Graphing Absolute Value on the TI-83

This page will demonstrate graphing problem Q16d on page 121:
To do this we will create the expression
which will have the value of -1 for all X where |6X+8|+26 is false, and the value 2 where that expression is true.

Figure 1
We start this sequence on the Y= screen by pressing the key. We start with the left parenthesis, . In order to insert the abs( we need to move first to the TEST screen, via the key sequence.
Figure 2
Fortunately for us, the abs( item is the very first in the CATALOG, as shown in Figure 2. The arrow pointer to the left of the abs( item could be moved with the up and down cursor keys, but we do not need to move it at all in this case. We press the key to select this item and put it into the expression we were creting.
Figure 3
Figure 3 shows the expression with the abs( pasted in from the CATALOG screen.
Figure 4
In Figure 4 we have added more of the expression. We now need to enter the symbol for "greater than or equal to". To do this we need to move to the TEST menu via the sequence.
Figure 5
Here is the test menu. The item we want is item number 4. We need only press the to select the item, close the screen and insert the item into the expression.
Figure 6
Figure 6 confirms the insertion of the symbol.
Figure 7
We finish the expression in Figure 7. Each time this expression is evaluated it will produce a 0 if it is false, and a 1 if it is true. We wanted to change those values so that the graph does not fall on the X-axis. The opening and closing parentheses allow us to multiply the expression by 3 and then we can subtract 1. This will change false values to -1 and true values to 2.
Figure 8
Figure 8 shows the completed expression. We are ready to graph the expression. To do this we press the key.
Figure 9
Here is the graph of the solution. We can see that the expression is true everywhere except from about -2 to about -2/3. We can focus in on this region by changing the WINDOW settings. To do this we press the key.
Figure 10
Figure 10 shows the WINDOW settings after we have changed them to concentrate our graph on the region from -2.5 to 0.5. At the same time we have changed the Y-values to range from -4 to 6.
Figure 11
Pessing the key returns us to the GRAPH screen shown in Figure 11.
Figure 12
For Figure 12 we have pressed the key to move into the TRACE mode of the calculator, and we have used the key to move to the right end of the lower segement of the graph. Note that the X-value is -.6808511. If we had moved one more pixel to the right we would jump up to the upper portion of the graph, and the X-value would be greater than -2/3. If we wanted to, we could return to the WINDOW settings and change them so that we examine just the region from -.68 to -.66, giving us an even better reading of the X-values at the ends of the upper and lower segments. For this example we will not do this. The graph here is meant to confirm, to check, the algebraic solution in the text.
Figure 13
We can use the key to move the TRACE pointer to the left. In Figure 13 we have moved it until it has jumped to the left upper portion of the graph. Again, we can see the X-value of the particular pixel.

The final three figures on this page are meant to demonstrate the relation between what we say we are doing, and what we are really doing. The original problem was to solve

Looking at this problem, here or in the text, we might notice that the variable Y does not appear in the problem. The original problem is an inequality in one variable, X. In the Figures above, we have worked to create a two-level graph of this problem, showing where the inequality is true and where it is false. We really created a two variable function, namely,
Now we will look at the related problem
where we have simply moved the 6 to the left side of the sign, giving us an expression on the left that is greater than or equal to 0. If we look at
then we are really asking, where is Y2 greater than or equal to 0.

Figure 14
In Figure 14 we have returned to the Y= screen and we have entered the new, unsimplified, equation.
Figure 15
We return to the GRAPH screen and we can see not only the original two level function, but also the graph of our new function
which comes out as a "V". Notice that the "V" is at or above the X-axis for all X-values where the two level function is at the upper level, i.e., is true. And, the "V" is below the X-axis for those values of X where the two level function is at -1, i.e., is false. In order to better see this, Figure 15 has been augmented and reproduced as Figure 16.
Figure 16
The change from Figure 15 to Figure 16 is that we have drawn two red and two blue vertical lines. The red lines are above and below the endpoints of the lower level segment of the two level function. That is, they mark the left and right ends of the region where the original expression was false. The blue lines are above and below the endpoints of the upper level segments of the two level function. That is, they mark the right and left ends of the region where the original expression was true.

These red and blue lines were added to make it easier to see where the "V" graph of the second function is above the X-axis and where it is below the X-axis.

©Roger M. Palay
Saline, MI 48176
February, 1999