Charlie does the following problem:
2 answers:
You might be interested in
Step-by-step explanation:
1. You'll have to express the linear inequality into an equality statement (in slope-intercept form, y = mx + b) in order to graph. However, you'd have to use a dashed line because of the inequality symbol, ">"
y = – 3x + 1
2. Next, you'll have to choose two points to use for graphing. It's always safe to start with the y-intercept, (0, 1).
3. From there, you could use the slope (m = -3/1) and do the "rise over run" technique.
- The only difference is going 3 units <u>down</u> (rise), and 1 unit to the <u>right</u> (run).
- You'll also go 3 units <u>up</u> (rise)<u>,</u> 1 unit to the <u>left</u> (run).
- You need at least 2 points to create a line.
4. Once you create a line, you'll have to choose a test point to see which region of the plane you need to shade. Let's choose the point of origin, (0, 0).
5. Plug these values into the inequality statement to see if it satisfies the inequality:
y > 1 – 3x
0 > 1 – 3(0)
0 > 1 – 0
0 > 1 (<u>False statement,</u> because 0 is not greater than 1).
Therefore, you'll have to shade the region where it doesn't contain the point, (0, 0) (shade the right side of the region).
Attached is the screenshots of the graph I created. The first one shows how I did step #3, where I started with plotting the y-intercept, (0, 1) and used the slope to plot other points and create the line. The 2nd screenshot is the final graph, where it shows the shaded region.
Hope this helps :)
