Problem 41 - Precalculus: Mathematics for Calculus

Question

Graph the solution set of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{array}{l} y<9-x^{2} \\ y \geq x+3 \end{array}\right.$$

Step-by-Step Solution

Answer

Vertices are \((-3,0)\), \((2,5)\), \((0,3)\). The solution set is bounded.

Step 1: Identify the inequalities

The system consists of two inequalities: 1. \( y < 9 - x^2 \) which is a downward-opening parabola with vertex at \((0,9)\).2. \( y \geq x + 3 \) which is a line with a slope of 1, intersecting the y-axis at 3.

Step 2: Graph the quadratic inequality

Sketch the graph of the parabola \( y = 9 - x^2 \). This parabola opens downwards with its vertex at \((0,9)\) and intersects the x-axis at \((3,0)\) and \((-3,0)\). The solution region for \( y < 9 - x^2 \) is below the parabola.

Step 3: Graph the linear inequality

Graph the line \( y = x + 3 \). This line passes through the points \((0,3)\) and \((3,6)\). Since the inequality is \( y \geq x + 3 \), the solution region is above or on this line.

Step 4: Identify the solution region

The solution to the system will be where the region below the parabola and above the line overlap. This means the area between the curves of \( y < 9 - x^2 \) and \( y = x + 3 \).

Step 5: Find the intersection points

Find where the parabola \( y = 9 - x^2 \) intersects the line \( y = x + 3 \). Set the equations equal: \[ 9 - x^2 = x + 3 \]Solving gives \[ x^2 + x - 6 = 0 \].Factor to get \((x - 2)(x + 3) = 0\), so \( x = 2 \) and \( x = -3 \), giving points \((2,5)\) and \((-3,0)\).

Step 6: Determine boundedness of the solution set

The parabola, being downward opening, creates a finite region above the line \( y = x + 3 \) limiting the solution set. The points of intersection \((-3,0)\) and \((2,5)\) along with \((0,3)\) (where the line crosses the y-axis) create a bounded triangular region.

Step 7: List the vertices

The vertices of the bounded region are where the parabola \(9-x^2\) intersects with the line \(x+3\) and the y-axis. The intersection points found earlier were \((-3,0)\) and \((2,5)\). We also include the point where the line crosses the y-axis \((0,3)\). These are the vertices: \((-3,0)\), \((2,5)\), \((0,3)\).

Key Concepts

Graphing Inequalities

Graphing inequalities is a crucial skill in understanding systems of inequalities. An inequality such as \( y < 9 - x^2 \) represents a set of points that fall below a certain curve—in this case, a parabola. Meanwhile, \( y \geq x + 3 \) describes points that lie above or on a straight line.
To graph these inequalities:

• Start by treating the inequalities as equations (\( y = 9 - x^2 \) and \( y = x + 3 \)) and graph these curves.
• For \( y < 9 - x^2 \), shade the region below the parabola.
• For \( y \geq x + 3 \), shade the area above the line.

Where these shaded regions overlap indicates the solution set of the system. It's visually representing the set of possible solutions.

Parabola

A parabola is a symmetrical, u-shaped curve, and is often represented by equations like \( y = ax^2 + bx + c \). In the system of inequalities problem, we deal with the equation \( y = 9 - x^2 \). It opens downwards due to the negative sign before \( x^2 \).
Key features of this parabola:

• The vertex is the highest point, located at \((0,9)\).
• It intersects the x-axis, where \( y = 0 \), at points \((3,0)\) and \((-3,0)\).

The vertex and intercepts help in sketching the parabola accurately. For inequalities like \( y < 9 - x^2 \), we are interested in the region below this parabola.

Intersection Points

Intersection points occur where two graphs meet. In our system of inequalities, they show where the parabola \( y = 9 - x^2 \) intersects the line \( y = x + 3 \).
To find these points:

• Set the equations equal: \( 9 - x^2 = x + 3 \).
• Solve for \( x \) to find the roots: \( x^2 + x - 6 = 0 \).
• Factoring gives \((x - 2)(x + 3) = 0\), leading to \( x = 2 \) and \( x = -3 \).

Substitute these \( x \) values back into either original equation to find respective \( y \) values, yielding intersection points \((2,5)\) and \((-3,0)\). These are crucial in determining the region enclosed by the system.

Bounded Region

A bounded region is the area enclosed entirely by the lines and curves on the graph. For the system of inequalities \( y < 9 - x^2 \) and \( y \geq x + 3 \), the solution set is a bounded triangular region.
Determining boundedness involves identifying if the graph forms a closed shape. Here:

• The downward-opening parabola limits the region from the top.
• The line \( y = x + 3 \) runs under this curve, creating a boundary.
• The intersection points \((-3,0)\), \((2,5)\), and \((0,3)\) serve as vertices of the triangle.

These constraints ensure the solution set is finite and enclosed, as opposed to stretching infinitely in any direction.