Explorations with Linear Functions
by
Bradley Johnson
In this exploration we will explore the effects of addition, multiplication, division and composition of two linear functions. Below is the original exploration prompt for reference.
i. h(x) = f(x) + g(x)
Let f(x) = -2x +1 and g(x) = 5x - 2. Graphically we see the following:
In this case, we notice that in h(x) both the slope and the intercepts differ from f(x) and g(x).
In particular we notice that the sum of two linear functions is again a linear function, whose slope and and intercepts are determined from the two summand functions. In the general case where f(x) = ax +b and g(x) = cx+d we have h(x) = f(x) + g(x) given by h(x) = (a+b)x + (c+d).
ii. h(x) = f(x) * g(x)
Let f(x) = x + 1 and g(x) = x -1
the first thing that we notice is that the graph of the resulting function, h(x), is quite different from the graphs of f(x) and g(x); That is the fact that h(x) is in an entirely different function family. We do see that h(x) crosses the x-axis at the same points as f(x) and g(x), which tells us that the roots of h(x) are identical to the roots of its factors. Algebraically h(x) can be written as (x+1)(x-1) or expanded as
.
iii. h(x) = f(x) / g(x)
Let f(x) = x+1 and g(x) = x
When we divide two linear function the resulting function belongs to the family of rational functions. The graph of h(x) shows up in two pieces; one piece on either side of the y-axis. The graph of h(x) does not have a numeric value for x = 0, as this results in division by zero which is undefined. As the value of x approaches 0 from the left the graph can be seen to be decreasing towards negative infinity. Likewise as x approaches zero from the right the graph is seen increasing towards positive infinity. The function has one root, given by the same root as f(x).
iv. h(x) = f(g(x))
Let f(x) = 2x - 1 and g(x) = x - 2
The composition f(g(x)) takes the range values of the function g(x) as the domain of f(x), the result is the function seen in green below. we can think of the algebra of the composition as restricting the domain of f(x) to the function g(x). Working this out we have f(g(x)) = 2(x-2) - 1 = 2x - 5.
