Looking at problem 2, it states that f(x) and g(x) are linear functions. Exploration of h(x) with different pairs of f(x) and g(x) when

I.
h(x)
= f(x) + g(x)

II.
h(x)
= f(x) * g(x)

III.
h(x)
= f(x)/g(x)

IV.
h(x)
= f(g(x))

In
the above examples we will generalize f(x) = m_{1}x +
b_{1} and g(x) = m_{2}x + b_{2}.

In
the first example in which h(x) = f(x) + g(x) we now have the
equation

h(x)
= (m_{1}x + b_{1}) + (m_{2}x + b_{2}),

and
after grouping like terms would give us

h(x) = (m_{1} + m_{2})x + (b_{1}
+ b_{2})

Ex.
1

f(x)
= 2x +3 (purple) and g(x) = x 2 (red) so h(x) = 3x
+ 1(blue)

Ex.
2

f(x) = -x + 1 (purple) and g(x) = x + 2 (red) so h(x) =
3 (blue)

In
the second exploration h(x) = f(x) * g(x) or

h(x) = (m_{1}x + b_{1}) * (m_{2}x
+ b_{2}).

Both
f(x) and g(x) have a degree of one so that when they are multiplied
h(x) will have a degree of two, so f(x) and g(x) are both linear
functions, h(x) will be a quadratic function.

Ex.
1

f(x)
= 2x +3 (purple) and g(x) = x 2 (red) so h(x) = 2x^{2}
x 6 (blue)

Ex.
2

f(x) = -x + 1 (purple) and g(x) = x + 2 (red) so h(x) =
-x^{2} x + 2 (blue)

In
the third exploration h(x) = f(x)/g(x) or

h(x) = (m_{1}x
+ b_{1})/(m_{2}x + b_{2}).

The
function h(x) will be undefined at -b_{2}/m_{2}
and there will be a vertical asymptote at this point (unless f(x)
= g(x)). There will
also be a horizontal asymptote at m_{1}/m_{2}
(unless f(x) = g(x)).

Ex.
1

f(x)
= 2x +3 (purple) and g(x) = x 2 (red) so h(x) = (2x-3)/(x-2)
(blue)

Ex.
2

f(x) = -x + 1 (purple) and g(x) = x + 2 (red) so h(x) =
(-x+1)/(x+2) (blue)

Ex.
3

In
the final exploration we will look at h(x) = f(g(x)) or

h(x) = m_{1}(m_{2}x + b_{2}) +
b_{1}.

The function h(x) will be a linear function since the degree will be one. If we look at f(x) and g(x) to see if these functions are increasing or decreasing we can determine if h(x) will be increasing or decreasing.

The first example has f(x) and g(x) both increasing

Ex. 1

f(x) = 2x + 3 (purple) and g(x) = x 2 (red)so h(x) = 2(x 2) + 3 (blue)

The second example has g(x) increasing and f(x) decreasing

Ex. 2

f(x) = -x + 1 (purple) and g(x) = x + 2 (red) so h(x) = -(x + 2) + 1 (blue)

The third example has f(x) and g(x) both decreasing

Ex. 3

f(x) = -2x + 1 (purple) and g(x) = -x (red) so h(x) = -2(-x) + 1 (blue)