To graph the linear equation \( y = 2x - 3 \), let's identify some points on the line by substituting values for \( x \):
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When \( x = 0 \): \[ y = 2(0) - 3 = -3 \] This gives the point \( (0, -3) \).
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When \( x = 1 \): \[ y = 2(1) - 3 = -1 \] This gives the point \( (1, -1) \).
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When \( x = 2 \): \[ y = 2(2) - 3 = 1 \] This gives the point \( (2, 1) \).
Now we have three points that lie on the line: \( (0, -3) \), \( (1, -1) \), and \( (2, 1) \).
Once plotted on the coordinate plane:
- The point \( (0, -3) \) is on the y-axis.
- The point \( (1, -1) \) is one unit to the right and one unit up from the origin.
- The point \( (2, 1) \) is two units to the right and one unit up from the previous point.
A line drawn through these points will be the graphical representation of the equation. The line has a positive slope of 2 (indicating it rises 2 units for every 1 unit it moves to the right) and crosses the y-axis at -3.
Given this information, the correct illustration of the graph for \( y = 2x - 3 \) will show a line passing through the points \( (0, -3) \) and \( (1, -1) \), and will extend infinitely with arrows at both ends.
The options you provided include incorrect representations of the graph. None of the points or slopes match \( y = 2x - 3 \). The correct graphical representation will not be among the options you've listed.