Product of Two Quadratics


By Courtney Cody


For Assignment #1, I chose to explore question #4 for which I wish to find two quadratic functions f(x) and g(x) such that their product h(x) = f(x) * g(x) is tangent to each of f(x) and g(x) at two different points.


Let us begin by observing two simple quadratic functions f(x)=x^2 and g(x)=2x^2, which have a different leading coefficient, and their product.


Graph A


By looking at the graph we see the product of these two quadratic functions, which is pictured in blue, does not appear to be tangent to either of f(x) or g(x).   The blue curve only appears wider at its base when compared to f(x) and g(x) and intersects each of the functions at two points.  Hence, we need to change f(x) and g(x) such that there is more of a difference in the equations than just leading coefficient.


Suppose we change the two quadratic functions such that the signs of the leading coefficient are opposite.  For example, let f(x)=x^2 and g(x)=-2x^2.


Graph B


Since the graph does not appear to be getting closer to satisfying the conditions of the assignment since h(x) is not tangent to either f(x) and g(x) at any points, more changes need to be made to the equations of f(x) and g(x).


In order to have f(x) and g(x) intersect at a couple of points, let us decrease f(x) by a constant and increase g(x) by a constant.  For example, let f(x)=x^2-1 and g(x)=-x^2+1 and consider the graph of these two functions and their product, h(x), below:


Graph  C


This graph looks much better!  The function h(x) in blue is already tangent to f(x) at one point.  If we look at the red curve and notice that if we change it such that the blue curve is completely contained and touching it, then the h(x) will be tangent to g(x) on each of the sides.  Hence, let us change g(x) such that it has a wider slope in order to just slightly contain h(x). For example, let the leading coefficient of g(x) be -1.


Graph D


While changing g(x) such that its slope is wider brought us closer to satisfying the conditions of the assignment, the blue curve still lies just outside of g(x). What would happen if we moved the red curve up by a unit?  Let us explore this change. 


Graph E


Wow!  This looks exactly like what we are looking for to answer the assignment.  Not only does the red curve contain the blue curve such that they are touching as we desired, but the blue curve also touches the purple curve at exactly two points as well!  Hence, when f(x)=x^2-1 and g(x)=-x^2+2, the product h(x)=f(x)*g(x) is tangent to each of f(x) and g(x) at two different points. 


Is this the only solution to the problem?  What if f(x) and/or g(x) contain a linear term in their equations?  Let us explore.


Suppose we take f(x) and g(x) as they are in the previous graph and add a linear term.  We now obtain the following graph:


Graph F


This does not look right!  By adding a linear term, the graphs of the functions became skewed and asymmetrical. Additionally, h(x) is no longer tangent to either f(x) or g(x) at any points.  It seems important for f(x) and g(x) to be centered on the same x-value.  In the previous example, by adding an x term to each of f(x) and g(x), we actually shifted f(x) to the left one unit and g(x) to the right one unit.  What if we made the equations such that both f(x) and g(x) were shifted to the left?  By changing the coefficient of the linear term in g(x) to be -1 and keeping f(x) the same, we obtain the following graph:


Graph G


Once again, we have satisfied the conditions of the assignment since h(x) is tangent to f(x) and g(x) at two different points.


What would happen if the coefficients of the quadratic and linear terms were not 1?  Is there a way to manipulate the equations such that the conditions of the assignment still hold?  Consider the following graph:


Graph H



Let us declare a general formula for f(x) and g(x).


In general, there are infinitely many solutions of the form f(x) = a1x^2 + b1x -1 and g(x)=a2x^2+b2x+2, where  a1 = -a2 and b1=-b2.  Also, the sign of a1 must equal the sign of b1 and the sign of a2 must equal the sign of b2.