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| 8 Theory slides |
| 10 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
x | y=∣3x−6∣+3 | y |
---|---|---|
0 | y=∣3(0)−6∣+3 | 9 |
1 | y=∣3(1)−6∣+3 | 6 |
2 | y=∣3(2)−6∣+3 | 3 |
3 | y=∣3(3)−6∣+3 | 6 |
4 | y=∣3(4)−6∣+3 | 9 |
Plot the points and draw the boundary line. Since the given inequality is strict, the boundary line will be dashed.
x=0, y=0
Zero Property of Multiplication
Subtract terms
∣-6∣=6
The test point is not a solution, so the region that does not contain the test point — inside the V-shaped graph — represents the solution set for the inequality. Finally, the appropriate region can be shaded.
Analyze the graph of the given absolute value inequality in two variables to determine the proper inequality symbol.
LHS+4=RHS+4
a=22⋅a
Factor out 25
∣a⋅b∣=∣a∣⋅∣b∣
∣∣∣∣∣25∣∣∣∣∣=25
a=-(-a)
x | -25∣x+2∣+4 | y=-25∣x+2∣+4 |
---|---|---|
-4 | -25∣-4+2∣+4=-1 | -1 |
-3 | -25∣-3+2∣+4=1.5 | 1.5 |
-2 | -25∣-2+2∣+4=4 | 4 |
-1 | -25∣-1+2∣+4=1.5 | 1.5 |
0 | -25∣0+2∣+4=-1 | -1 |
Finally, plot the points and draw the absolute value equation.
As can be seen, the mirror has a V-shape with a vertex of (-2,4) in the coordinate plane.
x | y=0.4∣x−4∣+3 | y |
---|---|---|
0 | y=0.4∣0−4∣+3 | 4.6 |
2 | y=0.4∣2−4∣+3 | 3.8 |
4 | y=0.4∣4−4∣+3 | 3 |
6 | y=0.4∣6−4∣+3 | 3.8 |
8 | y=0.4∣8−4∣+3 | 4.6 |
Plot the points and draw the boundary line. Since the inequality is non-strict, the boundary line will be solid. Note that only positive values of x and y makes sense in the context of the situation.
In the same auction, Emily was also interested in the object that Ignacio bought. Emily knows that Ignacio will sell it sooner or later. The maximum amount of money Emily can allocate to buy Ignacio's object is represented by the following graph.
To write the equation of the boundary line, use the vertex form of an absolute value equation y=a∣x−h∣+k, where (h,k) is the vertex of the absolute value equation.
On the given graph, x represents the years after the auction and y represents the price of the object in thousands of dollars. Since Emily can allocate any amount less than $3800 for the object 4 years after the auction, these values correspond to point (4,3.8), which is the vertex of the boundary line.
x=10, y=5
Subtract term
∣6∣=6
LHS−3.8=RHS−3.8
LHS/6=RHS/6
Rearrange equation
less than,<.
Any point in this region is the solution to both inequalities. In the context of the last two examples, Ignacio and Emily can make a deal as long as the values stay in that region.
Which graph represents the solution set of the inequality ∣y∣<∣x∣?
We can write the given inequality as a compound inequality. |y| < |x| ⇓ y < |x| and y > - |x| We will draw the inequalities separately. Then, we will find the overlapping region of the inequalities, which represents the solution set the given inequality.
Let's start by determining the boundary line of the inequality by changing the inequality symbol to an equal sign. Inequality & Boundary Line y < |x| & y = |x| We have seen that an absolute value equation y=|x| gives us a vertical V-shaped graph with the vertex at the origin. Since the inequality is strict, the boundary line should be dashed.
We can now test a point to determine the region to be shaded. Let's use (1,0). y < |x| [( 1, 0)]Substitute 0 < | 1| ✓ Because (1,0) created a true statement, we will shade the region that contains this point.
The boundary line of the second inequality is y = - |x|. Inequality & Boundary Line y > - |x| & y = - |x| The graph of this boundary line is an inverted V-shaped graph with vertex at the origin. The boundary line is also dashed because of the strict inequality symbol.
We can use (1,0) again to test which side of the line to shade. y > - |x| [( 1, 0)]Substitute 0 > - | 1| ✓ Substituting the point created a true statement, so the region that contains this point must be shaded.
Let's draw the graphs of the inequalities on the same coordinate plane.
Finally, we remove the shading from the unnecessary regions. We are only interested in the overlapping regions of the two inequalities, so we will consider only this area.
We obtained the graph from option C. Therefore, this is the correct answer.