We'll graph a line using the slope-intercept form of a linear equation, y equals m x plus b. Here, m represents the slope of the line, and b is the y-intercept. In our case, the slope m is one-half, and the y-intercept b is negative one. So our specific equation is y equals one-half x minus one.
The y-intercept is the point where the line crosses the y-axis, which occurs when x equals zero. To find this point, we substitute x equals zero into our equation y equals one-half x minus one. This gives us y equals one-half times zero minus one, which simplifies to y equals negative one. Therefore, our y-intercept is the point zero, negative one. Let's plot this point on our coordinate plane.
The slope of one-half can be interpreted as rise over run equals one over two. This means that for every two units we move to the right along the x-axis, we move one unit up along the y-axis. Starting from our y-intercept point zero, negative one, if we move two units to the right and one unit up, we arrive at the point two, zero. This movement is fundamental to plotting additional points on our line.
Now that we understand the slope, let's find more points on our line. Starting from our y-intercept point zero, negative one, we apply the slope of one-half. Moving two units right and one unit up from zero, negative one brings us to the point two, zero. We can repeat this process: from two, zero, moving two units right and one unit up takes us to the point four, one. These three points zero, negative one; two, zero; and four, one are all on our line.
To draw our complete line, we connect the three points we've plotted: zero, negative one; two, zero; and four, one. Since a line extends infinitely in both directions, we add arrows at both ends to indicate this. The equation y equals one-half x minus one is written next to the line to show the relationship it represents. This line correctly shows all points that satisfy our linear equation.
Let's verify our work by checking that our plotted points satisfy the equation y equals one-half x minus one. For the point zero, negative one: substituting x equals zero gives y equals one-half times zero minus one, which equals negative one. This checks out. For two, zero: one-half times two minus one equals zero. Correct again. And for four, one: one-half times four minus one equals one. Perfect. Key features of our graph include the positive slope, shown by the line rising from left to right, the y-intercept at zero, negative one, and the x-intercept at two, zero, where the line crosses the x-axis.
To summarize our process, we started with the slope-intercept form of a linear equation, y equals m x plus b. We identified our slope as one-half and our y-intercept as negative one. We plotted the y-intercept point zero, negative one, then used the slope to find additional points by moving two units right and one unit up each time. Finally, we connected these points with a straight line extending infinitely in both directions. The resulting graph accurately represents all solutions to our equation y equals one-half x minus one.