Problem 32 - Precalculus Mathematics for Calculus

Question

Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{aligned} y & \geq x^{2} \\ x+y & \geq 6 \end{aligned}\right.$$

Step-by-Step Solution

Answer

Vertices are (-3, 9), (2, 4), and (0, 6). The solution set is unbounded.

Step 1: Identify Each Inequality

The given system consists of the inequalities: 1. \( y \geq x^2 \), which is a parabola opening upwards with its vertex at the origin (0,0).2. \( x+y \geq 6 \), which can be rewritten as \( y \geq -x + 6 \), a line with a slope of -1 and y-intercept at 6.

Step 2: Graph the Parabola

Draw the parabola represented by the inequality \( y \geq x^2 \). Shade the region above the parabola since we are considering the values where \( y \) is greater than or equal to \( x^2 \). The parabola is a standard graph opening upwards.

Step 3: Graph the Line

Plot the linear inequality \( y \geq -x + 6 \). Start at (0,6) and use the slope to plot another point (1,5). Draw the line through these points. Since we need \( y \) values that are greater than or equal to \(-x + 6\), shade the region above this line.

Step 4: Determine Intersection Points

Find where the parabola \( y = x^2 \) intersects with the line \( y = -x + 6 \). Set the equations equal to each other:\[ x^2 = -x + 6 \]Rearrange to get\[ x^2 + x - 6 = 0 \]Solve the quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = 1 \), and \( c = -6 \). This gives\[ x = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2} \]Thus, \( x = 2 \) or \( x = -3 \). Substitute back to find \( y \):- For \( x = 2 \), \( y = 4 \). - For \( x = -3 \), \( y = 9 \). So, the points of intersection are (2, 4) and (-3, 9).

Step 5: Shade the Solution Region

Mark the solution region that satisfies both inequalities by shading the area where the shaded regions for each inequality overlap. This area covers the region above the parabola up to where it meets and goes beyond the line \( y = -x + 6 \).

Step 6: Identify All Vertices of the Feasible Region

The vertices of the bounded region where the shaded areas overlap are the points where the boundaries intersect. These vertices are (-3, 9), (2, 4), and on the line, the y-intercept point (0, 6). The parabola's vertex (0,0) is not a boundary vertex.

Step 7: Determine If the Solution Set Is Bounded

The solution set is not bounded because the shaded region extends infinitely in the positive direction along the line and upwards. The region is open and unbounded.

Key Concepts

System of Inequalities

A **system of inequalities** consists of multiple inequalities that we solve together. This means we are looking for values that satisfy all inequalities at the same time. Here, each inequality represents a different condition, and when combined, they define a more constrained solution region.
The given system is:

• <code>\( y \geq x^2 \)</code>: This inequality represents a parabola with a vertex at (0,0) and opens upwards. The area above this parabola includes all points \( (x, y) \) where the value of \( y \) is greater than or equal to the value of \( x^2 \).
• <code>\( x+y \geq 6 \)</code>: After rearranging, it can be rewritten as \( y \geq -x + 6 \). This represents a straight line with a negative slope. The shading is above this line, including all points where \( y \) is larger than \( -x + 6 \).

In graphing this system, your goal is to find the solution region, which is where the shadings of both inequalities overlap.

Solution Region

The **solution region** is the area on the graph where both inequalities are true simultaneously. It is determined by the overlap of the shaded areas from each inequality.
For this system, it involves:

• The area above the parabola \( y \geq x^2 \), which includes all possible \( y \) values that satisfy being equal to or greater than \( x^2 \).
• The area above the line \( y \geq -x + 6 \), which encompasses all \( y \) values that are equal to or greater than \( -x + 6 \).

The intersecting shaded areas create the solution region on the graph. This region is where both conditions exist together. It extends infinitely in the positive direction because the line continues without bounds. Therefore, the solution set is not bounded.

Intersection Points

Intersection points are crucial for understanding where different parts of the solution region connect. These points occur where the equations of the curves meet.
To find the intersection points of \( y = x^2 \) and \( y = -x + 6 \), we solve the equation:

• \( x^2 = -x + 6 \)
• Rearranging gives \( x^2 + x - 6 = 0 \).
• This quadratic equation is solved using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = -6 \).
• The solutions are \( x = 2 \) and \( x = -3 \).

For these \( x \) values:

• Substituting \( x = 2 \) gives \( y = 4 \).
• Substituting \( x = -3 \) gives \( y = 9 \).

The intersection points here are (2, 4) and (-3, 9). These points mark the boundaries of the solution region along the curves they belong to.