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Joshua DuMont

Finding two linear functions f(x)=ax+b and g(x)=cx+d, a,c ≠0 such that their product h(x) = f(x).g(x)

is tangent to each of f(x) and g(x) at two distinct points.

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If we have two linear functions f(x)=ax+b and g(x)=cx+d their product

h(x) = f(x)*g(x)=(ax+b)(cx+d)=(ac)x²+(ad+bc)x+bd is a parabola.

We would not expect the lines to be parallel to the parabola in most cases, but there are times where this happens.

Functions intersect where they have equal values. So, to start off we set h(x) and f(x) equal to each other.
h(x)=f(x)
(ax+b)(cx+d)=ax+b
(ax+b)(cx+d)-(ax+b)=0
(ax+b)(cx+d-1)=0
Thus either: 
cx+d-1=0
cx=1-d
x=
or
ax+b=0
x=

If there is only one intersection between these functions then these x values must be the same.
=

a-ad=-bc
a*) a-ad+bc=0

Similarly, we can use the same arguments to look at the relationship between h(x) and g(x).
Functions intersect where they have equal values. So, to start off we set h(x) and g(x) equal to each other.
h(x)=g(x)
(ax+b)(cx+d)=cx+d
(ax+b)(cx+d)-(cx+d)=0
(cx+d)(ax+b-1)=0
Thus either: 
ax+b-1=0
ax=1-b
x=
or

cx+d=0
x=
If there is only one intersection between these functions then these x values must be the same.
=
c-bc=-ad
b*) c-bc+ad=0

Notice that a*) can be rearranged to say bc=ad-a.
Plugging this into b*) we can see that:
c-(ad-a)+ad=0
c+a=0
c=-a
Replacing c with -a in equation a*) yields:
a-ad+b(-a)=0
1-d-b=0
b+d=1
These are the constraints our lines must meet to produce the desired parabola.

We now have f(x)=ax+b and g(x)=cx+d=-ax+1-b.

Graphing the functions f(x), g(x), h(x) we notice a couple of interesting things:

First, the intersection of f(x) and g(x) happens at a fixed height as a and b change.
We can find this height by setting the two functions equal to each other.
ax+b=-ax+1-b
2ax=1-2b
x=
Plugging this back in we get the fixed height is:
y=+b=-b+b=

Secondly, the points of tangency appear to be on the x-axis. Notice that if f(x)=0 then h(x) must also be 0 since h(x)=f(x)*g(x). Similarly, if g(x)=0 then h(x) must also be 0 since h(x)=f(x)*g(x). We can conclude that the points of tangency are in fact on the x-axis. The x-intercepts of the two lines occur at x= and x=== .

Third, the vertex of parabola h(x) seems to be at a fixed height of as a and b change. The x-coordinate of the vertex is at the midpoint of the segment connecting the two x-intercepts: x= = .

If we plug this x-value into h(x), we get that h( )=(a()+b)(-a()+1-b)=(+b)(-+1-b)

= (+b)(-+1-b)= ()(-+1)= ()()=