Assignment 1: Operating with Functions
By:
The problem is to make up linear functions f(x) and g(x). Explore, with different pairs of f(x) and g(x) the graphs for
i. h(x) = f(x) + g(x)
ii. h(x) = f(x) x g(x)
iii. h(x) = f(x) / g(x)
iv. h(x) = f(g(x))
Summarize, explain and illustrate.
First we will come up with 3 different pairs of functions for f(x) and g(x).
1. f(x) = 2x + 3, g(x) = 3x - 2
2. f(x) = 3x + 1, g(x) = -4x - 2
3. f(x) = -3x + 3, g(x) = x + 1
i. Now, we will add each pair of linear functions and graph them.
1. h(x) = 5x + 1 *(Purple)
2. h(x) = -x - 1 *(Red)
3. h(x) = -2x + 4 *(Blue)
Observe, when we add each of our pairs of functions they remain a linear function.
ii. Now we will multiply each pair of linear functions and graph them.
1. h(x) = (2x + 3)(3x - 2) *(Purple)
2. h(x) = (3x + 1)(-4x - 2) *(Red)
3. h(x) = (-3x + 3)(x + 1) *(Blue)
Observe, when we multiply each of our pairs of functions they will turn into quadratic functions. The parabola in each quadratic function acts as we expect it to. When the coefficient of our 
term is positive the parabola is opening upwards, and when the coefficient of our term is negative the entire function is reflected about the x-axis and it is opening downwards. Our x-intercepts also occur where we would expect them to. If we set each factor equal to zero, that will give us each of the x-intercepts.
iii. Now we will divide each pair of linear functions and graph them.
Observe, when we divide each of our paris of functions they will turn into a Hyperbola.
We can see that each hyperbola has a vertical asymptote. I decided to look at one pair of functions and see if there was a relationship between the denominator and where the vertical asymptote will be. I will explore the denominator of one of our functions.
Lets work with the purple function from our original graph,
we see that our vertical asymptote is defined by the equation 
Lets change the denominator to 5x - 2,
we can see that the vertical asymptote is defined by the equation 
After graphing a couple of functions I realized that our vertical asymptote would be whatever x value I could plug into the denominator and get zero.
iv. Now we will solve for h(x) = f(g(x)).
1. h(x) = 2(3x - 2) + 3 *(Purple)
2. h(x) = 3(-4x - 2) + 1 *(Red)
3. h(x) = -3(x + 1) + 3 *(Blue)
This will create another linear function for each pair. We know that this will be a linear function since f(x) is a linear function and for each x value of f(x) we are inputting g(x) which is another linear function. Therefore this will always create a linear function.
