To graph the linear equation \( y = \frac{2}{3}x - 1 \), we can follow these steps:
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Identify the slope and y-intercept:
- The equation is in slope-intercept form \( y = mx + b \), where:
- \( m = \frac{2}{3} \) (the slope)
- \( b = -1 \) (the y-intercept)
- The equation is in slope-intercept form \( y = mx + b \), where:
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Plot the y-intercept:
- Start by plotting the point where the line crosses the y-axis. This occurs at \( (0, -1) \).
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Use the slope:
- From the y-intercept \( (0, -1) \), the slope \( \frac{2}{3} \) means you can go up 2 units and to the right 3 units. This will give you another point:
- Starting from \( (0, -1) \), move up 2 to \( (0, 1) \) and right 3 to \( (3, 1) \). So, plot the point \( (3, 1) \).
- From the y-intercept \( (0, -1) \), the slope \( \frac{2}{3} \) means you can go up 2 units and to the right 3 units. This will give you another point:
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Draw the line:
- Draw a straight line through the points \( (0, -1) \) and \( (3, 1) \), and extend it with arrows on both ends.
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Check for other points:
- You can verify by substituting additional values for \( x \) (negative or beyond the plotted points) to ensure the line is accurate.
In the response options, you should look for a plot that accurately reflects the points \( (0, -1) \) and \( (3, 1) \), following the slope of \( \frac{2}{3} \).
If the graphed points and line correctly show this behavior, that is your answer. The correct description is likely:
- A coordinate plane shows the x-axis ranging from negative 9 to 9 in increments of 1 and the y-axis ranging from 11 to negative 11 in increments of 1. A line with arrows at both ends joins two plotted points. The coordinates of the plotted points are as follows: \( (0, -1) \) and \( (3, 1) \).
This description accurately reflects the graph for the equation \( y = \frac{2}{3}x - 1 \).