Problem 44
Question
41–44 Use a graphing calculator to graph the solution of the system of inequalities. Find the coordinates of all vertices, correct to one decimal place. $$\left\\{\begin{array}{l}{y \geq x^{3}} \\ {2 x+y \geq 0} \\ {y \leq 2 x+6}\end{array}\right.$$
Step-by-Step Solution
Verified Answer
The vertices are approximately at (0,0), (-1.5,3), and (1.6,4.1).
1Step 1: Understand the Inequalities
The system of inequalities is:\( y \geq x^3 \), \( 2x + y \geq 0 \), \( y \leq 2x + 6 \). We want to shade the region where all these inequalities intersect on the graph.
2Step 2: Graph the First Inequality
Plot the graph of \( y = x^3 \). The region of interest is the area above the curve since \( y \geq x^3 \).
3Step 3: Graph the Second Inequality
Plot the line \( y = -2x \). The inequality \( 2x + y \geq 0 \) indicates shading the region above this line.
4Step 4: Graph the Third Inequality
Plot the line \( y = 2x + 6 \). The inequality \( y \leq 2x + 6 \) requires shading the region below this line.
5Step 5: Find Regions of Overlap
Identify the region where all shaded areas from Step 2, 3, and 4 overlap on the graph. This is the solution region for the system of inequalities.
6Step 6: Identify the Vertices
Calculate the coordinates where the boundaries intersect to find vertices in the overlap region. Solve for intersections: \( y = x^3 \) and \( y = -2x \), \( y = x^3 \) and \( y = 2x + 6 \), and \( y = -2x \) and \( y = 2x + 6 \).
7Step 7: Calculate Intersections
Solve the following equations to find intersection points:- \( x^3 = -2x \): This gives points at \( (0,0) \), \( (\sqrt{2}, -2\sqrt{2}) \), and \( (-\sqrt{2}, 2\sqrt{2}) \).- \( x^3 = 2x + 6 \): Requires numeric methods or graphing calculator for accurate decimals, intersection approximately around \( (1.6, 4.1) \).- \( -2x = 2x + 6 \): Solve to get the point \( (-1.5, 3) \).
8Step 8: List Vertices
The vertices of the overlapping solution region, rounded to one decimal place, are approximately: \( (0,0) \), \( (-1.5, 3) \), and \( (1.6, 4.1) \).
Key Concepts
Graphing CalculatorIntersection PointsInequalitiesVertices Calculation
Graphing Calculator
A graphing calculator is an incredibly useful tool when working with systems of inequalities.
- It allows you to visually represent equations and inequalities.
- You can observe where different regions overlap.
- The calculator will visualize the curves and lines.
- It shades the regions where inequalities are true.
Intersection Points
Intersection points are where two or more lines or curves meet on a graph. Finding these points helps determine the vertices of the solution region for a system of inequalities.To find these points:
- Use your calculator to trace the curves.
- The calculator often automatically marks intersection points.
Inequalities
Inequalities in mathematics are expressions involving the symbols \( \geq \), \( > \), \( \leq \), and \( < \).
- They describe a range of values rather than a single value.
- Graphically, they represent areas or regions.
Vertices Calculation
Finding vertices involves calculating the points where the boundaries of the inequality regions intersect. These are crucial for outlining the shape of the solution region. To determine vertices:
- Identify and solve for intersections between boundaries using algebra and your calculator.
- Ensure the points lie within the shaded overlap region.
Other exercises in this chapter
Problem 43
35–46 Solve the system of linear equations. $$\left\\{\begin{aligned} x \qquad +z+w=4 \\ y-z \qquad =-4 \\ x-2 y+3 z+w =12 \\ 2 x \qquad -2 z+5 w=-1 \end{aligne
View solution Problem 43
Use the graphical method to find all solutions of the system of equations, correct to two decimal places. \(\left\\{\begin{array}{l}{\frac{x^{2}}{9}+\frac{y^{2}
View solution Problem 44
Determine \(A, B, C,\) and \(D\) in terms of \(a\) and \(b :\) $$\frac{a x^{3}+b x^{2}}{\left(x^{2}+1\right)^{2}}=\frac{A x+B}{x^{2}+1}+\frac{C x+D}{\left(x^{2}
View solution Problem 44
\(29-44\) Use Cramer's Rule to solve the system. $$ \left\\{\begin{array}{l}{x+y=1} \\ {y+z=2} \\ {z+w=3} \\\ {w-x=4}\end{array}\right. $$
View solution