For this assignment we are looking to find two linear equations whose product creates a quadratic equation that is tangent to both of the original linear equations. If we name the linear equations f(x) and g(x), we are looking for a quadratic, h(x), where h(x) = f(x)*g(x), and where h(x) is tangent to both f(x) and g(x). We start by using a graphing program and use several attempts at a solution to get a feel for the problem. The first graph we will look at is very basic; both of the linear functions will be the same. This should give us some idea of how to eventually change the linear functions in order to satisfy the problem. For the remaining diagrams the graph of f(x) will be displayed in purple, g(x) in red, and h(x) in blue.
Here we see that the purple and red function lie right on top of one another because they are the same equation. After looking at the graph, it appears as though the slope of either f(x) or g(x) will need to be negative while the other will have to remain positive. Below we change the equation for g(x) from having a positive slope to having a negative slope.
This definitely looks better than what we saw before, but it is very clear that h(x), the blue graph, is not tangent to either f(x) or g(x). However, it does appear that if we were to shift the red and purple graphs up, we might be able to achieve tangency.
We see above that changing both y-intercepts of the linear functions has done nothing but shift all three of the graphs up. After attempting several different combinations of y-intercepts we find one that works. A solution to the original problem is shown below.
The image above shows the equations and graph of one possible solution, but there are others. Below are two other solutions to the original problem.
After many trials of manipulating all of the variables of the two linear equations we begin to see a pattern forming for the solutions. For the linear equations the slopes must be opposite of one another, one negative and one positive. In other words if 5 is the slope for one of the equations, then -5 must be the slope for the other. Along with this the y-intercepts of both linear functions always has a sum of 1.
If we generalize the functions we can come to the following conclusions based on all of the solutions that we have graphed thus far.
Given the variables above we have observed in our solutions thus far that and that .
Because of the fact that f(x) is going to be tangent to h(x) we know that the two graphs will intersect at exactly one point. To find that point we must first find all of the situations in which f(x) = h(x).
We now have two equations that we can set equal to 0 and solve for x.
Because we are trying to find a point of tangency, the two intersections we see above will have to be at the same point. We find the point that the two intersections above occur at by setting them equal to one another.
If you’ll remember we had made the observation earlier that it appeared a = -c, so in order to show that we will try and isolate c in much the way we did for a above.
We now have an equation for a and an equation for c, let’s see how they are related.
This proves our conjecture from before that any solution to the original problem will have slopes that are opposite one another.
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