Problem 25 - Algebra and Trigonometry

Question

19–40 Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{array}{l}{y<9-x^{2}} \\ {y \geq x+3}\end{array}\right.$$

Step-by-Step Solution

Answer

Vertices: (2, 5), (-3, 0), and possibly (0, 3). The solution set is bounded.

Step 1: Graph the First Inequality

Start by graphing the inequality \( y < 9 - x^2 \). This is a downward-opening parabola with vertex at (0, 9). Since the inequality is strictly less than, we use a dashed line to represent the boundary of this region. Shade the area below the parabola.

Step 2: Graph the Second Inequality

Now graph the inequality \( y \geq x + 3 \). This is a straight line with a slope of 1 and a y-intercept of 3. Since the inequality includes equal to, we use a solid line to represent the boundary. Shade the area above the line.

Step 3: Identify the Intersection

The solution to the system of inequalities is the region where the shaded areas from both inequalities overlap. This requires finding common points where the parabola \( y = 9 - x^2 \) and the line \( y = x + 3 \) intersect by setting them equal and solving for \( x \).

Step 4: Solve for Intersection Points

Set the equations \( 9 - x^2 = x + 3 \) and solve:1. Move all terms to one side: \( -x^2 - x + 6 = 0 \).2. Rearrange: \( x^2 + x - 6 = 0 \).3. Factor the quadratic: \( (x - 2)(x + 3) = 0 \).4. Solve for \( x \): \( x = 2 \) and \( x = -3 \).5. Substitute back to find \( y \) values: for \( x = 2, y = 5 \); for \( x = -3, y = 0 \).

Step 5: Coordinate of Vertices

The vertices derived from intersections are (2, 5) and (-3, 0). Additional vertex is at (0, 3) from directly analyzing changes in line to parabola below it.

Step 6: Determine If Bounded

A solution set is bounded if it does not extend to infinity in any direction. Here, the solution is bounded in an area enclosed by the intersection points (2, 5), (-3, 0) and segment on line \( y = x+3 \) with parabola.

Key Concepts

Graphing Inequalities

Graphing inequalities involves visualizing the regions in a coordinate plane that satisfy an inequality.
Here, we use inequalities involving a parabola and a linear equation.

• Begin with the inequality:
  \(y < 9 - x^2\).
  This is a downward-opening parabola with its vertex at (0, 9). Since our inequality uses "less than", we draw a dashed boundary to represent the parabolic curve. Shade the region below this boundary because the inequality is strictly less.
• Next, for the inequality \(y \geq x + 3\), we graph a line with a slope of 1 and a y-intercept at 3.
  Since the inequality includes "equal to", the line is solid, representing that points on the line are part of the solution. Shade the region above the line, indicating all y-values at or above this line satisfy the inequality.

Solution Set Boundedness

Determining the boundedness of the solution set involves deciding if the solution region extends infinitely or remains enclosed within a certain area. In the context of inequalities:

• A solution is bounded if it forms a closed region on the graph, meaning it does not reach towards infinity in any direction.
  
• Unbounded solutions, by contrast, extend infinitely in one or more directions. To determine boundedness, examine whether the overlapping (shaded) area from multiple inequalities encloses a finite space.

For our system, after graphing, since the region is enclosed by the parabola and the line, the solution set is bounded. The intersections and vertex within the graph contain this encompassed region.

Intersection Points

Intersection points are crucial as they define where the graphs of two equations meet. These points are solutions to the system of equations formed by equating the equations from the inequalities:

• For our parabola \(y = 9 - x^2\) and line \(y = x + 3\), we set them equal:
  \(9 - x^2 = x + 3\).
• This simplifies into the quadratic equation \(-x^2 - x + 6 = 0\), which rearranges to \(x^2 + x - 6 = 0\). Factor this as \((x - 2)(x + 3) = 0\).
• Solve for \(x\): \(x = 2\) and \(x = -3\). Substitute these values back into the linear equation to find corresponding \(y\):
  For \(x = 2\), \(y = 5\); for \(x = -3\), \(y = 0\).

These intersection points (2, 5) and (-3, 0) highlight where solutions overlap, forming part of our bounded solution region.

Parabola

A parabola is a symmetrical curve on a graph, representing quadratic functions, and described by the equation \(y = ax^2 + bx + c\). In this exercise, the parabola has the equation \(y = 9 - x^2\):

• It opens downward due to the negative coefficient of \(x^2\).
• The vertex of this parabola, being the highest point because it opens downwards, is at (0, 9).
• The vertex serves as a starting reference for graphing, utilizing its symmetry to complete the structure of the parabola.

Understanding the parabola's orientation and vertex helps in accurately shading the correct region that fits the inequality \(y < 9 - x^2\). It also plays a key role in determining the bounded solution area.

Linear Equation

Linear equations are equations of straight lines, represented as \(y = mx + b\), where \(m\) is the slope and \(b\) the y-intercept. In our system, we have the linear inequality \(y \geq x + 3\):

• The slope \(m = 1\) indicates the line rises one unit vertically for every unit it moves to the right horizontally.
• The y-intercept \(b = 3\) tells us it crosses the y-axis at 3.
• This line serves as one boundary of the solution set, dictating that all points on or above it satisfy the inequality.

Knowing how to graph linear equations allows one to quickly draw accurate boundaries and assess the overlap with other graph features, enabling a complete understanding of where the solution set lies within the coordinate plane.