Problem 45 - 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. $$\left\\{\begin{array}{c} x^{2}-y \leq 0 \\ 2 x^{2}+y \leq 12 \end{array}\right.$$

Step-by-Step Solution

Answer

The vertices are \((2, 4)\) and \((-2, 4)\), and the solution set is bounded.

Step 1: Rewrite Inequalities in Standard Form

First, rearrange the inequalities to make them easier to graph. The first inequality is \(x^2 - y \leq 0\), which can be rewritten as \(y \geq x^2\). The second inequality, \(2x^2 + y \leq 12\), can be rearranged to \(y \leq 12 - 2x^2\).

Step 2: Graph the Parabola Boundary Lines

Graph the boundary line of \(y = x^2\), which is a parabola opening upwards. Then graph \(y = 12 - 2x^2\), a parabola opening downwards. These are the boundary lines for the inequalities.

Step 3: Determine the Regions of the Plane

Determine which regions satisfy each inequality. For \(y \geq x^2\), shade the region above or on \(y = x^2\). For \(y \leq 12 - 2x^2\), shade the region below or on \(y = 12 - 2x^2\). The solution set is the overlapping shaded area.

Step 4: Find the Intersection Points

To find the vertices of this region, solve for the intersections of these boundary lines. Set \(x^2 = 12 - 2x^2\) to find intersection points: rearrange as \(3x^2 = 12\), then solve \(x^2 = 4\). So, \(x = \pm 2\). Calculate \(y\) for these: at \(x = 2\), \(y = 4\) for both; at \(x = -2\), \(y = 4\) for both. The intersection points are \((2, 4)\) and \((-2, 4)\).

Step 5: Analyze Boundedness

The solution set consists of points that satisfy both inequalities. The region is bounded since it is contained within the overlapping sections of the two parabolas and does not extend indefinitely in any direction.

Key Concepts

Bounded Solution Sets

When we discuss bounded solution sets in graphing systems of inequalities, we're talking about whether the solution to the system of inequalities fits within a limited region on the graph or if it extends infinitely in one or more directions. In our exercise, we have a situation where the solution set is indeed bounded. Let's delve into why that's the case.

For the given system of inequalities:

• First, rearrange the equations: from the inequalities, we know it's bounded within two parabolas.
• The parabola given by \(y = x^2\) constrains the solution from one side.
• The other parabola given by \(y = 12 - 2x^2\) constrains it from the opposite side and above.

The overlapping shaded region where both inequalities are satisfied is enclosed by these curves, forming a shape similar to a lens. This implies that all solutions lie within a finite space, creating a bounded region.

Thus, this region or set of possible solutions does not extend indefinitely, as it would if there were no upper or lower bounds set by our parabolic inequalities. A bounded solution set is compact and easier to analyze, especially when determining limits or confronting geometric constraints.

Parabolas

Parabolas are essential in this exercise, serving as critical elements of the boundary lines for the inequalities.

Points to consider:

• The first parabola derives from the inequality \( y \geq x^2 \), where \( y = x^2 \) is the boundary line. This parabola opens upwards, forming a U-shape.
• The second parabola comes from \( y \leq 12 - 2x^2 \), where the boundary is \( y = 12 - 2x^2 \). This parabola opens downwards, creating an inverted U-shape.

Graphing these parabolas helps determine where the inequalities intersect, which leads to pinpointing our solution set.

A parabola is symmetrical about its axis, meaning each side mirrors the other along a line known as the axis of symmetry. Recognizing this symmetry helps visualize the solution set. In our graphing process, these characteristic shapes direct the shading for regions that fulfill the inequalities up to the intersection points, where the solution is confirmed.

Intersection Points

Finding intersection points in a system of inequalities is crucial as they often denote vertices of the solution set's boundary.

For our specific system, intersection points arise where the two parabola equations equal each other:

• Equating \( y = x^2 \) and \( y = 12 - 2x^2 \) leads to solving \( x^2 = 12 - 2x^2 \).
• This rearranges to \( 3x^2 = 12 \), or \( x^2 = 4 \).
• Solutions for \( x \) are \( x = 2 \) and \( x = -2 \).

These values of \( x \) provide the y-coordinates when plugged back into either equation, giving us \( y = 4 \). Therefore, the intersection points are

1. (2, 4) and
2. (-2, 4)

Each intersection point serves as a vertex of the solution set's boundary. Understanding intersection points is vital because they often signify where solutions shift from one inequality to the other, outlining key structure in the solution set. Knowing this, students can effectively graph the intersection and verify boundedness, ensuring a full grasp of the solution's spatial domain.