To draw the boundary lines, we will treat them as equations instead of inequalities.
|c|c|
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Inequality & Equation -2pt
[-0.7em]
y≥ |x|-3 & y= |x|-3 [0.5em]
[-0.7em]
y≤- |x|+5 & y= - |x|+5 [0.5em]
To graph an absolute value function, we should consider its general form.
y= a|x-h|+k
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Horizontal Translation:& h
Vertical Translation:& k
Vertical Stretch:& aAs we can see, the constants and the coefficient control different transformations of the parent function.
y=|x|
Let's start with the graph's translation. The first function is translated down by 3 units. The second function is translated up by 5 units. Notice that both inequalities are non-strict which means we keep the lines solid.
Now we have the boundary line of the first inequality. To complete the graph of the second inequality's boundary line, we have to reflect the translated graph in the horizontal line through its locator point at (0,5).
Shading
To draw the inequalities, we have to shade the correct side of the boundary lines. We can do that by testing a point that does not lie on any of the boundary lines. The easiest point we can choose is the origin.
|c|c|c|
-3pt (x,y) -3pt & Inequality & Evaluate
( 0, 0) & 0? ≥| 0|-3 & 0≥-3 ✓ [0.5em]
( 0, 0) & 0? ≤- | 0|+5 & 0 ≤ 5 ✓ [0.5em]
Since the origin produces a true statement for both inequalities, we should shade the side of the boundary line that contains the origin.
What Polygon is Formed?
The polygon resembles a square. If it is, adjacent sides are perpendicular and all sides have the same length. To determine if adjacent sides are perpendicular, we will zoom in the graph and determine the slope och each individual segment. We will also determine the coordinates of the polygon's vertices.
As we can see, adjacent sides have a slope of m= 1 and m= -1. If the sides are perpendicular, the product of the slopes equals -1.
1( -1)=-1
Adjacent sides are perpendicular which means this is a rectangle. Finally, we will calculate the length of each side using the Distance Formula.
|c|c|c|
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Points & -3pt sqrt((x_2-x_1)^2+(y_2-y_1)^2) -3pt & -3pt d -3pt [0.2em]
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(-4,1) (0,-3) & sqrt(( -4- 0)^2+( 1-( -3))^2) & 4sqrt(2) [0.2em]
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(-4,1) (0,5) & sqrt(( -4- 0)^2+( 1- 5)^2) & 4sqrt(2) [0.2em]
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(4,1) (0,-3) & sqrt(( 4- 0)^2+( 1-( -3))^2) & 4sqrt(2) [0.2em]
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(4,1) (0,5) & sqrt(( 4- 0)^2+( 1- 5)^2) & 4sqrt(2) [0.2em]
As we can see, the rectangle's sides all have the same length. This means the shape is in fact a square.
b
The vertices of the polygon are located at the intersections of the two absolute value functions and at their locator points. By examining the graph we can identify the coordinates.
The vertices are located at (-4,1), (0,5), (4,1), and (0,-3).
c
From Part A, we know that this is a square with a side of 4sqrt(2) units. To calculate the area of a square we have to square its side.
