Problem 76 - A Graphical Approach to Precalculus with Limits

Question

Graph the solution set of each system of inequalities by hand. $$\begin{aligned}&y \geq \frac{1}{x+2}\\\&y \geq x-2\end{aligned}$$

Step-by-Step Solution

Answer

Shade the region where the areas above \( y = \frac{1}{x+2} \) and \( y = x-2 \) overlap.

Step 1: Understand the inequalities

The system consists of two inequalities. The first inequality is \( y \geq \frac{1}{x+2} \), a rational function. The second inequality is \( y \geq x-2 \), a linear function. The solution set is the region where both inequalities are satisfied.

Step 2: Graph the rational inequality

Start by graphing the line \( y = \frac{1}{x+2} \). This is a hyperbola that shifts leftward by 2 units. Point out that the graph does not exist for \( x = -2 \) because it results in division by zero, so there is a vertical asymptote here. Shade the region above or on this curve, representing \( y \geq \frac{1}{x+2} \).

Step 3: Graph the linear inequality

Now, graph the line \( y = x-2 \). This is a straight line with a slope of 1 and a y-intercept at \( (0, -2) \). Shade the area above this line, including points on the line, because the inequality is \( y \geq x-2 \).

Step 4: Find the solution region

The solution region of the system is where the shaded areas from both inequalities overlap. Review both graphes and identify this common region. This region is a part of the plane where any point would satisfy both inequalities simultaneously.

Key Concepts

Rational Functions

A rational function is a type of function defined by the ratio of two polynomials. In the inequality \(y \geq \frac{1}{x+2}\), the expression \(\frac{1}{x+2}\) represents a rational function.
Understanding rational functions helps in graphing them and finding the solution region for inequalities. The function \(\frac{1}{x+2}\) represents a hyperbola, which means the graph has two disconnected curves, or branches.
One key feature of rational functions is asymptotes:

• Vertical Asymptote: Occurs where the denominator is zero. Here, \(x+2=0\) or \(x=-2\) is a vertical asymptote. The function is undefined at \(x=-2\).
• Horizontal Asymptote: Indicates behavior as \(x\) approaches infinity. For \(\frac{1}{x+2}\), the horizontal asymptote is \(y = 0\) as \(x\) becomes very large or very negative.

In graphing the inequality \(y \geq \frac{1}{x+2}\), the area above or on this hyperbolic curve is shaded. Each point in this region satisfies the inequality.

Linear Functions

Linear functions form the backbone of many mathematical and real-world models. The inequality \(y \geq x-2\) involves a linear function.
A linear function is represented by the equation \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. For \(y = x-2\):

• Slope: The coefficient of \(x\) is 1, meaning the graph rises one unit for every unit it moves to the right.
• Y-intercept: Occurs at \(b = -2\), so the point \((0, -2)\) is where the line crosses the y-axis.

To graph \(y \geq x-2\), draw a straight line through the y-intercept \((0, -2)\) with a slope of 1. Then, shade the region above this line, which includes all points that satisfy the inequality. Points on the line itself also satisfy the inequality \(y \geq x-2\), so they are part of the shaded region.

Solution Region

In graphing systems of inequalities, finding the solution region is crucial. It represents where all the conditions are true simultaneously.
For the system \(y \geq \frac{1}{x+2}\) and \(y \geq x-2\), the solution region is where the shaded regions of the two inequalities overlap on the graph. Steps to identify this:

• Shade the rational function region: Graph \(y = \frac{1}{x+2}\), mark the vertical asymptote at \(x = -2\), and shade above the hyperbolic curve.
• Shade the linear function region: Graph \(y = x-2\), drawing a line with a slope of 1 from the y-intercept \((-2)\), and shade above the line.
• Identify overlap: The overlapping shading on the graph is the solution region where both inequalities are true.

In this shared region, any point will satisfy both \(y \geq \frac{1}{x+2}\) and \(y \geq x-2\). This area highlights solutions that meet all constraints of the given system.