Problem 15
Question
In Exercises 15-18, complete the table. Use the resulting solution points to sketch the graph of the equation. \( y = -2x + 5 \)
Step-by-Step Solution
Verified Answer
The solution points of the equation \( y = -2x + 5 \) are (-2,9), (0,5), and (2,1). By plotting these points and drawing a straight line through them, a sketch of the function \( y = -2x + 5 \) is obtained.
1Step 1: Identify the slope and y-intercept
The equation is in the form \( y = mx + c \), where 'm' is the slope and 'c' is the y-intercept. In the equation \( y = -2x + 5 \), the slope 'm' is -2 and the y-intercept 'c' is 5.
2Step 2: Choose values for x
Choose a few values for 'x'. The selection can be arbitrary, but often it's easiest to choose small numbers, including negatives, zero, and positives. Let's choose -2, 0, and 2.
3Step 3: Solve for corresponding y values
Use the chosen x-values and the equation \( y = -2x + 5 \) to solve for the corresponding y-values. For x = -2, y = -2(-2) + 5 = 9. For x = 0, y = -2(0) + 5 = 5. For x = 2, y = -2(2) + 5 = 1. So, the solution points are (-2,9), (0,5), and (2,1).
4Step 4: Sketch the graph
Plot the points on a two-dimensional coordinate system and draw a straight line through them. This line represents the function \( y = -2x + 5 \). Ensure the line crosses the y-axis at the y-intercept point which is 5, validating the graph.
Key Concepts
Understanding the SlopeDefining the Y-InterceptFinding Solution PointsPlotting in the Coordinate System
Understanding the Slope
The slope of a line in a linear equation helps us understand how the line tilts in a coordinate system. It is represented by the letter 'm' in the equation of a line, usually written as \( y = mx + c \). In our given example, \( y = -2x + 5 \), the slope \( m \) is -2.
The numerical value of the slope indicates how steep the line is:
The numerical value of the slope indicates how steep the line is:
- A positive slope means the line rises as you move to the right.
- A negative slope indicates the line falls as you move to the right.
- A larger absolute value means a steeper line.
- When \( m = 0 \), the line is horizontal, displaying no tilt.
Defining the Y-Intercept
The y-intercept is where the line crosses the y-axis on a graph. In our equation \( y = -2x + 5 \), the y-intercept is 5, represented by 'c'.
The y-intercept tells us the starting point of the line on the y-axis when \( x \) is zero. This means if you plug \( x = 0 \) into the equation, you will find \( y = 5 \).
Understanding the y-intercept:
The y-intercept tells us the starting point of the line on the y-axis when \( x \) is zero. This means if you plug \( x = 0 \) into the equation, you will find \( y = 5 \).
Understanding the y-intercept:
- It is the point (0, c) on the graph.
- The y-intercept is essential for sketching the graph accurately.
- It tells us where the line starts rising or falling along the y-axis.
Finding Solution Points
Solution points are specific coordinates that lie on the line of a linear equation. For the equation \( y = -2x + 5 \), we can find these points by plugging in different values for \( x \) and solving for \( y \).
Let's use the calculated values:
Let's use the calculated values:
- For \( x = -2 \), \( y = -2(-2) + 5 = 9 \) resulting in the point (-2, 9).
- For \( x = 0 \), \( y = -2(0) + 5 = 5 \) resulting in the point (0, 5).
- For \( x = 2 \), \( y = -2(2) + 5 = 1 \) resulting in the point (2, 1).
Plotting in the Coordinate System
The coordinate system is a two-dimensional plane where we can visualize linear equations as lines. It is composed of two axes: the horizontal axis (x-axis) and the vertical axis (y-axis).
For graphing our equation \( y = -2x + 5 \):
This graph visually represents the relationship between x and y as described by the equation.
For graphing our equation \( y = -2x + 5 \):
- Start by marking the y-intercept at (0, 5).
- Next, plot the solution points: (-2, 9), (0, 5), and (2, 1).
- Draw a straight line through these points.
This graph visually represents the relationship between x and y as described by the equation.
Other exercises in this chapter
Problem 15
In Exercises 11-18, (a) write the linear function \(f\) such that it has the indicated function values and (b) sketch the graph of the function. \(f(-5) = -1\),
View solution Problem 15
In Exercises 15 and 16, which sets of ordered pairs represent functions from \(A\) to \(B\)? Explain. \(A = \\{0, 1, 2, 3\\}\) and \(B = \\{-2, -1, 0, 1, 2\\}\)
View solution Problem 15
In Exercise 15-24, determine the quadrant(s) in which \( (x, y) \) is located so that the condition(s) is (are) satisfied. \( x > 0 \) and \( y
View solution Problem 16
In Exercises 9-16, find (a) \((f+g)(x)\), (b) \((f-g)(x)\), (c) \((fg)(x)\), and (d) \((f/g)(x). What is the domain of \)f/g\(? \)f(x) = \frac{x}{x+1}\(, \)g(x)
View solution