Does H & R Block Really Get You More Money

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The Department of Mathematics Instruction

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

by

James West. Wilson
University of Georgia

and

David Barnes
Academy of Missouri

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This may have been an attempt to write a paper with a longer title than the newspaper itself. In fact, this "paper" is a discussion of our examining a detail trouble that having tools like part graphers available might make possible unlike approaches.

The trouble is:

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

This problem was posed by a group of teachers during a workshop in which the use of function graphers was beingness explored. Our analysis is presented equally a sort of stream of consciousness business relationship of how one might explore the problem with the tools at hand. In fact, we came up with two different streams of consciousness and and so we accept two senarios that are parallel in that they cover culling approaches to the problem. The senarios represent a blended of several discussions of the problem with teachers, students, and colleagues. Annotation, the goal here is using this trouble context, non only to solve the problem posed, but to sympathise the concepts and procedures underlying the trouble.

Function graphers are available for well-nigh whatsoever computing platform or graphic estimator. Such tools make information technology possible to look at new topics in the mathematics curriculum or to expect at electric current topics in dissimilar means. This problem has some elements of each. In general it would not exist included in the school curriculum, only there is no reason it should not. Further, the use of technological tools to examine visualizations of the functions makes for a different approach to the trouble.

Is information technology possible to find two linear functions, f(x) = mx + b and chiliad(x) = nx + c, such that the part h(10) = f(x).m(10) is tangent to each. A traditional approach would begin with algebraic manipulation. This is useful considering it keeps the students occupied, but what practice they larn from it? Clearly, h(x) = (mx + b)(nx + c) is a polynomial of degree 2 and h(ten) has ii roots. The corresponding roots are when f(x) = 0 and g(ten) = 0. This ways the graph of h(x) crosses the x-axis at the same two points as f(x) and g(x). Thus, if there are points of tangency then they must occur at these common points on the x-axis. Experienced students, very bright students, and skillful problem solvers could whittle this data and a lot more than data out of this algebraic assay. Novice students would be more than tentative.

Senario I

Lets open the function grapher and explore with some specific f(10), g(x), and the resulting h(x). Lets effort

The graphs on the same set up of axes are

For some novices, seeing the graph of the product h(x) = (3x + 2)(2x+1) and the graphs of the 2 straight lines from the factors on the aforementioned coordinate axes provides a new experience. This particular graph has i of the two lines "close" to existence tangent to the production curve merely the other one is non shut. How could the flick be inverse?

Ane idea is to spread the two lines so that i has negative slope. Try

The graphs are

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

The graphs are

Still not besides practiced just at least the graph of h(x) was "moved downwardly."

Try smaller y-intercepts, such as

and effect in graphs of

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

and the graphs are

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

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

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. 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 k(10) each share exactly one betoken in common with h(x). Again the tradition is to do so algebraically, just it might exist instructive to look at some graphs of h(x) - f(x) and h(x) - thou(x), such as the following:

Can nosotros immediately generate graphs of other f(ten), g(x), and h(x) satisfying the weather condition of the trouble? Reviewing the graphs and the strategy, its seems that the slopes of the lines can vary, the simply condition being that they are thousand and -thou. So, a simpler example might be to let the slopes be 1 and -1, giving

and

It is as well of interest to meet both of the solutions on the same graph:

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

gives the graphs

for which the equations simplify to

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

What do these lines tell you nigh the parabola? What points exercise y'all know the bend will go through? Why? What causes it to open up the way it does? Now let's add together the graph of the parabola and compare it with our sketch.

How is it like our sketch? How is it different? Is there anything nosotros should notice or consider?

It appears that all 3 graphs seem to intersect at 1 on the y axis. Lets zoom for a closer expect.

Perchance changing 1 of the functions will help with the explanation.
Consider

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

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

It seems that if f(x) = one then h(x) = grand(x) and if g(x) = 1 then h(x) = f(ten). Lets test this past trying to generate h(ten) from a new f(x) and g(x). Let

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

.

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

Now the goal is to get one line tangent to the parabola. The office k(x) is shut to being tangent. If nosotros could but become the two points to slid together then they would become i point -- the betoken of tangency. (If a line intersects a parabola in exactly one point, what is true nigh the line?)

Since f(x) takes on a value of 1 when 10 = 1, then lets try to alter k(x) and then that g(1) = 0. Lets see we could change the slope or change g(x)'southward position up and down.

Lets try them both.

To modify slope, m(x) = -2x + 2 and exam to see if m(ane) = -2(1) + 2 = 0

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

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

What would we have to do to become both f and g tangent to h? That would mean that when f(ten) = 0, then yard(ten) = 1; and when g(10) = 0, and then f(10) = one. Lets beginning first with an like shooting fish in a barrel function for f and so try to generate a g(x) which satisfies what we want. Lets begin with f(x) = x.

Now when f(10) = 0, and then g(10) should accept a value of 1. In other words if f(0) = 0, and then grand(0) = 1. Likewise, when m(x) = 0, f(x) should have a value of i. Since f(ane) = 1 then m(1) = 0. We need a our linear office chiliad(10) to go through (0,1) and (ane,0). So our g(x) = -x +ane. Lets graph it to cheque.

That looks right! Now, add the graph of the product and then exam it to encounter 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 direction. Does information technology piece of work on the previous problem? When we left off, f(x) = x and thou(x) = -3x + iii. Tin we use our technique to find a different f(x) that works for g(x) to produce h(ten) = f(x).g(x) with f(x) and thousand(x) each tangent to h(10)? Nosotros have the following graph.

Since m(ane) = 0, then f(1) = i and since 1000(2/3) = 1 then f(2/3) = 0 will be necessary. So f(x) contains the points (1,ane) and (two/3, 0). Try g(x) = -3x - two.

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 actually tangent to the parabola? How might this be proved? Where else could the function f(x) possibly take on the aforementioned value every bit h(x) if h(x) = f(x).g(x)? And how can we interpret this on the graphs?

Summary

Many issues are hidden in these blended accounts of examination of this problem. Nosotros nonetheless have the additional problem of writing a curtailed argument of proof of the demonstration -- that the solution volition ever take the two lines of slope 1000 and -1000 crossing on y = i 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(ten) and g(x) such that their product role h(10) = f(x).g(x) has each tangent? The following graphs show such functions. How can they be generated?

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

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

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