the graphs of f(x) and g(x) are shown below: what are the solutions to the equation f(x) = g(x)?

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The Department of Mathematics Education

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The Product of Two Linear Functions
each of which is Tangent to the Product Role

by

James W. Wilson
University of Georgia

and

David Barnes
University of Missouri

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This may have been an attempt to write a paper with a longer championship than the paper itself. In fact, this "paper" is a discussion of our examining a item problem that having tools like role graphers available might make possible different approaches.

The problem is:

Find ii linear functions f(ten) and one thousand(x) such that the product
h(10) = f(x).g(ten) is tangent to each.

This trouble was posed by a group of teachers during a workshop in which the use of function graphers was being explored. Our analysis is presented as a sort of stream of consciousness account of how one might explore the problem with the tools at hand. In fact, we came upwardly with ii different streams of consciousness and and so nosotros have two senarios that are parallel in that they cover alternative approaches to the trouble. The senarios represent a composite of several discussions of the problem with teachers, students, and colleagues. Notation, the goal here is using this trouble context, not only to solve the trouble posed, but to understand the concepts and procedures underlying the problem.

Function graphers are available for almost any computing platform or graphic calculator. Such tools make it possible to expect at new topics in the mathematics curriculum or to look at current topics in different ways. This trouble has some elements of each. In general it would not be included in the schoolhouse curriculum, but at that place is no reason it should not. Further, the utilise of technological tools to examine visualizations of the functions makes for a dissimilar approach to the problem.

Is it possible to find 2 linear functions, f(x) = mx + b and m(x) = nx + c, such that the function h(x) = f(10).k(x) is tangent to each. A traditional approach would brainstorm with algebraic manipulation. This is useful considering it keeps the students occupied, simply what do they learn from it? Clearly, h(x) = (mx + b)(nx + c) is a polynomial of degree two and h(ten) has two roots. The corresponding roots are when f(x) = 0 and k(ten) = 0. This means the graph of h(x) crosses the 10-axis at the same two points equally f(10) and g(x). Thus, if at that place are points of tangency so they must occur at these common points on the 10-axis. Experienced students, very bright students, and good problem solvers could whittle this information and a lot more data out of this algebraic assay. Novice students would exist more than tentative.

Senario I

Lets open up the function grapher and explore with some specific f(x), g(ten), and the resulting h(x). Lets try

The graphs on the aforementioned prepare of axes are

For some novices, seeing the graph of the product h(x) = (3x + 2)(2x+i) and the graphs of the ii straight lines from the factors on the same coordinate axes provides a new feel. This particular graph has one of the ii lines "close" to being tangent to the production curve simply the other one is not close. How could the picture show be changed?

1 idea is to spread the two lines and so that one has negative gradient. Effort

The graphs are

This is better? What can exist observed? How can the graph of h(x) be "moved down?" What if the graphs of f(x) and g(x) had smaller y-intercepts? Endeavour

The graphs are

Notwithstanding not likewise good just at least the graph of h(x) was "moved down."

Effort smaller y-intercepts, such as

and outcome in graphs of

These graphs seem shut, but clearly the line with negative slope is not tangent to the graph of h(10). Looking back over the sequence of graphs (and peradventure generating some others) the graph of h(x) always has a line of symmetry parallel to the y-centrality. It seems that the pair of tangent lines will have to have this aforementioned symmetry. How? Try making the slopes three and -3. The functions are

and the graphs are

A zoom to the right paw side of the graph with give

showing tangency has not been accomplished. A zoom to the left hand side shows a similar trouble.

One could try adjusting the y-intercepts. In fact, if the y-intercepts were equal, the y-axis would be the line of symmetry. Try

The resulting graphs are

Worse. Endeavor

The graphs are

A zoom to the left shows

and to the right shows

The consequence seems on target. Information technology remains to ostend f(x) and chiliad(x) each share exactly one point in common with h(ten). Again the tradition is to exercise so algebraically, only it might be instructive to look at some graphs of h(x) - f(ten) and h(ten) - g(x), such as the following:

Can we immediately generate graphs of other f(x), g(x), and h(ten) satisfying the conditions of the problem? Reviewing the graphs and the strategy, its seems that the slopes of the lines can vary, the merely condition being that they are m and -m. Then, a simpler example might exist to let the slopes be 1 and -1, giving

and

Information technology is also of interest to see both of the solutions on the same graph:

Other solutions could exist generated by making some other vertical line the axis of symmetry. Indeed substituting

gives the graphs

for which the equations simplify to

First consider the graphs of f(10) and g(x) and try to sketch in h(10). The graph is

What do these lines tell you well-nigh the parabola? What points do you know the curve will go through? Why? What causes it to open the mode information technology does? Now let'southward add together the graph of the parabola and compare information technology with our sketch.

How is it similar our sketch? How is information technology different? Is in that location annihilation nosotros should find or consider?

It appears that all three graphs seem to intersect at one on the y axis. Lets zoom for a closer expect.

Possibly changing one of the functions will aid with the explanation.
Consider

The three functions no longer intersect at 1 on the y-axis. However, the changed function, f(x), does intersect the bend at its y-intercept.

When thou(x) = ane the parabola intersects f(ten). Is the opposite true? Lets graph and run across.

It seems that if f(x) = 1 then h(x) = 1000(10) and if g(x) = one then h(x) = f(10). Lets examination this by trying to generate h(x) from a new f(x) and m(x). Allow

Then add the sketch h(x) and compare with . . .

.

In this process we seem to have also noticed that the lines and the parabola intersect at the points when the lines cross the x-axis. Why would this be true?

At present the goal is to get one line tangent to the parabola. The function m(10) is shut to being tangent. If we could just get the two points to slid together then they would become one point -- the betoken of tangency. (If a line intersects a parabola in exactly ane betoken, what is true about the line?)

Since f(x) takes on a value of 1 when x = 1, then lets try to alter m(ten) so that thou(i) = 0. Lets run into we could change the slope or change m(x)'s position up and down.

Lets effort them both.

To modify slope, thousand(x) = -2x + 2 and test to run into if grand(1) = -2(1) + 2 = 0

Or change position, k(x) = -3x + 3 and test g(one) = -iii(1) + iii = 0

This seems to imply that f(10) and h(x) are tangent at the f(x) and h(ten)'s common root, if the office m(ten) takes on the value 1 at this root. Or in other words if f(a) = 0 and g(a) = 1 then f(x) is tangent to h(x) at a.

What would nosotros have to do to become both f and g tangent to h? That would mean that when f(10) = 0, then g(x) = i; and when g(x) = 0, then f(x) = i. Lets start first with an easy function for f and then try to generate a g(10) which satisfies what we want. Lets begin with f(x) = x.

Now when f(x) = 0, then g(x) should accept a value of one. In other words if f(0) = 0, so g(0) = 1. Likewise, when g(x) = 0, f(x) should have a value of 1. Since f(1) = 1 then g(one) = 0. We need a our linear function g(x) to get through (0,i) and (1,0). So our 1000(x) = -ten +one. Lets graph information technology to check.

That looks correct! Now, add the graph of the production and then test information technology to see if the curves are tangent.

That looks good. (What is the coordinates of the vertex?) Lets zoom in at the roots.

and .

This seems to exist a useful management. Does it work on the previous problem? When we left off, f(x) = x and g(x) = -3x + three. Can nosotros use our technique to find a dissimilar f(10) that works for g(x) to produce h(10) = f(x).g(x) with f(x) and thousand(x) each tangent to h(x)? We take the post-obit graph.

Since m(1) = 0, and then f(1) = 1 and since g(2/three) = 1 and then f(ii/3) = 0 will be necessary. And then f(10) contains the points (ane,1) and (ii/3, 0). Try k(ten) = -3x - 2.

Zoom in for a closer expect.

What are the coordinates of the upper vertex of the triangle? What are the coordinates of the vertex of the parabola? Are the lines really tangent to the parabola? How might this exist proved? Where else could the function f(x) peradventure take on the same value as h(x) if h(ten) = f(x).chiliad(10)? And how can we interpret this on the graphs?

Summary

Many issues are hidden in these composite accounts of examination of this problem. Nosotros still have the additional problem of writing a curtailed statement of proof of the demonstration -- that the solution will always have the two lines of slope grand and -m crossing on y = 1 and the vertex of the parabola on y = 1/2.

Each senario presents a somewhat different approach. Which would be most helpful in finding two quadratic functions f(x) and g(x) such that their production function h(x) = f(10).1000(x) has each tangent? The following graphs show such functions. How tin can they be generated?

What are some generalizations of the trouble (and the solutions)?

evansloseen.blogspot.com

Source: http://jwilson.coe.uga.edu/Texts.Folder/tangent/f(x).g(x)%3Dh(x).html