Presented By:

Dana TeCroney


The challenge in this assignment is to find two functions f(x) and g(x) such that their product:

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


is tangent at two distinct points.


Not only will I give an example to solve the given question, but this method will be generalized to show that when give h(x) = ax2 +bx +c, there are an infinite number of lines that could produce such a quadratic.


To begin, consider the polynomial h(x) = -x2 + 6x - 5


Choose a slope for f(x), say m = 3.  Because of the symmetry of parabolas, the slope of g(x) will be m = -3.  Now, values for the constants in f(x) and g(x) must be determined. 


Consider f(x) = 3x + d


As 'd' changes, the line shifts such that sometimes f(x) and h(x) intersect in two places, sometimes in one place, and at other times not at all.  For a graph of this, press HERE. 


We are concerned with where f(x) and h(x) are tangent, i.e. intersect once.  To find this value we want to know where h(x) = f(x) at only one point, or, where h(x) - f(x) = 0 has only one root.  To see the effects of changing d press HERE.  The graph would look like this:


Another way to look at this is when the discriminate of h(x) – f(x) = 0:


h(x) – f(x) = -x2 + 6x – 5 – (3x + d) =0


= -x2 + 3x + (-5 - d) = 0


The discriminate would be equal to zero where


32 – 4(-1)(-5 - d) = 0


Or, when d = -11/4


Therefore, f(x) = 3x – 11/4


Here is a graph of f(x) (red) and h(x) (purple):



The same process can be followed to find g(x).  Remember, the slope of g(x) was -3:


g(x) = -3x + e


What should our e be?


Again we want h(x) and g(x) to be tangent, so we want the discriminate of h(x)- g(x) = 0:

-x2 + 6x – 5 – (-3x + e) = 0

=> -x2 + 9x + (-3 – e) = 0


The discriminate of this would be equal to zero where:


92 – 4(-1)(-3 – e) = 0


Or where e = 61/4


Therefore g(x) = -3x + 61/4


Here is the resulting graph:




In general, consider a quadratic of the form h(x) = ax2 + bx + c. Choose an arbitrary slope for f(x), say m.  The slope for g(x) must be –m.


h(x) = ax2 + bx + c

f(x) = mx + d

g(x) = -mx + e


To find values for constants d and e, one must find where h(x) = f(x) and where h(x) = g(x) tangentially.


In other words the discriminates of the function h(x) – f(x) = 0 is zero and where the discriminate h(x) – g(x) = 0 is zero.