MAT 222 week 5 discusion

MAT 222 Week 5 Discussion

For Week 5’s Discussion, we are to solve problems according to the last letter of our last name. I will be solving problems #16 and #50 on pages 708-711.

#16) h(x) = |-2x|

x |-2x|
-5 |-2(-5)| = 10
-4 |-2(-4)| = 8
-3 |-2(-3)| = 6
-2 |-2(-2)| = 4
-1 |-2(-1)| = 2
0 |-2(0)| = 0
1 |-2(1)| = 2
2 |-2(2)|= 4
3 |-2(3)| = 6
4 |-2(4)| = 8
5 |-2(5)| = 10

This is a graph of an absolute value function. Plotting these points in a graph, they would form a v-shape. The v-shape would fall in the first and second quadrant. Since any real number can be used for x in h(x) =|-2x| and because the graph could keep extending to the left and right when adding more points, the domain is ().

If we were looking at the graph, we would see that the vertex of the graph is (0, 0). Because |-2x| is never negative, the graph does not go below the x-axis. This means that the range of the graph would be [0,).

This equation is a function. We can test this using the vertical line test. If we were to draw a straight line through any point it would only touch the graph once, so it passes the vertical line test. However, if the line touched the graph in more than one place it would not be a function.

Select one of your graphs and assume it has been shifted upward three units and four units to the left. Discuss how this transformation affects the equation by rewriting the equation to incorporate those numbers.

Let’s look at the function from #16 and shift it three units upwards and four units to the left. How will this affect the equation?

#50) x= -3

  1. Three units upwards means that we will +3 to the outside of the absolute value bars.
  2. When we move four units to the left this means we will +4 inside the absolute value bars.
  3. The function will now look like: h(x) = |-2x+4| +3. This reflects a transformation of the function three units up and four units to the left. After shifting the graph, it would still fall in the first and second quadrants.
x y
-3 1
-3 -1
-3 2
-3 -2

The graph of this relation would be a vertical line going through (-3, 0). This vertical line would fall in the second and third quadrants. In this relation x an only be equal to -3 so the domain would be {-3}. For the y values, we could input any real number. So the range for x= -3 would be (-). When I graph the points above, all of them fall on the vertical line.

This is not a function because x = -3 on this graph goes to points y=1, y=-1, y=2 and y=-2. This would fail the vertical line test because it touches more than one point on the line. Also, the input value -3 has more than two different output values which shows us this cannot be a function.

CheshireMath. (2012, August 1). 21a Representing relations and functions [Web Video]. Retrieved from

Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY: McGraw-Hill Publishing.

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