Combinations of Linear Functions
by
Rob Walsh
The intention here is to take two specified linear equations and combined them by addition, multiplication, division and composition for the purposes of analyzing the resulting graphs. The goal is to come to a conclusion about what types of graphs are produced in making these combinations. To begin my observations, I begin with two simple linear functions, namely, f(x) = mx + b and g(x) = nx + d. Let's look at the sum of two linear functions, namely, a function h(x) = f(x) + g(x). Consider a general scenario. Using the notion of "combining like terms," we quickly see that for any two linear functions, f(x) = mx + b and g(x) = nx + d, we get the follow:
h(x) = f(x) + g(x)
= (mx + b) + (nx + d)
= (mx + nx) + (b + d)
= (m + n)x + (b + d)
What we find is that this combination results in a new linear function, having a slope of m + n and a y-intercept of b + d. To further illusrate, let's take some specific examples and assume certain cases involving the slopes and y-intercepts:
1. Two lines are parallel (m = n). Take that f(x) = x + 1 and g(x) = x + 2
We see that h(x) = f(x) + g(x) = (x + 1) + (x + 2) = 2x + 3.
2. Two lines are perpendicular (m = -1/n). Take that f(x) = 2x - 3 and g(x) = -1/2x + 2...
We see that h(x) = f(x) + g(x) = (2x - 3) + (-1/2x + 2) = 3/2x - 1.
3. Two lines are coincidental (m = n and b = d). Take that f(x) = -6x + 1/3 and g(x) = -6x + 1/3
We see that h(x) = f(x) + g(x) = (-6x + 1/3) + (-6x + 1/3) = -12x + 2/3.
4. Lines that intersect through the y-axis (b = d). Take that f(x) = -2x - 8 and g(x) = 3x - 8
We see that h(x) = f(x) + g(x) = (-2x - 8) + (3x - 8) = x - 16.
Again, we see that in combining linear graphs with addition, our result is a new linear function. It is minimally interesting to coinsider these specifications, short of the notion that the resulting line has altered steepness and position. In fact, this lead me to think about the composition of two linear functions and the fact that a similar result will occurr. Let's look at this next, using each of our example pairs from above:
1. Take that f(x) = x + 1 and g(x) = x + 2... f(g(x)) = (x + 2) + 1 = x + 3
2. Take that f(x) = 2x - 3 and g(x) = -1/2x + 2... f(g(x)) = 2(-1/2x + 2) - 3 = -x + 1
3. Take that f(x) = -6x + 1/3 and g(x) = -6x + 1/3... f(g(x)) = -6(-6x + 1/3) + 1/3 = 36x -5/3
4. Take that f(x) = -2x - 8 and g(x) = 3x - 8... f(g(x)) = -2(3x - 8) - 8 = -6x + 8
Similar to function addition, function composition results in new linear functions. We will not take anymore time to graph these, but let's generalize by using f(x) = mx + b and g(x) = nx + d:
h(x) = f(g(x))
= m(nx + d) + b
= mnx + md + b
= (mn)x + (md + b)
k(x) = g(f(x))
= n(mx + b) + d
= nmx + nb + d
= (nm)x + (nb + d)
We quickly abandon analysis of these scenarios in relation to function addition and composition in order to look at more interesting combinations. This time, we take one function pair and consider multiplication. Perharps implications associated with varying slopes and intercepts will show themselves after the fact...
Let's take our first pair of linear functions, f(x) = x + 1 and g(x) = x + 2.
f(x)g(x) = (x + 1)(x + 2) = x2 + 3x +2
We see that our resulting function is quadratic in nature, which if we generalize for f(x) = mx + b and g(x) = nx + d, it is readily obvious that this is the case:
f(x)g(x) = (mx + b)(nx + d)
= mxnx + mxd + bnx + bd
= mnx2 + mdx + bd
With coefficients a = mn, b = md and c = bd, find that the product of two linear functions is a quadratic. We save further discoveries of the graph for later analysis of quadratics, but before we move forward, take a look at how changing one positive slope, one negative slope or the y-intercept of one of the lines effects the parabola.
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Now, we consider the quotient of two linear functions.
Let's take our two functions, f(x) = x + 1 and g(x) = x + 2, and graph f(x)/g(x)...
Right away, we can see from the graph that the resulting graph is hyperbolic in nature. This brings about three things, vertical asymptotes, horizontal asymptotes and position. let's disect our new function:
y = (x + 1)/(x + 2)
The denominator will tell us where the function will not exist and therefore, where we have a vertical asymptote. Let x + 2 = 0, which cannot happen since we cannot have division by zero, we get an asymptote at x = -2. Therefore, the linear function that makes up our new function also gives us this boundary line.
We can also solve this equation to find its inverse:
y = (x + 1)/(x + 2)
y(x + 2) = (x + 1)
xy + 2y = x + 1
xy - x = 1 - 2y
x (y - 1) = (1 - 2y)
x = (1 - 2y)/(y -1)
The denominator of our inverse tells us where our hyperbolic function follows a horizontal boundary line. Again, letting the denominator equal zero, we get y - 1 = 0 and so, y = 1. This is our horizontal asymptote.
Finally, if we just look at the general nature of our hyperbolic function, our leading coefficients can tell us about the approximate location of the two curves in the hyperbola will be:
y = (x + 1)/(x + 2)
Since our leading coefficients are both positive (1 and 1), we look at them as a fraction 1/1 and since this is positive, we know our hyperbola will fall predominantly in the second and fourth quadrants. Conversely, if one of them would have been negative, the graph would have fallen into quadrants one and three.
Check our a GCF file and see what happens as our leading coefficients change...
Let's generalize:
f(x) = mx + b
g(x) = nx + d
h(x) = f(x)/g(x) = (mx + b)/(nx + d)
Vertical Asymptote
nx + d = 0
x = -d/n
Horizontal Asymptote
y = (mx + b)/(nx + d)
y(nx + d) = (mx + b)
ynx + yd = mx + b
ynx - mx = b - yd
x(yn - m) = (b - yd)
x = (b - yd)/(yn - m)
yn - m = 0
y = m/n
Position
m/n > 0; quadrants 2 and 4
m/n < 0; quadrants 1 and 3
Interesting how our horizontal asymptote is quickly found in the same way as position! We could have used this right away! In any case, hopefully we have a better sense of what results when combining linear functions in various ways!
