Chapter 8: Problem 97
Write an equation for a linear function whose graph has the given characteristics. See Example 7. Slope \(\frac{1}{5},\) passes through \((10,1)\)
Short Answer
Expert verified
The equation is \( y = \frac{1}{5}x - 1 \).
Step by step solution
01
Understand the Linear Function Format
A linear function can be generally expressed in the slope-intercept form as \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. Given the slope \( m = \frac{1}{5} \), we will use this in our equation.
02
Use the Point-Slope Form
Since we know a point \((10,1)\) on the line and the slope \( m = \frac{1}{5} \), we can use the point-slope form \( y - y_1 = m(x - x_1) \) where \((x_1, y_1)\) is a point on the line. Here \((x_1, y_1) = (10, 1)\).
03
Substitute Values into the Point-Slope Equation
Substitute the values into the equation: \( y - 1 = \frac{1}{5}(x - 10) \). This uses \( x_1 = 10 \), \( y_1 = 1 \), and \( m = \frac{1}{5} \).
04
Simplify the Equation
Simplify the equation from the point-slope form to the slope-intercept form. Expand the equation: \( y - 1 = \frac{1}{5}(x - 10) \) becomes \( y - 1 = \frac{1}{5}x - 2 \).
05
Solve for \( y \)
Add 1 to both sides to isolate \( y \): \( y = \frac{1}{5}x - 2 + 1 \), which simplifies to \( y = \frac{1}{5}x - 1 \). This is the slope-intercept form of the equation.
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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 popular way of writing linear equations. It allows you to directly identify the two main features of a line: its slope and its y-intercept. The equation is expressed as \( y = mx + b \).
In this format, the \( m \) stands for the slope of the line. This tells us how much the line rises (or falls) vertically for every unit it moves horizontally. The \( b \) is the y-intercept, which indicates where the line crosses the y-axis.
In the given exercise, once we simplified the equation from the point-slope form to the slope-intercept form, we ended up with \( y = \frac{1}{5}x - 1 \). Here, \( \frac{1}{5} \) is the slope, showing how steep or flat the line is, and \( -1 \) is the y-intercept, indicating the line crosses the y-axis at -1.
In this format, the \( m \) stands for the slope of the line. This tells us how much the line rises (or falls) vertically for every unit it moves horizontally. The \( b \) is the y-intercept, which indicates where the line crosses the y-axis.
In the given exercise, once we simplified the equation from the point-slope form to the slope-intercept form, we ended up with \( y = \frac{1}{5}x - 1 \). Here, \( \frac{1}{5} \) is the slope, showing how steep or flat the line is, and \( -1 \) is the y-intercept, indicating the line crosses the y-axis at -1.
- Slope (\( m \)): \( \frac{1}{5} \) – The line rises one-fifth of a unit for each unit of increase in \( x \).
- Y-intercept (\( b \)): \( -1 \) – The point where the line hits the y-axis.
Point-Slope Form
While the slope-intercept form is great for identifying the slope and y-intercept quickly, the point-slope form helps when we know a specific point on the line and its slope. The point-slope form equation is written as \( y - y_1 = m(x - x_1) \).
Using this format, \((x_1, y_1)\) represents a specific point on the line, while \( m \) is still the slope.
In the exercise, the given point \((10, 1)\) and slope \( \frac{1}{5} \) were plugged into the point-slope equation, resulting in \( y - 1 = \frac{1}{5}(x - 10) \). This form shines when you have a point but do not yet know the y-intercept.
Using this format, \((x_1, y_1)\) represents a specific point on the line, while \( m \) is still the slope.
In the exercise, the given point \((10, 1)\) and slope \( \frac{1}{5} \) were plugged into the point-slope equation, resulting in \( y - 1 = \frac{1}{5}(x - 10) \). This form shines when you have a point but do not yet know the y-intercept.
- Point (\((x_1, y_1)\)): \((10, 1)\)
- Slope (\( m \)): \( \frac{1}{5} \)
Graphing Linear Functions
Graphing a linear function involves plotting points on a coordinate plane and drawing a straight line through them. This visual representation is significant for understanding the behavior of the function.
To graph the equation obtained in the exercise, \( y = \frac{1}{5}x - 1 \), we can start by plotting the y-intercept \((0, -1)\). This is the point where the line crosses the y-axis.
From the y-intercept, use the slope to find another point. Since the slope is \( \frac{1}{5} \), we can go up 1 unit and over 5 units to the right starting from the y-intercept to plot the next point.
Once at least two points are plotted, draw a straight line through them to extend infinitely in both directions.
To graph the equation obtained in the exercise, \( y = \frac{1}{5}x - 1 \), we can start by plotting the y-intercept \((0, -1)\). This is the point where the line crosses the y-axis.
From the y-intercept, use the slope to find another point. Since the slope is \( \frac{1}{5} \), we can go up 1 unit and over 5 units to the right starting from the y-intercept to plot the next point.
Once at least two points are plotted, draw a straight line through them to extend infinitely in both directions.
- Y-intercept: Start point for graphing, \((0, -1)\)
- Slope: Guides how to move from one point to another (rise over run)
- Line: Extending through plotted points, representing function behavior