Problem 38 - Algebra and Trigonometry

Question

19–40 Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{array}{c}{x^{2}-y \geq 0} \\ {x+y<6} \\\ {x-y<6}\end{array}\right.$$

Step-by-Step Solution

Answer

Vertices: (2, 4), (-3, 9), (6, 0). The solution set is bounded.

Step 1: Analyze Each Inequality

First, analyze each inequality separately. The inequality \(x^2 - y \geq 0\) can be rewritten as \(y \leq x^2\), which describes the region below the parabola \(y = x^2\). The inequality \(x + y < 6\) represents the region below the line \(y = -x + 6\), and \(x - y < 6\) represents the region to the left of the line \(y = x - 6\).

Step 2: Graph the Inequalities

Plot each inequality on the coordinate plane. The parabola \(y = x^2\) opens upwards, and the solution is the region below it. The line \(y = -x + 6\) is a straight line with slope -1, solution region is below this line. The line \(y = x - 6\) is also straight with slope 1, and the solution region is to the left. Shade these areas.

Step 3: Find the Intersection Points

To find the vertices of the feasible region, solve the system of equations at the boundaries of the shaded regions. For instance, set \(x^2 = -x + 6\) and \(x^2 = x - 6\), and solve for \(x\) and \(y\) to find exact intersection coordinates.

Step 4: Solve \(x^2 = -x + 6\)

Solve the quadratic equation \(x^2 + x - 6 = 0\) to find \(x = 2, x = -3\). Plug these into \(y = x^2\) to get the intersection points (2, 4) and (-3, 9).

Step 5: Solve \(x^2 = x - 6\)

Solve the quadratic equation \(x^2 - x + 6 = 0\). This equation has no real solutions, as the discriminant \(b^2 - 4ac = 1 - 24 = -23\) is less than zero. Hence, there are no intersections with this line.

Step 6: Solve \(y = -x+6\) and \(y = x-6\)

Solve \(-x + 6 = x - 6\) for \(x\), yielding \(x = 6\). Substitute into either line to find \(y = 0\). Thus, the intersection is at (6, 0).

Step 7: Determine the Bounds of the Solution

The vertices are (2, 4), (-3, 9), and (6, 0). The region is bounded by the parabola and the two lines, making the solution set bounded. Create a triangle-like feasible region within these boundaries.

Key Concepts

Graphing Inequalities

Graphing inequalities is all about representing solutions that satisfy given conditions on a mathematical graph. A key part is understanding the two sides of an inequality: for example, the inequality \( y \leq x^2 \) represents a region below the parabola \( y = x^2 \). Similarly, \( y < -x + 6 \) describes everything below the straight line with a negative slope, and \( y < x - 6 \) everything below the line with a positive slope. This involves:

• Drawing the boundary, which can be a line or curve (like a parabola).
• Identifying the region to shade, which indicates the possible solutions to the inequality.

After plotting each boundary and shading the appropriate regions, the graph should clearly show all areas where the inequalities overlap.

Bounding Solution Sets

A solution set is a collection of all possible solutions that satisfy a set of inequalities simultaneously. Determining whether this set is bounded means checking if it's enclosed on all sides, forming a closed shape. In our problem:

• The intersection of a parabola \( y = x^2 \) with the lines \( y = -x + 6 \) and \( y = x - 6 \) creates a triangular region where all solutions meet.
• This resulting triangular area indicates a bounded solution set as it is enclosed by the parabola and lines.

Visualizing bounded sets is important because it tells us the practical limits of solutions in real-world applications, ensuring no possible solution extends indefinitely in any direction.

Quadratic Equations

Understanding quadratic equations is crucial when dealing with curves, such as parabolas in our inequality system. A typical quadratic equation is in the form \( ax^2 + bx + c = 0 \). Solving these equations helps find intersection points:

• The solutions for \( x^2 = -x + 6 \) were found by rearranging to \( x^2 + x - 6 = 0 \) and solving using the quadratic formula, resulting in \( x = 2 \) and \( x = -3 \).
• The discriminant \( b^2 - 4ac \) determines the nature of the solutions: if it's negative, as in \( x^2 = x - 6 \), no real solution exists.

Being confident in solving these equations lets you accurately generate intersection points, critical for defining the limits of solution regions.

Intersection Points

Intersection points occur where the boundaries of different inequalities meet on a graph, marking the vertices of the solution set. To find them, you need to solve multiple equations simultaneously, such as:

• Solving for \( x^2 = -x + 6 \) identified intersection points at (2, 4) and (-3, 9).
• For the lines \( y = -x + 6 \) and \( y = x - 6 \), calculating the meeting point at \( x = 6 \), with \( y = 0 \), found the point (6, 0).

These points are pivotal in outlining the feasible region's shape and size, providing clear coordinates that anchor the bounded solution set. Recognizing and computing these intersection points can illuminate the geometry of your solution region, making it easier to graph and analyze.