Problem 47 - 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. $$\left\\{\begin{array}{l} x^{2}+y^{2} \leq 9 \\ 2 x+y^{2} \leq 1 \end{array}\right.$$

Step-by-Step Solution

Answer

The vertices of the bounded region are approximately \((4, 1)\) and \((-2, 0)\). The solution set is bounded.

Step 1: Identify the forms of given inequalities

The given system of inequalities is \(x^2 + y^2 \leq 9\) and \(2x + y^2 \leq 1\). The first inequality \(x^2+y^2 \leq 9\) represents a circle centered at the origin with radius 3. The second inequality \(2x + y^2 \leq 1\) is a parabolic region.

Step 2: Graph the Circle

Graph the circle with center at the origin, \((0,0)\), and radius 3. The inequality \(x^2 + y^2 \leq 9\) includes the interior of the circle, so shade the region inside and on the boundary of this circle.

Step 3: Graph the Parabolic Region

Graph the equation \(2x + y^2 = 1\) which forms a parabola. To find intercepts, set \(x = 0\) to get \(y^2=1\), so \(y=\pm 1\). Similarly, set \(y=0\) resulting in \(2x=1\), so \(x=0.5\). The region satisfying the inequality \(2x + y^2 \leq 1\) is inside or on this parabola.

Step 4: Find the Intersection Points

To find the vertices of the overlapping region, solve the system \(x^2+y^2=9\) and \(2x+y^2=1\). Subtract the second equation from the first: \((x^2 - 2x) + (y^2 - y^2) = 9 - 1\) yielding \(x^2 - 2x = 8\). Solving gives \(x=4 , x=-2\). Checking these \((x,y)\) in \(x^2 + y^2 = 9\) provides \((4,1)\) and \((-2,0)\). Verify for \(2x + y^2 = 1\) to confirm these are within the solution set.

Step 5: Analyze Boundedness

The bounded region lies within both the circle and parabola. The circle \(x^2 + y^2 \leq 9\) ensures that the region cannot have points where \(x^2 + y^2\) exceeds 9, hence the solution is bounded.

Key Concepts

System of Inequalities

A system of inequalities consists of two or more inequalities that share the same set of variables. The objective is to find the set of solutions that satisfy all inequalities simultaneously. These solutions can often be represented as a region on the coordinate plane, where every point in this region meets all the conditions outlined by the inequalities.

• The first inequality given is \(x^2 + y^2 \leq 9\), which is a circle centered at the origin with a radius of 3.
• The second, \(2x + y^2 \leq 1\), defines a parabolic region.

On a graph, the solution to a system of inequalities represents the area where the solutions to all included inequalities overlap. It's essential to shade the correct region for each inequality to accurately illustrate the solution set of the system.

Intersection Points

Intersection points are where the graphs of the inequalities meet. These points can be found by solving the included equations simultaneously. For the system \(\{x^2 + y^2 \leq 9, 2x + y^2 \leq 1\}\), we need to work with their respective equalities:

• The circle defined by \(x^2 + y^2 = 9\).
• The parabola given by \(2x + y^2 = 1\).

By equating or substituting these expressions, we can find exact intersection points - solutions that satisfy both equations are critical, as they define the edges of the overlapping region. To solve:

• Subtract the second equation from the first: \((x^2 - 2x) + (y^2 - y^2) = 9 - 1\), simplifying to \(x^2 - 2x = 8\).
• This yields solutions for x: \(x = 4\) and \(x = -2\).
• These x-values can be plugged back into one of the equations to find corresponding y-values.

Vertex Coordinates

Vertices are the corner points of the region enclosed by the system of inequalities. These coordinates are significant because they define the limits of the solution set. When working with systems of inequalities, the vertices are often found at the intersection of the boundary lines or curves of the graphs.For the current exercise:

• Using the x-values acquired from solving \(x^2 - 2x = 8\), \(x=4\) and \(x=-2\), we substitute back into \(x^2 + y^2 = 9\) to find y-values.
• If \(x = 4\), then substitute into \(x^2 + y^2 = 9\) to receive no real solution, indicating an error possibly or rounding has occurred elsewhere.
• If \(x = -2\), substituting back yields \(y = 3\), resulting in the vertex coordinate \((-2, 3)\).
• For accuracy, more methods such as graphical checks or numerically resolving may be required to confirm the correct points.

Capturing these vertices allows for a complete picture of the bounded region's shape and size.

Bounded Region

A bounded region in a graph is a closed, finite area where the solutions to a system of inequalities exist. Understanding whether a solution is bounded or unbounded is crucial as it impacts the nature of solutions contained within a finite or infinite area.In the present problem:

• \(x^2 + y^2 \leq 9\) depicts a circle, which inherently creates a boundary because it limits solutions to within its radius.
• \(2x + y^2 \leq 1\) shapes a parabolic region, and its intersection with the circle confines the solutions further.

The region of overlap between the two inequalities forms the bounded solution set. Since both the circle and parabola inclusively restrict points that satisfy both inequalities, the solution space is finite and bounded. This characteristic ensures the solution set has definite edges and does not extend infinitely in any direction.