Problem 41 - Precalculus: Mathematical for Calculus

Question

Use a graphing calculator to graph the solution of the system of inequalities. Find the coordinates of all vertices, correct to one decimal place. $$\left\\{\begin{array}{l}y \geq x-3 \\\y \geq-2 x+6 \\\y \leq 8\end{array}\right.$$

Step-by-Step Solution

Answer

The vertices are (3, 0), (-1, 8), and (11, 8).

Step 1: Understanding the inequalities

We have three inequalities: \( y \geq x-3 \), \( y \geq -2x+6 \), and \( y \leq 8 \). These inequalities define a region on the coordinate plane, bounded by the lines \( y = x-3 \), \( y = -2x+6 \), and the horizontal line \( y = 8 \).

Step 2: Graphing the lines

Use a graphing calculator to enter the equations \( y = x-3 \), \( y = -2x+6 \), and \( y = 8 \). Note the intersections and orientation of the lines. This helps visualize where the solutions to the inequalities lie relative to these lines.

Step 3: Shading the feasible region

Shade the region that satisfies all three inequalities. For \( y \geq x-3 \), shade above the line \( y = x-3 \). For \( y \geq -2x+6 \), shade above the line \( y = -2x+6 \). For \( y \leq 8 \), shade below the line \( y = 8 \). The overlapping shaded region is the feasible solution area.

Step 4: Finding intersection points (vertices)

Calculate the intersection points of the lines. Solve \( y = x-3 \) and \( y = -2x+6 \) to get their intersection: equating gives \( x-3 = -2x+6 \), solving it, \( 3x = 9 \), \( x = 3 \) and \( y = 0 \). Similarly, find the intersection of \( y = x-3 \) with \( y = 8 \) by substituting \( y \) to get \( 8 = x-3 \) giving \( x = 11 \). Last, set \( y = -2x+6 \) with \( y = 8 \), giving an intersection at \( x = -1 \).

Step 5: Listing the coordinates

The vertices of the solution region (where the shaded areas overlap) are at the points of intersection: \( (3, 0) \), \( (-1, 8) \), and \( (11, 8) \).

Key Concepts

graphing calculator

A graphing calculator is an essential tool for visualizing systems of inequalities, such as the one in the original exercise. It allows you to input multiple equations and inequalities. This helps to see how they interact on a coordinate plane. Here is how this tool works and why it's useful:

• Input each equation or inequality separately into the graphing calculator. For the given exercise, you would input the lines: \( y = x-3 \), \( y = -2x+6 \), and \( y = 8 \).
• The calculator will plot each equation as a distinct line on the graph, making it easier to identify their positions relative to each other.
• It can shade regions above or below the line based on the inequality (e.g., \( y \geq x-3 \) means shading above the line).

Breaking down these inequalities by graphing them visually helps to comprehend where the intersection and feasible regions lie. Graphing calculators simplify the complex task of handling multiple equations at once, turning a potentially confusing numerical problem into a clear visual representation.

intersection points

Intersection points are crucial in determining the vertices of the solution area of a system of inequalities. These are the points where the lines intersect each other, defining the corners of the feasible region.

• To find an intersection point, solve two equations simultaneously. For instance, to find where \( y = x-3 \) intersects \( y = -2x+6 \), equate them: \( x-3 = -2x+6 \). Solving this gives \( x = 3 \) and substituting back gives \( y = 0 \), providing the intersection point \( (3, 0) \).
• The next intersection between \( y = x-3 \) and \( y = 8 \) comes from substituting: \( 8 = x-3 \), giving \( x = 11 \), thus \( (11, 8) \).
• Similarly, solving for where \( y = -2x+6 \) meets \( y = 8 \) gives another point: set \( 8 = -2x+6 \), leading to \( x = -1 \), resulting in the point \((-1, 8)\).

Understanding intersection points is key for identifying the vertices of the feasible region. These calculated coordinates help define the boundaries of the solution set.

feasible region shading

Shading the feasible region involves identifying areas on the graph that satisfy all given inequalities. This is the solution region, where every point within it meets all the conditions set by the inequalities.

• For \( y \geq x-3 \), shade the area above the line \( y = x-3 \).
• For \( y \geq -2x+6 \), shade above this line as well.
• For \( y \leq 8 \), shade below the horizontal line at \( y = 8 \).

The overlapping section where all these shaded areas meet is the feasible region. This region is the only area where all inequalities hold true simultaneously. Identifying this region is crucial as it represents all possible solutions to the system of inequalities. With vertices calculated and the region shaded, the graph provides a clear visual of the potential solutions.