $y - 6 = -\frac{1}{2}(x - 0)$, which is $y = -\frac{1}{2}x + 6$.

Since the region is below the line, the inequality is $y \leq -\frac{1}{2}x + 6$.

Or $2y \leq -x + 12$, so $x + 2y \leq 12$.

3. Final Answer

Part a: The graph is a parabola passing through $(-1, 0)$ and $(3, 0)$ with the vertex at $(1, -4)$.

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

For the second problem, the four inequalities are:

$x \geq -2$

$x \leq 4$

$y \geq 2$

$x + 2y \leq 12$