Graphing Absolute Value on the TI-86

This page will demonstrate graphing problem Q16d on page 121: |6x+8|+26 To do this we will create the expression (abs(6X+8)+26)*3-1 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(x)= screen by pressing the key, and then selecting the y(x)= menu option via the key. We start with the left parenthesis, . In order to insert the abs we need to move first to the MATH menu, via the key sequence.
Figure 2 Figure 2 shows the MATH menu displayed at the bottom. We need to select the NUM menu option so we press the key. This will open the NUM sub-menu, which we can see displayed at the bottom of Figure 3.

Figure 3 The abs option is in the fifth position, so we press to insert abs into the expression we are creating. Figure 3 illustrates the resulting text.

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 by pressing .

Figure 5 Figure 5 shows the expression as far as we have entered it, and the TEST menu at the bottom. We want the fifth item in the sub-menu. Therefore, we will press to insert the "greater than or equal to" symbol into our expression.

Figure 6 Figure 6 confirms the insertion of the symbol, along withour entry of the rest of the original problem statement. Note that as the expression gets too long for the line on the screen, the TI-86 scrolls the expression off to the left. Even though part of the expression disappears, it is all still there.

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 repeats the completed expression. However, in Figure 8, we have closed the TEST sub-menu by pressing the key. Now, to actually graph the function we need only press to select GRAPH from the top menu.

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 WIND 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 -.6666666667. 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

|6x+8|+26 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, y1=(abs(6x+8)+26)*3-1 Now we will look at the related problem |6x+8|+2-60 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 y2=abs(6x+8)+2-6 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(x)= 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 y2=abs(6X+8)+2-6 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.

Figure 1	We start this sequence on the y(x)= screen by pressing the key, and then selecting the y(x)= menu option via the key. We start with the left parenthesis, . In order to insert the abs we need to move first to the MATH menu, via the key sequence.
Figure 2	Figure 2 shows the MATH menu displayed at the bottom. We need to select the NUM menu option so we press the key. This will open the NUM sub-menu, which we can see displayed at the bottom of Figure 3.
Figure 3	The abs option is in the fifth position, so we press to insert abs into the expression we are creating. Figure 3 illustrates the resulting text.
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 by pressing .
Figure 5	Figure 5 shows the expression as far as we have entered it, and the TEST menu at the bottom. We want the fifth item in the sub-menu. Therefore, we will press to insert the "greater than or equal to" symbol into our expression.
Figure 6	Figure 6 confirms the insertion of the symbol, along withour entry of the rest of the original problem statement. Note that as the expression gets too long for the line on the screen, the TI-86 scrolls the expression off to the left. Even though part of the expression disappears, it is all still there.
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 repeats the completed expression. However, in Figure 8, we have closed the TEST sub-menu by pressing the key. Now, to actually graph the function we need only press to select GRAPH from the top menu.
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 WIND 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 -.6666666667. 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.

Figure 14	In Figure 14 we have returned to the y(x)= 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 y2=abs(6X+8)+2-6 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.