Problem 49
Question
\(47-50\) . Use a graphing calculator to graph the solution of the system of inequalities. Find the coordinates of all vertices, rounded to one decimal place. $$ \left\\{\begin{array}{c}{y \leq 6 x-x^{2}} \\ {x+y \geq 4}\end{array}\right. $$
Step-by-Step Solution
Verified Answer
Vertices of the intersection area, rounded to one decimal place, are approximately (0.7, 3.3) and (6.3, -2.3).
1Step 1: Understand the Inequalities
The system of inequalities consists of two inequalities: \( y \leq 6x - x^2 \) and \( x + y \geq 4 \). The graph will show the region where both inequalities overlap.
2Step 2: Graph the Parabola
Graph the quadratic inequality \( y \leq 6x - x^2 \). This is a downward-opening parabola because the coefficient of \( x^2 \) is negative. Since the inequality is \( \leq \), shade the area below the parabola.
3Step 3: Graph the Line
Graph the linear inequality \( x + y \geq 4 \). Rearrange to \( y \geq -x + 4 \) to identify the line's slope and y-intercept. Since the inequality is \( \geq \), shade above the line.
4Step 4: Determine the Feasible Region
Locate the intersection area where the shaded regions from both inequalities overlap. This area represents the solutions to the system of inequalities.
5Step 5: Find Intersection Points
To find the vertices of the intersection area, solve the equality system: \( y = 6x - x^2 \) and \( x + y = 4 \). Substitute \( y = 4 - x \) into the first equation: \( 4 - x = 6x - x^2 \). Simplify to get \( x^2 - 7x + 4 = 0 \).
6Step 6: Solve the Quadratic Equation
Solve the quadratic equation \( x^2 - 7x + 4 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -7, c = 4 \). Applying this, we get: \( x = \frac{7 \pm \sqrt{49 - 16}}{2} = \frac{7 \pm \sqrt{33}}{2} \). Calculate to find approximate values for \( x \).
7Step 7: Calculate y-coordinates
Substitute each value of \( x \) back into \( y = 4 - x \) to find the corresponding \( y \) values for the intersection points.
8Step 8: Identify Vertices
Using the roots from the quadratic equation, calculate and round the coordinates of the vertices to one decimal place.
Key Concepts
Graphing CalculatorQuadratic InequalitiesLinear InequalitiesFeasible Region
Graphing Calculator
A graphing calculator is a powerful tool in visualizing and solving systems of inequalities. It can graph complex equations, making it easier to identify solution areas.
Here's how it can help with inequalities:
Here's how it can help with inequalities:
- Input Equations: You can directly enter the inequalities into the calculator.
- Visual Representation: The graphing feature shows how different equations interact visually.
- Feasible Region: The overlapping shaded areas solution can be clearly seen.
- Precision: Vertices and points of intersections can be approximated precisely.
Quadratic Inequalities
Quadratic inequalities are equations where the variable is of the second degree, such as in the expression \( y \leq 6x - x^2 \). For these inequalities:
- The quadratic term \( x^2 \) influences the curve's shape.
- If negative, like in our case, it forms a downward-opening parabola.
- Graph the corresponding equation as if it were an equality.
- Determine which side to shade based on the inequality sign.
Linear Inequalities
Linear inequalities, like \( x + y \geq 4 \), involve linear equations expressed with inequality signs. These are straight lines in a graph.
- Transform the inequality to \( y \geq -x + 4 \) for easier graphing.
- Identify the slope (here \(-1\)) and y-intercept (here \(4\)).
- Plot the line: this is just like drawing the usual line, only that now the inequality sign will guide us where to shade.
Feasible Region
A feasible region in the context of systems of inequalities is where all constraints are satisfied. When graphing, it's the intersection of shaded areas from each inequality.
For problems like ours:
For problems like ours:
- Graph each inequality separately to identify individual shaded regions.
- The feasible region is where these shaded areas overlap.
- This region contains all possible solutions that satisfy both inequalities at the same time.
- Vertices of this region can be found by solving the equality of the system for intersection points.
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