Find two linear functions f(x) and g(x) such that the product
h(x) = f(x).g(x) is tangent to each.
The graphs on the same set of axes are
For some novices, seeing the graph of the product h(x) = (3x + 2)(2x+1)
and the graphs of the two straight lines from the factors on the same coordinate
axes provides a new experience. This particular graph has one of the two
lines "close" to being tangent to the product curve but the other
one is not close. How could the picture be changed?
One idea is to spread the two lines so that one has negative slope. Try
The graphs are
This is better? What can be observed? How can the graph of h(x) be "moved
down?" What if the graphs of f(x) and g(x) had smaller y-intercepts?
The graphs are
Still not too good but at least the graph of h(x) was "moved down."
Try smaller y-intercepts, such as
and result in graphs of
These graphs seem close, but clearly the line with negative slope is
not tangent to the graph of h(x). Looking back over the sequence of graphs
(and perhaps generating some others) the graph of h(x) always has a line
of symmetry parallel to the y-axis. It seems that the pair of tangent lines
will have to have this same symmetry. How? Try making the slopes 3 and -3.
The functions are
and the graphs are
A zoom to the right hand side of the graph with give
showing tangency has not been achieved. A zoom to the left hand side
shows a similar problem.
One could try adjusting the y-intercepts. In fact, if the y-intercepts
were equal, the y-axis would be the line of symmetry. Try
The resulting graphs are
The graphs are
A zoom to the left shows
and to the right shows
The result seems on target. It remains to confirm f(x) and g(x) each
share exactly one point in common with h(x). Again the tradition is to do
so algebraically, but it might be instructive to look at some graphs of
h(x) - f(x) and h(x) - g(x), such as the following:
Can we immediately generate graphs of other f(x), g(x), and h(x) satisfying
the conditions of the problem? Reviewing the graphs and the strategy, its
seems that the slopes of the lines can vary, the only condition being that
they are m and -m. So, a simpler case might be to let the slopes be 1 and
It is also of interest to see both of the solutions on the same graph:
Other solutions could be generated by making some other vertical line
the axis of symmetry. Indeed substituting
gives the graphs
for which the equations simplify to
First consider the graphs of f(x) and g(x) and try to sketch in h(x).
The graph is
What do these lines tell you about the parabola? What points do you know
the curve will go through? Why? What causes it to open the way it does?
Now let's add the graph of the parabola and compare it with our sketch.
How is it like our sketch? How is it different? Is there anything we
should notice or consider?
It appears that all three graphs seem to intersect at 1 on the y axis. Lets zoom for a closer look.
Maybe changing one of the functions will help with the explanation.
The three functions no longer intersect at 1 on the y-axis. However,
the changed function, f(x), does intersect the curve at its y-intercept.
When g(x) = 1 the parabola intersects f(x). Is the opposite true? Lets graph and see.
It seems that if f(x) = 1 then h(x) = g(x) and if g(x) = 1 then h(x)
= f(x). Lets test this by trying to generate h(x) from a new f(x) and g(x).
Then add the sketch h(x) and compare with . . .
In this process we seem to have also noticed that the lines and the parabola
intersect at the points when the lines cross the x-axis. Why would this
Now the goal is to get one line tangent to the parabola. The function g(x) is close to being tangent. If we could just get the two points to slid together then they would become one point -- the point of tangency. (If a line intersects a parabola in exactly one point, what is true about the line?)
Since f(x) takes on a value of 1 when x = 1, then lets try to change g(x) so that g(1) = 0. Lets see we could change the slope or change g(x)'s position up and down.
Lets try them both.
To change slope, g(x) = -2x + 2 and test to see if g(1) = -2(1) + 2 = 0
Or change position, g(x) = -3x + 3 and test g(1) = -3(1) + 3 = 0
This seems to imply that f(x) and h(x) are tangent at the f(x) and h(x)'s
common root, if the function g(x) takes on the value 1 at this root. Or
in other words if f(a) = 0 and g(a) = 1 then f(x) is tangent to h(x) at
What would we have to do to get both f and g tangent to h? That would mean that when f(x) = 0, then g(x) = 1; and when g(x) = 0, then f(x) = 1. Lets first start with an easy function for f and then try to generate a g(x) which satisfies what we want. Lets begin with f(x) = x.
Now when f(x) = 0, then g(x) should take a value of 1. In other words
if f(0) = 0, then g(0) = 1. Likewise, when g(x) = 0, f(x) should have a
value of 1. Since f(1) = 1 then g(1) = 0. We need a our linear function
g(x) to go through (0,1) and (1,0). So our g(x) = -x +1. Lets graph it to
That looks right! Now, add the graph of the product and then test it
to see if the curves are tangent.
That looks good. (What is the coordinates of the vertex?) Lets zoom in
at the roots.
This seems to be a useful direction. Does it work on the previous problem?
When we left off, f(x) = x and g(x) = -3x + 3. Can we use our technique
to find a different f(x) that works for g(x) to produce h(x) = f(x).g(x)
with f(x) and g(x) each tangent to h(x)? We have the following graph.
Since g(1) = 0, then f(1) = 1 and since g(2/3) = 1 then f(2/3) = 0 will
be necessary. So f(x) contains the points (1,1) and (2/3, 0). Try g(x) =
-3x - 2.
Zoom in for a closer look.
What are the coordinates of the upper vertex of the triangle? What are
the coordinates of the vertex of the parabola? Are the lines really tangent
to the parabola? How might this be proved? Where else could the function
f(x) possibly take on the same value as h(x) if h(x) = f(x).g(x)? And how
can we interpret this on the graphs?
Many issues are hidden in these composite accounts of examination of this problem. We still have the additional problem of writing a concise argument of proof of the demonstration -- that the solution will always have the two lines of slope m and -m crossing on y = 1 and the vertex of the parabola on y = 1/2.
Each senario presents a somewhat different approach. Which would be most helpful in finding two quadratic functions f(x) and g(x) such that their product function h(x) = f(x).g(x) has each tangent? The following graphs show such functions. How can they be generated?
What are some generalizations of the problem (and the solutions)?