Question

Graph the line corresponding to the equation \(y=-2 x+1\) by graphing the points corresponding to \(x=0,1,\) and 2 . Give the \(y\) -intercept and slope for the line. How is this line related to the line \(y=2 x+1\) of Exercise \(12.1 ?\)

Short answer

Answer: The two lines have the same y-intercept (1) but different slopes (-2 for the first line and 2 for the second line). This means that they are parallel to each other with respect to the y-axis, sloping in opposite directions.

Step-by-step solution

Find the y values for x = 0, 1, 2

Use the equation \(y=-2x+1\) to find the corresponding \(y\) values for \(x = 0, 1, 2\). When \(x = 0\): \(y = -2(0) + 1 = 1\) When \(x = 1\): \(y = -2(1) + 1 = -1\) When \(x = 2\): \(y = -2(2) + 1 = -3\) Now we have the points \((0, 1)\), \((1, -1)\), and \((2, -3)\).

Plot the points and graph the line

Plot the points \((0, 1)\), \((1, -1)\), and \((2, -3)\) on a coordinate plane. These points should be collinear. Connect the points with a straight line, representing the line of the equation \(y = -2x + 1\).

Identify the y-intercept and slope

The \(y\)-intercept is the point where the line crosses the \(y\)-axis. In this case, it is the point \((0,1)\). So the \(y\)-intercept is 1. The slope is the coefficient in front of \(x\) in the slope-intercept form of the equation. For our line, \(y = -2x + 1\), the slope is -2.

Compare the line to the line from Exercise 12.1

Now, let's compare the given line, \(y = -2x + 1\), to the line given in Exercise 12.1, which is \(y = 2x + 1\). Both lines have the same \(y\)-intercept, as they both cross the \(y\)-axis at the point \((0,1)\). However, they have different slopes. The slope of our given line is -2 whereas the slope of the line in Exercise 12.1 is 2. This means that our given line is sloping downward from left to right while the line in Exercise 12.1 is sloping upward in the same direction. In conclusion, the two lines are parallel to each other with respect to the y-axis, and they have the same y-intercept but different slopes.