Graph each of the inequalities. The inequality that does not lie in Quadrant IV will have no solutions in it.
C
Practice makes perfect
To decide which one of the given inequalities has no solutions in Quadrant IV, we will graph each of them. The inequality that does not lie in Quadrant IV will have no solutions in it. Let's start with option A. To graph an inequality, we should first rewrite it in general form by isolating the y-variable.
Next, we will determine its boundary line by replacing the inequality sign with an equals sign.
Inequality & & Boundary Line
y ≥ |x-3|-2 & &y = |x-3|-2
To graph the boundary line, we will first determine its vertex and the slope of its branches.
&General Form &&Boundary Line
& y= a|x- h|+ k && y= 1|x- 3| -2
In the general form, ( h, k) is the vertex, a represents the slope of the branch to the right of the vertex, and x= h is the axis of symmetry. We know that the vertex is ( 3, -2), the slope to the right is 1. Let's plot the vertex and determine a second point on the right branch using the slope.
Next, we will draw a line that passes through the points to the right of the vertex. Notice that the boundary line will be solid, since the inequality is non-strict.
Now we will sketch the left branch by reflecting the right one in the axis of symmetry x= 3.
Now that we have the boundary line, there is one step left to complete the graph. We will test a point to decide which region we should shade. Let's use the point (3,1). If it satisfies the inequality, we will shade the region that contains the point. Otherwise, we will shade the other region.
Since the point satisfies the inequality, we will shade the region that contains the point.
As we can see, part of the inequality lies in Quadrant IV, so it has solutions in that quadrant. We will graph the other inequalities in the same way to see if they have any solutions in Quadrant IV.
Inequality
General Form
Boundary Line
Graph
Solutions in Quadrant IV
A. y+2 ≥ |x-3|
y ≥ 1|x- 3| -2
y= 1|x- 3| -2
Yes
B. y >3- |5-x|
y > - 1|x- 5|+ 3
y = - 1|x- 5|+ 3
Yes
C. y-1 >|2x+6|
y > 2|x+ 3|+ 0
y = 2|x+ 3|+ 0
No
D. y≤ |4x|-7
y ≤ 4|x+ 0| -7
y = 4|x+ 0| -7
Yes
We can see that the only inequality that does not have any solutions in Quadrant IV is y-1>|2x+6|. This corresponds to answer C.
