Tangent Product Funtions

Colleen Foy

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,  .

Observe:

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

and . Therefore. By the zero product property either  or .  Note the following graph:

Note the values of a,b,c, and d, For this particular example, we have the following:

 a 1.2 b 0.8 c -1.2 d 0.2

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:

Thus,              and

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 implies that the parabola opens down since the coefficient in front of is negative.

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.

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