Graphing an involves two main steps.
- Plotting the .
- Shading one half-plane to show the solution set.
To graph the inequality, we have to draw the boundary line. The equation of a boundary line is obtained by replacing the inequality symbol with an equals sign.
3x+2y=12
To draw this line, we will first rewrite the equation in .
3x+2y=12
Write in slope-intercept form
2y=-3x+12
y=2-3x+12
y=2-3x+212
y=-1.5x+6
Now that the equation is in slope-intercept form, we can identify the m and b.m=-1.5b=6
We will plot the y-intercept and then use the slope to plot another point on the line. We will then connect these points with a straight line. This is our boundary line. Since the inequality has the symbol ≥ the boundary should be included in the .
Now, to determine which region to shade we'll test a point to see if it's a solution to the inequality. Let's choose the origin as a test point, (0,0).
y≥-1.5x+6
0≥?-1.5⋅0+6
0≥?0+6
0≱6
Since the substitution of the test point did not create a true statement, we will shade the region that does not contain (0,0).
Determining the solutions
Finally, we will plot the given points to determine which ones belong to the solution set.
We can see that no of the points are a solution to the inequality.
To graph the inequality, we have to draw the boundary line. The equation of a boundary line is obtained by replacing the inequality symbol with an equals sign.
8−4y=2x
To draw this line, we will first rewrite the equation in .
8−4y=2x
Write in slope-intercept form
-4y=2x−8
y=-42x+8
y=-42x+-48
y=-42x−48
y=2x−2
Now that the equation is in slope-intercept form, we can identify the m and b.
y=2x−2⇔y=21x+(-2)
We will plot the y-intercept b=2, and then use the slope m=21 to plot another point on the line. We will then connect these points with a straight line. This is our boundary line. Since the inequality has the symbol ≥, the boundary line will be included in the solution set.
To decide which region to shade, we will test a point, that's not on the boundary line. If the substitution creates a true statement, we shade the region that includes the test point. Otherwise, we shade the opposite region. Let's use (0,0) as our test point.
8−4y≥2x
8−4⋅0≥?2⋅0
8−0≥?0
8≥0
Since the substitution of the test point created a true statement, we will shade the region that contains (0,0).
Finally, we will plot the given points to determine which ones belong to the solution set.
Points that lie within the shaded region are solutions to the inequality.
(-2,0),(0,0), and (1,4)