Tangent Product Funtions
Consider the functions f(x)=ax+b and g(x)=cx+d. Using geogebra we can create sliders a,b,c, and d.
Graph the functions f(x) and g(x):
Now, we can move the sliders a,b,c, and d to change the graphs of the function. We can also change the size of the sliders to anything we want. Note that moving a and c changes the slopes of f(x) and g(x) respectively. Changing b and d changes the y-intercept of f(x) and g(x) respectively.
Next, graph the product function, .
The product of f(x) and g(x) is a parabola. This is due to the fact that when you multiply two linear functions, we get a quadratic function, as was pointed out in the algebra above. We can manipulate sliders, a,b,c, and d until we get a parabola that appears to be tangent to f(x) and g(x) at two points. No matter what the values of a,b,c, and d are, the intersections of f(x) with h(x) and g(x) with h(x) are always on the x-axis. Thus the x-intercepts of f(x) and g(x) are the x-intercepts of h(x). The x-intercepts are found by setting the function equal to zero. Doing this we obtain
Note the values of a,b,c, and d, For this particular example, we have the following:
It seems that when and we get the desired graph. We can now explore this algebraically. Click here to download a geogebra file to experiement with the sliders.
Let f(x) = ax+b and g(x) = cx+d. Note the following:
Another way to solve for the roots is by using the quadratic formula. Observe:
We need h(x) to be tangent to f(x) and g(x) at and . Using calculus we can find the slope of the tangent lines at these two points. Taking the derivative of h(x) we obtain:
In order to find the slope of h(x) at
and , we substitute the two values into h(x) and obtain the following two values: (-ad+bc) and (ad-bc). Thus, at the point the slope m1of h(x) is (-ad+bc). At the point
the slope m2 of h(x) is (ad-bc). The slopes m1 and m2 the opposite of each other: and m1= -m2 and -m1 = m2 These values give us the following two equations for the tangent lines:
Since a + c = 0 we know that a = - c. Thus by substitution:
The following observations are also useful. Suppose g(x) = 0. Thus,
Similarly when f(x) = 0 we have:
So, when f(x) = 0, g(x) = 1 and when g(x) = 0, f(x) = 1.
In either case, the y coordinate of the intersection is 0.5. In the case where f(x) = 0 and g(x) = 1 we set the equations equal to each other and use the fact that a = - c and b = 1- d:
Substituting this value of x into f(x) we obtain: The second case can be left to the reader.
Another way to look at this problem would be to take the functions, y=ax+d and y=cx+d and set them equal to 0. Thus we have:
ax+b=0 and cx+d=0.
Multiplying the two expressions together we obtain an equation equal to 0. Observe:
When the coefficient is 0 and (a+c) = 0 and (b+d) = 1 we obtain the following equation: which is in the form of a quadratic equation. If we did not have the coefficient equal to zero we would not obtain a quadratic function resulting in a parabola. When we substitute (– c ) in for a into this equation we obtain:
This equation, is quadratic which is what we obtain if multiplying two linear equations together. If we did not have those constraints on (a+c) and (b+d), and on the coefficient, we would not have a quadratic equation.
LetŐs take a look at some possible options for f(x) and g(x) that produce a solution.
For example, a=-1, b=0 c=1 d=1 :
Or, a=0.5,b=-0.5, c=-0.5, d=1.5:
Or, a=-5, b=2, c=5, d=1:
As and get larger, the function compresses. As and get smaller, the function stretches.