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Chapter 9: Problem 46

Graph each compound inequality. \(y \leq \frac{1}{4} x+2\) and \(y \geq-1\)

Short Answer

Expert verified
To graph the compound inequality \(y \leq \frac{1}{4}x + 2\) and \(y \geq -1\), first graph each individual inequality. For \(y \leq \frac{1}{4}x + 2\), draw a solid line for the equation \(y = \frac{1}{4}x + 2\) and shade the region below the line. For \(y \geq -1\), draw a solid horizontal line at \(y = -1\) and shade the region above the line. The overlapping shaded region represents the solution to the compound inequality.

Step by step solution

01

Graph the first inequality: \(y \leq \frac{1}{4}x + 2\)

To graph the inequality \(y \leq \frac{1}{4}x + 2\), follow these steps: 1. Sketch the line \(y = \frac{1}{4}x + 2\) (dashed line if \(y < \frac{1}{4}x + 2\), solid line if \(y \leq \frac{1}{4}x + 2\)). In this case, it鈥檚 a solid line since the inequality is less than or equal to. 2. Shade the region below the line, since it is a less than or equal to inequality (less than would require shading below the line, greater than or equal to would require shading above the line).
02

Graph the second inequality: \(y \geq -1\)

To graph the inequality \(y \geq -1\), follow these steps: 1. Sketch the horizontal line \(y = -1\) (dashed line if \(y > -1\), solid line if \(y \geq -1\)). In this case, it鈥檚 a solid line since the inequality is greater than or equal to. 2. Shade the region above the horizontal line, since it is a greater than or equal to inequality (less than would require shading below the line, greater than would require shading above the line).
03

Find the region that satisfies both inequalities

Now, let's find the overlapping region that satisfies both inequalities: 1. Look for the region where both the shaded areas from the first and second inequality overlap. 2. Shade that overlapping region on the graph, as it represents the region that satisfies both inequalities \(y \leq \frac{1}{4}x + 2\) and \(y \geq -1\). The resulting graph and shaded region represent the solution to the compound inequality.

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