/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 \(-x+4 y-4 \leq 0\)... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 13

\(-x+4 y-4 \leq 0\)

Short Answer

Expert verified
The solution is the region below the line \(y = \frac{1}{4}x + 1\), including the line itself.

Step by step solution

01

Convert the Inequality to Slope-Intercept Form

First, we want to rearrange the given inequality \(-x + 4y - 4 \leq 0\) so that it resembles the slope-intercept form of a line, \(y = mx + b\). Add \(x\) to both sides to isolate terms involving \(y\): \[4y - 4 \leq x\]Next, add 4 to both sides:\[4y \leq x + 4\]Finally, divide every term by 4 to solve for \(y\): \[y \leq \frac{1}{4}x + 1\].
02

Determine the Boundary Line

The inequality \(y \leq \frac{1}{4}x + 1\) implies we can consider the line \(y = \frac{1}{4}x + 1\) as a boundary.The slope (\(m\)) of the line is \(\frac{1}{4}\), and the y-intercept (\(b\)) is 1. Thus, the boundary line is a solid line because the original inequality is \(\leq\), indicating the region on or below this line.
03

Graph the Inequality

To graph the inequality \(y \leq \frac{1}{4}x + 1\):1. Start at the y-intercept (0, 1) on the graph.2. Use the slope \(\frac{1}{4}\) to determine the next point by rising 1 unit and running 4 units to the right, so another point is (4, 2).3. Draw a solid line through the points since the inequality is \(\leq\).4. Shade the region below the line to represent all the points \((x, y)\) that satisfy the inequality \(y \leq \frac{1}{4}x + 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Slope-Intercept Form
The slope-intercept form is a straightforward and very useful way to express a linear equation. It is written as: \[ y = mx + b \] where:
  • \( m \) represents the slope of the line, showing the change in the y-value for a one-unit increase in the x-value.
  • \( b \) is the y-intercept, which is the point where the line crosses the y-axis.
In the exercise provided, you began with the inequality \(-x + 4y - 4 \leq 0\). To make it easier to graph, you rearrange it into slope-intercept form: \[ y \leq \frac{1}{4}x + 1 \].This transformation sets you up perfectly for graphing and clearly shows the slope (\( \frac{1}{4} \)) and the y-intercept (1). These two components are vital because they allow you to draw the line accurately on a graph.
Graphing Inequalities
Graphing inequalities involves more than just drawing a line. It also includes identifying the area of the graph that satisfies the inequality. Let's break this down:
  • Start by graphing the boundary line. For the inequality \( y \leq \frac{1}{4}x + 1 \), use the slope (\( \frac{1}{4} \)) and the y-intercept (1) to sketch the line.
  • Since the inequality sign is \( \leq \), you draw a solid line. If it were \( < \), a dashed line would be used instead to indicate that points on the line are not included in the solution.
  • Next, determine which area of the graph fulfills the inequality. Since you have \( y \leq \), you shade below the line, capturing all the points (x, y) that meet the condition.
By following these steps, you ensure that all solutions to the inequality are represented on the graph.
Boundary Line
A boundary line is essential in graphing inequalities because it defines where one side of the inequality changes to the other. Here are the key details:
  • For the given inequality \( y \leq \frac{1}{4}x + 1 \), the line \( y = \frac{1}{4}x + 1 \) acts as the boundary.
  • The boundary line is drawn using the slope and y-intercept from the slope-intercept form.
  • Since the inequality includes "equals" (\( \leq \) or \( \geq \)), the boundary is depicted as a solid line. This means that points on the line are also part of the solution set.
The role of the boundary line is to divide the graph into two regions鈥攐ne that satisfies the inequality and one that does not. This helps you accurately depict all potential solutions on the graph.

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Most popular questions from this chapter

Remember that we did some work with parallel lines back in the graphing calculator activities in Problem Set 7.1. Now let's do some work with perpendicular lines. Be sure to set your boundaries so that the distance between tick marks is the same on both axes. (a) Graph \(y=4 x\) and \(y=-\frac{1}{4} x\) on the same set of axes. Do they appear to be perpendicular lines? (b) Graph \(y=3 x\) and \(y=\frac{1}{3} x\) on the same set of axes. Do they appear to be perpendicular lines? (c) Graph \(y=\frac{2}{5} x-1\) and \(y=-\frac{5}{2} x+2\) on the same set of axes. Do they appear to be perpendicular lines? (d) Graph \(y=\frac{3}{4} x-3, y=\frac{4}{3} x+2\), and \(y=-\frac{4}{3} x+2\) on the same set of axes. Does there appear to be a pair of perpendicular lines? (e) On the basis of your results in parts (a) through (d), make a statement about how we can recognize perpendicular lines from their equations.

The problem of finding the perpendicular bisector of a line segment presents itself often in the study of analytic geometry. As with any problem of writing the equation of a line, you must determine the slope of the line and a point that the line passes through. A perpendicular bisector passes through the midpoint of the line segment and has a slope that is the negative reciprocal of the slope of the line segment. The problem can be solved as follows: Find the perpendicular bisector of the line segment between the points \((1,-2)\) and \((7,8)\). The midpoint of the line segment is \(\left(\frac{1+7}{2}, \frac{-2+8}{2}\right)\) \(=(4,3)\). \(=(4,3)\). The slope of the line segment is \(m=\frac{8-(-2)}{7-1}\) \(=\frac{10}{6}=\frac{5}{3}\) Hence the perpendicular bisector will pass through the point \((4,3)\) and have a slope of \(m=-\frac{3}{5}\). $$ \begin{aligned} y-3 &=-\frac{3}{5}(x-4) \\ 5(y-3) &=-3(x-4) \\ 5 y-15 &=-3 x+12 \\ 3 x+5 y &=27 \end{aligned} $$ Thus the equation of the perpendicular bisector of the line segment between the points \((1,-2)\) and \((7,8)\) is \(3 x+5 y=27 .\) Find the perpendicular bisector of the line segment between the points for the following. Write the equation in standard form. (a) \((-1,2)\) and \((3,0)\) (b) \((6,-10)\) and \((-4,2)\) (c) \((-7,-3)\) and \((5,9)\) (d) \((0,4)\) and \((12,-4)\)

The cost (c) of playing an online computer game for a time \((t)\) in hours is given by the equation \(c=3 t+5\). Label the horizontal axis \(t\) and the vertical axis \(c\), and graph the equation for nonnegative values of \(t\).

Verify that the points \((-3,1),(5,7)\), and \((8,3)\) are vertices of a right triangle. [Hint: If \(a^{2}+b^{2}=c^{2}\), then it is a right triangle with the right angle opposite side \(c .]\)

Now let's use a graphing calculator to get a graph of \(\mathrm{C}=\frac{5}{9}(\mathrm{~F}-32)\). By letting \(\mathrm{F}=x\) and \(\mathrm{C}=y\), we obtain Figure 7.15. Pay special attention to the boundaries on \(x\). These values were chosen so that the fraction \(\frac{\text { (Maximum value of } x \text { ) minus (Minimum value of } x \text { ) }}{95}\) would be equal to 1 . The viewing window of the graphing calculator used to produce Figure \(7.15\) is 95 pixels (dots) wide. Therefore, we use 95 as the denominator of the fraction. We chose the boundaries for \(y\) to make sure that the cursor would be visible on the screen when we looked for certain values. Now let's use the TRACE feature of the graphing calculator to complete the following table. Note that the cursor moves in increments of 1 as we trace along the graph. \begin{tabular}{l|lllllllll} F & \(-5\) & 5 & 9 & 11 & 12 & 20 & 30 & 45 & 60 \\ \hline C & & & & & & & & & \end{tabular} (This was accomplished by setting the aforementioned fraction equal to 1 .) By moving the cursor to each of the F values, we can complete the table as follows. \begin{tabular}{r|rrrrrrrrr} F & \(-5\) & 5 & 9 & 11 & 12 & 20 & 30 & 45 & 60 \\ \hline C & \(-21\) & \(-15\) & \(-13\) & \(-12\) & \(-11\) & \(-7\) & \(-1\) & 7 & 16 \end{tabular} The \(\mathrm{C}\) values are expressed to the nearest degree. Use your calculator and check the values in the table by using the equation \(\mathrm{C}=\frac{5}{9}(\mathrm{~F}-32)\).
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