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Chapter 2: Problem 19

For each of the following, find the slope and the vertical intercept, then sketch the graph. (Hint: Find two points on the line.) a. \(y=0.4 x-20\) b. \(P=4000-200 C\)

Short Answer

Expert verified
a. Slope: 0.4, Vertical Intercept: -20. b. Slope: -200, Vertical Intercept: 4000.

Step by step solution

01

Title - Identify the slope and vertical intercept for part a

The given equation is in the form of the slope-intercept form: \( y = mx + b \) where \( m \) is the slope and \( b \) is the vertical intercept. For the equation \( y = 0.4x - 20 \), the slope \( m = 0.4 \), and the vertical intercept \( b = -20 \).
02

Title - Find two points on the line for part a

To find two points on the line, choose any two values for \( x \) and solve for \( y \). If \( x = 0 \), then \( y = 0.4(0) - 20 = -20 \). The point is (0, -20). If \( x = 10 \), then \( y = 0.4(10) - 20 = 4 - 20 = -16 \). The point is (10, -16).
03

Title - Sketch the graph for part a

Plot the points (0, -20) and (10, -16) on the graph, then draw a straight line through these points. This is the graph of the equation \( y = 0.4x - 20 \).
04

Title - Identify the slope and vertical intercept for part b

The given equation is in the form of the slope-intercept form: \( P = mx + b \) where \( m \) is the slope and \( b \) is the vertical intercept (though in this case, \( x \) is represented by \( C \)). For the equation \( P = 4000 - 200C \), the slope \( m = -200 \), and the vertical intercept \( b = 4000 \).
05

Title - Find two points on the line for part b

To find two points on the line, choose any two values for \( C \) and solve for \( P \). If \( C = 0 \), then \( P = 4000 - 200(0) = 4000 \). The point is (0, 4000). If \( C = 10 \), then \( P = 4000 - 200(10) = 4000 - 2000 = 2000 \). The point is (10, 2000).
06

Title - Sketch the graph for part b

Plot the points (0, 4000) and (10, 2000) on the graph, then draw a straight line through these points. This is the graph of the equation \( P = 4000 - 200C \).

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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 way of writing linear equations. In this form, the equation of a line is written as: \[ y = mx + b \]Here,
  • \( m \) represents the slope of the line. The slope indicates how steep the line is and the direction it goes.
  • \( b \) is the y-intercept, which is the point where the line crosses the y-axis.
For example, in the equation \( y = 0.4x - 20 \), the slope \( m \) is 0.4, and the y-intercept \( b \) is -20. This means the line goes up 0.4 units for each 1 unit it goes to the right, starting from the point (0, -20). Understanding the slope-intercept form makes it easier to visualize and graph linear equations. When the equation is in this form, you only need the slope and the y-intercept to sketch the graph.
linear equations
Linear equations represent straight lines when graphed on a coordinate plane. These equations come in different forms, but the most common is the slope-intercept form: \( y = mx + b \).Some key points to understand about linear equations include:
  • They have constant rates of change, represented by their slopes.
  • They create straight lines because the relationship between the variables is linear.
  • Each solution to the equation represents a point on the line.
For instance, in the equation \( P = 4000 - 200C \), setting different values for \( C \) results in different values for \( P \), but they all fall on a straight line when plotted. This consistency is the defining feature of linear equations, making them predictable and easy to work with. Knowing how to identify and work with linear equations is a fundamental skill in algebra and beyond.
graphing lines
Graphing lines involves plotting points on a coordinate plane and then connecting them to form a straight line. With the slope-intercept form \( y = mx + b \), there are specific steps you can follow to graph a line:
  • Identify the slope \( m \) and the y-intercept \( b \) from the equation.
  • Plot the y-intercept point \((0, b)\) on the graph.
  • Use the slope to find another point. For a slope of \( m = \frac{\text{rise}}{\text{run}} \), move up/down (rise) and left/right (run) from the y-intercept to locate a second point.
  • Draw a line through the two points, extending it across the graph.
For example, given \( y = 0.4x - 20 \), with slope 0.4 and y-intercept -20:
  • First, plot (0, -20).
  • From there, move up 0.4 units and to the right 1 unit to find another point (1, -19.6). (However, more practical is to choose integer values for simplicity, like (10, -16)).
  • Then draw the line through these points.
Repeat this process for any linear equation to visualize it on a graph. Graphing helps you understand the behavior of linear equations beyond just the numbers.

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Most popular questions from this chapter

The \(y\) -axis, the \(x\) -axis, the line \(x=6,\) and the line \(y=12\) determine the four sides of a 6 -by- 12 rectangle in the first quadrant (where \(x>0\) and \(y>0\) ) of the \(x y\) plane. Imagine that this rectangle is a pool table. There are pockets at the four corners and at the points (0,6) and (6,6) in the middle of each of the longer sides. When a ball bounces off one of the sides of the table, it obeys the "pool rule": The slope of the path after the bounce is the negative of the slope before the bounce. (Hint: It helps to sketch the pool table on a piece of graph paper first.) a. Your pool ball is at (3,8) . You hit it toward the \(y\) -axis, along the line with slope \(2 .\) i. Where does it hit the \(y\) -axis? ii. If the ball is hit hard enough, where does it hit the side of the table next? And after that? And after that? iii. Show that the ball ultimately returns to \((3,8) .\) Would it do this if the slope had been different from \(2 ?\) What is special about the slope 2 for this table? b. A ball at (3,8) is hit toward the \(y\) -axis and bounces off it at \(\left(0, \frac{16}{3}\right) .\) Does it end up in one of the pockets? If so, what are the coordinates of that pocket? c. Your pool ball is at (2,9) . You want to shoot it into the pocket at \((6,0) .\) Unfortunately, there is another ball at (4,4.5) that may be in the way. i. Can you shoot directly into the pocket at (6,0)\(?\) ii. You want to get around the other ball by bouncing yours off the \(y\) -axis. If you hit the \(y\) -axis at \((0,7),\) do you end up in the pocket? Where do you hit the line \(x=6 ?\) iii. If bouncing off the \(y\) -axis at (0,7) didn't work, perhaps there is some point \((0, b)\) on the \(y\) -axis from which the ball would bounce into the pocket at (6,0) Try to find that point.

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