# Assignment 1.1

By

Ken Montgomery

An investigation of

Equation 1:

Letting a = 4, we have Equation 2.

Equation 2:

Plotting this relation (Figure 1) yields a curve that does not pass the vertical line test and is therefore not a function. We notice that the y–intercepts are at –1, and 1 and that the x–intercepts are at –2 and 2 and that the graph passes through the origin.

Figure 1: Graph of  , with a = 4

By increasing the value of a, from a = 4 to a = 5, we obtain Equation 3.

Equation 3:

Again, plotting this relation we obtain a curve of the same general shape, passing through the origin and through the y–intercepts –1 and 1. However, we notice that the absolute value of the x–intercepts has increased (i.e. the x–intercepts have moved away from the origin, Figure 2). From the graph in Figure 2, we approximate the x-intercepts to be -2.25 and 2.25 as points of reference for comparison.

Figure 2: Graph of  , with a = 5

Again we change the value of a, from a = 5 to a = 3, yielding Equation 4.

Equation 4:

Again, also we observe that the curve continues to pass through the origin and through the y-intercepts, –1 and 1 (Figure 3). However, the absolute value of the x-intercepts has decreased from 2.25 to 1.75.

Figure 3: Graph of  , with a = 3

With a = 2, we have Equation 5 and we see from the graph (Figure 4) that the x – intercepts are approximately –1.4 and 1.4 respectively. The “bends” in the curve (located in the areas of (-1.25, -3) and (1.24, 3)) have become more pronounced and seem to be approaching x = 0, as a approaches zero.

Equation 5:

Figure 4: Graph of  , with a = 2

Finally, we let a = 1, obtaining the graph of Equation 6, in Figure 5.

Equation 6:

Figure 5: Graph of  , with a = 1

The figure seems to be a graph of an ellipse and the minor axis, y = x. The composite graph of these combined functions is given in Figure 6, by Equation 7.

Equation 7:

Figure 6: Graph of

We can show that these graphs are in fact identical by factoring Equation 6 into Equation 7.

Ň

We wish to examine the graph for values of a close to 1, so we first let a = 1.1, obtaining Equation 8. We then let a = 0.9, to obtain Equation 8, and compare the graphs in Figure 7 and in Figure 8.

Equation 8:

Figure 7: Graph of  , with a = 1.1

We see that as a decreases, the vertical openings in the graph at a = 1.1 close at a = 1 and then open horizontally at a = 0.9.

Equation 9:

Figure 8: Graph of  , with a = 0.9

We then explore the effect on the graph of negative a values. Letting a = -3, we have Equation 10, and the graph in Figure 9. We still have a graph passing through the origin and with y-intercepts at y = -1 and y = 1. The curves are much less pronounced now, with the openings horizontal.

Equation 10:

Figure 9: Graph of  , with a = -3

For a = -1 (Equation 11, Figure 10) the curve becomes more pronounced.

Equation 11:

Figure 10: Graph of  , with a = -1

As a approaches zero from the left, with smaller and smaller negative values, we observe behavior similar to that in Figures 1-3, but about the y – axis, as opposed to the x – axis.

Equation 12:

Figure 11: Graph of  , with a = -0.1

An animation of the relation was created for a = n, with –10 < n < 10.

Here’s a link to the movie Assign1KM.avi

It may then be apparent that for each value of a, we are viewing a different cross-section, or contour map of a relation which could be plotted in three dimensions, for Equation 13.

Equation 13:

Here’s a link to the movie Assign1KM3D.avi

We next add a constant to the left side of Equation 2, to obtain Equation 14.

Equation 14:

We animate the graph by varying n between its bounds: –10 < n < 10.

Here’s a link to the Assign1xKM.gcf file for this equation

Investigating the same behavior for the function with a constant added to the left side, we have Equation 15.

Equation 15:

Again the graph is animated, with –10 < n < 10.

When n = 2, we have Equation 16, the graph of which is given in Figure 12.

Equation 16:

Figure 12: Graph of  , with a = 4 and n = 2

Here’s a link to the Assign1yKM.gcf file for this equation