Product of Two Linear Functions

Doug Griffin

Find two linear functions f(x) and g(x) such that their product h(x)=f(x).g(x) is tangent to each of the lines at two distinct points.

Each of the linear functions would take on the form y =ax+b for some a and b. It follows then that the product of these two linear functions would be of the form

The graph of the product of the two linear functions would be parabolic.

Lets begin by graphing two simple linear equations and see what the graph of the product looks like.

Although this appears to be a poor choice for the solution to our problem, quite often poor choices lead us to revised choices which are much better after we carefully reflect on the results. With respect to this case, it becomes apparent that for any two linear equations which both have a positive slope (or for that matter which both have a negative slope) the solution does not exist. The product produces a parabola with a vertical axis of symmetry opening up. A simple visualization of the vertex at any point in this plane reveals that the parabola must intersect the lines in two points or not at all. So we revise our next guess to include one line with a positive slope and one with a negative slope.

Now this appears to be a much more promising choice. In every case the product of the two linear functions will now produce a parabola which opens down (the coefficient of the squared term being a negative number). If by manipulating the coefficients in the linear equations we could move the vertex point of the parabola beneath the point of intersection of the lines we can see that we would have a hopeful situation. However one difficulty remains as a result of our studying closely the graph.

If we move the the vertex beneath the point of intersection, since the axis of symmetry does not contain the point of intersection of the lines, at any y value of the parabola it is a greater distance to one of the lines than it is the other one. Therefore it would be impossible to construct a parabola which intersected each of the two lines in a tangent point. How may we overcome this difficulty? How may we construct the slope of the two lines such that the point of intersection lies on the axis of symmetry of the parabola?

Careful thought about the graph above encourages us to try slopes which are negatives of each other.


Now we have coeeficients that seem to hold promise for a solution. Further inspection seems to indicae that we need to bring our point of intersection higher (so as to get the vertex point beneath it).Lets raise each of the lines by increasing their y intercept.


After some playing with graphics calculator we arrive at this solution. Lets try to get another solution using the tools we have learned to see if we have a generalizable solution generator.


Comparing these two solution graphs we see they have one characteristic in common. The y value at the point of intersection of the two lines is 1/2. Lets try this one more time.

Lets take the equation y=x+1 and see if we can construct the solution equation for it. The solution equation will take the form y=-x+b. For what x value does my first equation have a y value of 1/2? If x+1=1/2, then x=-1/2. Substituting in the second equation then we get 1/2=-(-1/2) +b. B must be zero.

So we wind up with the two equations y=x+1 and y=-x. Lets graph and check our generated solution.

So for any equation we may now generate the equation that will satisfy the problem. What is the general solution expressed algebraically? Let the initial equation be of the form y=ax+b. Since y must be 1/2 then 1/2=ax+b. Therefore ax=1/2-b, so x=(1/2-b)/a.

Now our solution equation must be of the form y=-ax + c (since the slopes are negatives of each other). Substituting we get 1/2=-a[(1/2-b)/a]+c. So 1/2=b-1/2+c. Therefore b+c=1. So for any equation y=ax+b we get a solution equation y=-ax+ (1-b).