The goal of this assignment is to find two
distinct linear functions f(x) and g(x) so that they are tangent
to their product h(x)=f(x)g(x) in exactly two places (like the
I started off using a method of trial and error
by graphing two linear functions and then trying to make observations
that would lead to a general case.
For the first try, I selected two equations
(at random) f(x)=x+2 and g(x)=x+4 making the product h(x)=(x+4)(x+2)
and the graphs looks like this:
What can we note from our first try?
- From our graph it looks as if a linear function
can intersect a quadratic equation at most at two points and
in this case since the linear function and quadratic function
are related to one another at a minimum of one place.
- The next question to ask is how are f(x),
g(x), and h(x) related to one another. We know that f(x) and
h(x) will intersect each other when they are both equal to zero.
For if f(x)=0 then h(x)=0*g(x) and similarly this will be true
for g(x) and h(x). To describe this in a graphical sense we know
that the x-intercepts of the linear functions f(x) and g(x) will
be sure to intersect h(x) where h(x) crosses the x axis.
- On a further note, this implies that the
quadratic h(x) in question must have two real roots which means
that it must cross the x-axis twice and (from the second statement)
these two points must be points of interesection with the linear
functions f(x) and g(x). Since we are looking for the linear
and quadratic functions to be tangent in exactly two places and
we know that their graphs must intersect at the x-axis we can
assume that these two points of tangency must be at the x-axis.
- Finally from the first three statements,
we can conclude that one of the slopes of the linear functions
must be negative...so lets try again.
Let's use similar values and generate a couple
sets of equations. On the right f(x) = x+2, g(x) = -2x+4, and
h(x) = (x+2)(-2x+4). On the left f(x) = x+2 and g(x) = -x+4 and
h(x) = (x+2)(-x+4):
What can we note from our second try?
- The first thing to note is that the parabola
has changed direction in both graphs. This is because if one
of the slopes of the linear equations is negative it makes the
leading term of the parabola -x^2, making the parabola open downward.
- The second thing to notice is that a parabola
is symetric about the x point of its vertex co-ordinate. This
means that it appears that when the slopes of the two lines are
not equal (the graph on the left) there would be no way to get
two points of tangency. The graph of the linear functions on
the right are symmetric and therefore it appears that there is
the possibility of getting two points of tangency. The graph
on the left, however, does have one benefit--the parabola is
symmetric about the y-axis (the line x = 0). Let's try to combine
these two things (symmetry in the y-axis and equal and opposite
slopes) for the next trial.
We can get the parabola to be symmetric about
the y-axis by making the roots of it eqaul and opposite. So let's
graph the equations f(x) = x+2, g(x) = -x+2, and h(x) = (x+2)(-x+2)
What can we note from the third try?
x+2 = -x+2
2x+2 = 2
2x = 0
x = 0
- It appears that the x-coordinate of the vertex
is the same as the x-coordinate of the intersection point of
the two lines. Let's check by setting the two linear equations
equal to one another to find the intersection point:
Now let's find the x coordinate of the vertex
of the parabola (the formula is x = -b/2a where h(x) = ax^2+bx+c)
h(x) = (x+2)(-x+2)
h(x) = -x^2+4
Vertex = -0/-2 = 0
So the x-coordinates of the intersection point
and the vertex are equal. This in fact turns out to be true in
the general case.
To see this click here
for Proof 1.
- The second observation that we can make is
that the vertex of the parabola must be below the point of intersection
of the lines ie the y coordinate of the parabola must be less
than the y coordinate of the intersection point. Students would
probably just use trial and error here but it is possible to
show what y values will give you a vertex below the intersection
point. To see this click here for Proof
Now I will try to modify the above values of
the y intercepts of the linear equations to see if I can get them
to be tangent to the parabola (for details on the numerical values
I choose look at proof 2). Let's try f(x) = x+3/4, g(x) = -x+3/4,
and h(x) = (x+3/4)(-x+3/4).
What can we note from this try?
- Were getting closer...How can we show the
conditions that will give us the general case? To
find out click here for proof 3. To continue pluging and
chugging continue downward on the page.
Trial 5: Let's try again
with the equations f(x) = x+1/2, g(x) = -x+1/2, and h(x) = (x+1/2)(-x+1/2).
What can we note from trial 5?
x+1/2 = (x+1/2)(-x+1/2)
(x+1/2)/(x+1/2) = ((x+1/2)(-x+1/2))/(x+1/2)
1 = -x+1/2
x = -1/2
- It looks like we got it!!! Let's make sure
by setting the equations equal to each other.
Great they intersect at -1/2 and 1/2 (you can
get 1/2 by letting g(x) = h(x)). These are the values of the x-intercepts
of the linear equations and the two roots of the quadratic.
Let's change the slope of the linear equations
and see if our functions still work. Let f(x) = 2x+1/2, g(x) =
-2x+1/2, and h(x) = (2x+1/2)(-2x+1/2).
What can we note from trial 6?
- The vertex point and the point of intersection
of the lines stays the same but the points of tangency between
the parabola and the lines have different x values. As shown
in the proofs page the slope of the equations doesn't matter
as long as they are equal and opposite.
Now let's check and see what happens when we
change the y-intercpets of the linear equations. Let f(x) = x+3,
g(x) = -x-2, and h(x) = (x+3)(-x-2) (left hand graph) and f(x)
= x+3, g(x) = -x-1, and h(x) = (x+3)(-x-1) (right hand graph).
What can we note from trial 7?
- The left hand graph looks good the right
hand one doesn't. Also, we have slid both graphs down the x-axis
making the axis of symmetry different. Why does one graph work
and the other doesn't? The answer lies in the fact that the y
coordinate of the vertex of the parabola must always be 1/4 and
the point of intersection of the linear equations must always
be 1/2. To see this go to the proof page.
From that, we get that when you add the y-intercepts of the linear
equation they should equal 1. So if f(x) = ax+b and g(x) = -ax+c.
This implies that b+c = 1 or that c = 1-b.
- The general equations that will give two
points of tangency are f(x) = ax+b and g(x) = -ax+(1-b).
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