What is the effect on the graph of the function f(x) = x2 then f(x) is changed to 7f(x)?

---

The Section of Mathematics Didactics

---

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

past

James W. Wilson
University of Georgia

and

David Barnes
University of Missouri

---

This may have been an attempt to write a paper with a longer title than the paper itself. In fact, this "paper" is a give-and-take of our examining a particular trouble that having tools like function graphers available might make possible different approaches.

The problem is:

Find 2 linear functions f(x) and g(10) such that the product
h(x) = f(ten).thousand(x) is tangent to each.

This problem was posed by a group of teachers during a workshop in which the use of role graphers was beingness explored. Our analysis is presented as a sort of stream of consciousness account of how ane might explore the problem with the tools at hand. In fact, we came up with 2 different streams of consciousness and and then nosotros have 2 senarios that are parallel in that they embrace alternative approaches to the problem. The senarios correspond a blended of several discussions of the problem with teachers, students, and colleagues. Note, the goal here is using this trouble context, not only to solve the trouble posed, simply to understand the concepts and procedures underlying the problem.

Part graphers are available for almost whatever computing platform or graphic calculator. Such tools brand it possible to look at new topics in the mathematics curriculum or to await at electric current topics in unlike means. This problem has some elements of each. In full general it would not be included in the school curriculum, but there is no reason it should not. Farther, the employ of technological tools to examine visualizations of the functions makes for a different arroyo to the problem.

Is it possible to notice two linear functions, f(10) = mx + b and m(ten) = nx + c, such that the function h(ten) = f(x).g(x) is tangent to each. A traditional arroyo would begin with algebraic manipulation. This is useful because it keeps the students occupied, but what do they larn 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(10) = 0 and one thousand(x) = 0. This means the graph of h(x) crosses the x-centrality at the same 2 points every bit f(x) and yard(x). Thus, if at that place are points of tangency then they must occur at these common points on the x-axis. Experienced students, very vivid students, and skillful problem solvers could whittle this information and a lot more information out of this algebraic analysis. Novice students would be more than tentative.

Senario I

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

The graphs on the same ready of axes are

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

One idea is to spread the ii lines then that one has negative slope. Endeavor

The graphs are

This is better? What tin can be observed? How can the graph of h(x) be "moved downwardly?" What if the graphs of f(10) and chiliad(x) had smaller y-intercepts? Attempt

The graphs are

Still non too skilful merely at least the graph of h(10) was "moved downwards."

Effort smaller y-intercepts, such equally

and result in graphs of

These graphs seem close, but clearly the line with negative slope is not tangent to the graph of h(ten). Looking back over the sequence of graphs (and perchance generating some others) the graph of h(10) always has a line of symmetry parallel to the y-axis. It seems that the pair of tangent lines will have to accept this same symmetry. How? Endeavor making the slopes 3 and -3. The functions are

and the graphs are

A zoom to the correct manus side of the graph with give

showing tangency has not been achieved. A zoom to the left manus side shows a like problem.

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

The resulting graphs are

Worse. Try

The graphs are

A zoom to the left shows

and to the right shows

The result seems on target. It remains to confirm f(x) and g(x) each share exactly one bespeak in common with h(x). Again the tradition is to do so algebraically, but it might be instructive to look at some graphs of h(ten) - f(x) and h(ten) - 1000(10), such as the following:

Tin we immediately generate graphs of other f(x), m(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 only condition being that they are m and -m. And then, a simpler case might exist to let the slopes be ane and -1, giving

and

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

Other solutions could exist generated by making another vertical line the centrality of symmetry. Indeed substituting

gives the graphs

for which the equations simplify to

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

What do these lines tell you about the parabola? What points practice you know the curve will go through? Why? What causes it to open the style it does? Now let'southward add the graph of the parabola and compare it with our sketch.

How is it like our sketch? How is it dissimilar? Is there anything we should notice or consider?

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

Perchance irresolute one of the functions will help with the explanation.
Consider

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

When g(x) = 1 the parabola intersects f(10). Is the opposite true? Lets graph and see.

Information technology seems that if f(x) = 1 then h(x) = yard(ten) and if g(10) = 1 and then h(x) = f(x). Lets test this by trying to generate h(x) from a new f(x) and g(x). Allow

And then add the sketch h(x) and compare with . . .

.

In this procedure nosotros 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 part g(ten) is close to being tangent. If we could just get the two points to slid together then they would get one indicate -- the indicate of tangency. (If a line intersects a parabola in exactly one point, what is truthful nearly the line?)

Since f(x) takes on a value of 1 when x = one, then lets try to change g(x) so that g(one) = 0. Lets see we could change the slope or modify yard(x)'s position upwards and downward.

Lets try them both.

To change gradient, g(x) = -2x + 2 and test to encounter if g(1) = -2(1) + 2 = 0

Or modify position, g(ten) = -3x + 3 and exam g(ane) = -3(1) + 3 = 0

This seems to imply that f(10) and h(10) are tangent at the f(x) and h(ten)'southward common root, if the function g(x) 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 we have to practise to get both f and grand tangent to h? That would hateful that when f(x) = 0, and so g(10) = 1; and when m(x) = 0, then f(x) = 1. Lets beginning showtime with an like shooting fish in a barrel function for f and then try to generate a g(x) which satisfies what we want. Lets begin with f(x) = 10.

Now when f(ten) = 0, then g(x) should take a value of one. In other words if f(0) = 0, then g(0) = 1. As well, when k(10) = 0, f(x) should have a value of 1. Since f(one) = ane then g(1) = 0. We demand a our linear function thousand(x) to become through (0,1) and (i,0). Then our grand(x) = -x +i. Lets graph it to check.

That looks right! At present, add the graph of the production and and then test it to see if the curves are tangent.

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

and .

This seems to exist a useful direction. Does it work on the previous trouble? When nosotros left off, f(x) = x and 1000(x) = -3x + 3. Tin can we use our technique to find a unlike f(x) that works for g(x) to produce h(x) = f(x).g(x) with f(x) and chiliad(x) each tangent to h(ten)? We have the post-obit graph.

Since yard(1) = 0, then f(one) = i and since g(2/3) = one then f(two/3) = 0 volition be necessary. So f(10) contains the points (ane,one) and (2/3, 0). Endeavour m(x) = -3x - ii.

Zoom in for a closer look.

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 be proved? Where else could the function f(10) peradventure take on the aforementioned value as h(x) if h(x) = f(x).thousand(10)? And how can we interpret this on the graphs?

Summary

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

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

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

reederficket.blogspot.com

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

Share this post