By: Melissa Wilson

Investigation One

In this investigation we will look at rational equations of the form:

We will look at how each parameter affects the graph. We will begin with a and have b, c, and d set equal to 1.

Notice that all of the graphs have an vertical asymptote at x= -1. This is because a vertical asymptote is formed when the denominator is zero, which makes our function undefined at that point. Since both c and d are 1, then the denominator is zero when x = -1. Changing a does not change this. The horizontal asymptote does change with a. Notice how as a increases the horizontal asymptote move further up or down from the x-axis. Also, the asymptote appears to be at the value for a.

The domain and range for this graph can also be discussed. The domain is the set of all x-values for this function. Therefore, we will have all of the x-values below the vertical asymptote and all of the values above it:

Also, we can define the range by including all values except for the horizontal asymptote as we did before. Since this is not unique for the whole collection we will say that it is at the a value for now. This will be modified later.

Now, let's look at the

bvalue. We will change thebvalue and keepa, c,anddas equal to 1.

Since our denominator has not changed since the previous example, we know that our vertical asymptote is the same. Also, this means that our domain is also still the same. Notice how the horizontal asymptote is now the same for all graphs, at y = 1. Notice how our a value is 1 in all the examples. We can find the range in the same way as we did previously.

Now, let's change the

cparameter and have all the others be equal to 1.

The first thing that you will notice is that the blue graph where c = 0 is now a straight line. This is because we made the equation into y = x+1. This equation has no asymptotes and the domain and range are both all real numbers.

For the other graphs we can look at the asymptotes. Let's zoom in on the graph.

The vertical asymptotes can be found for each graph by setting the denominator to zero. In a general form this will be at:

and the domain will also be:

Now we can also amend our previous formula for the horizontal asymptote. Now that c is not 1, we can see that the horizontal asymptote changes as c changes. The horizontal asymptote is a function of a and c and looks like:

And the range can be written as:

Lastly, while we already know the equations for the asymptotes and domain/range, we will still look at what happens with changing the d parameter.

This confirms our earlier general equations for the vertical and horizontal asymptotes and the domain and range values.

We can extend these rational functions to the following, with f(x) and g(x) being polynomials:

For the vertical asymptotes, we would use the same procedure as before and solve g(x) =0 for values of x. The domain will then be all real numbers minus the values for the vertical asymptotes.

The horizontal asymptotes can be found by dividing the coefficient of the highest degree x of f(x) and g(x). In our previous examples, this is the value of a divided by c since they are the coefficients for the x term. The range will be all real numbers minus the values for the horizontal asymptotes.

Investigation Two:

A square with side s is inscribed in a semicircle. The length a is on each side of the square along the diameter. You can construct a figure like the one shown below using this GSP file. We want to find the ratio of s to a.

We can see that the midpoint of the side of the square along the diameter is also the midpoint of the diameter. Therefore, we can say that the distance from this midpoint to one of the opposite corners of the square is a radius of the semicircle.

We know that the triangle formed by the radius, side of the square, and the half of the side of the square is a right triangle. Therefore, we can use the Pythagorean Theorem to get an expression for r.

By looking at the diagram, we can also notice that the radius is also half of the diameter.

Since the radius must the same in each case, we can set them equal to each other and solve for the desired ratio.

This gives an approximate ratio of 1.62. If we do some rearranging, we can see the following:

This is called the Golden Ratio.

Investigation Three

For this assignment I looked at problem 2 from this previous assignment.

Given these three medians, j, k, l. We need to construct the triangle that they come from.

First, we will use these three lengths to construct a triangle, ABC, shown below.

We know that these medians will be concurrent at some point that is 1/3 of the distance from the median. So our first step is to trisect AB, and we will call this point x. Then we will trisect AC and call this point y.

Next, construct a line through x that is parallel to AC. Use the radius length of xy and construct a circle centered at C. The intersection of the circle and the parallel line will be C'..

Now I will do the same as before and construct a circle at A with radius xy. The intersection will be A'.

Now I can construct A'C'.

At point x we have two of our medians, AB and A'C', intersecting at 1/3 of the distance from the medians. Now, we need segment BC to intersect at point x. First we will trisect segment BC and call the new point z.

As before I will construct a line through x that is parallel to BC. Using the radius xz, I will construct a circle centered at C. The intersection of the circle and the parallel line will be C'.

Now, I will do the same with a circle at B with radius xz. The intersection will be B'.

Now we can construct B'C'

Now we can connect points A', C', and B to form our triangle. We can check this construction by making sure our 3 original lines are the actual medians of the new triangle.

Here is a GSP script that you can use to draw three segments and it will make the triangle using those segments as the medians.