Desmos Advanced Algebra Exploration -- Power Functions, Inverses and Graphs

If link doesn't work try this twitter link..

NOTE: CLICKING ON THE GRAPH ABOVE SHOULD NOW LOAD THE DESMOS activity...


Hope you enjoy this new Common Core Investigation...

Students will examine the relationships among the graphs of y = kx^n, the inverse x = ky^n and their graphs. Beginning with particular values k=2, n=2 students will observe how the graphs are reflection images of each other over the line y=x. They will be asked to observe how the number of points of intersection vary over positive integer values of n, according to whether n is even or odd.

They will then determine the coordinates of these points first by estimating from the graphs, then by obtaining exact values using a system of equations.

Finally, they will use more advanced algebra tools to solve in terms of k and n. Some students will recognize the BIG IDEA that the points of intersection must lie on the graph of y=x, therefore the algebra is simplified by using y=x to replace x=ky^n when solving. This is crucial.

Thus, there is a blend of discovery and application of exponent skills. This is to me is the best use of technology - to enhance not replace instruction.

Desmos empowers the student to probe deep beneath the surface but the teacher must carefully plan and guide this process, otherwise many students will make pretty graphs and not get beyond moving sliders left and right. My opinion of course...

Desmos is a powerful teaching/learning tool because it enables students to discover important mathematical relationships and formulate key concepts for themselves. However, it is the expertise of the instructor which will determine WHAT they are learning. This is GUIDED self-discovery!

In this activity I included a detailed overview and guide for the instructor but I left it to the professional to tailor the investigation to the students and the curriculum. In other words, I did not provide a student worksheet. I encourage the professional to modify as he/she sees fit.

Your feedback is very important to me as I continue to develop these. Feel free to comment below or contact me directly using the new Blogger contact form. Also follow me on Twitter @dmarain.


Use new contact form at top of right sidebar to contact me directly!





 If interested in purchasing my NEW 2012 Math Challenge Problem/Quiz book, click on BUY NOW at top of right sidebar. 175 problems divided into 35 quizzes with answers at back and DETAILED SOLUTIONS/STRATEGIES for the 1st 8 quizzes. Suitable for SAT I, Math I/II Subject Tests, Common Core Assessments, Math Contest practice and Daily/Weekly Problems of the Day. Includes multiple choice, case I/II/III type and constructed response items. Price is $9.95. Secured pdf will be emailed when purchase is verified.

A Classic Algebra Challenge

Once students learn the strategy for doing these kinds of questions, the SAT and other standardized tests seem rather easy!

x+y=10
x^2+y^2=10
Find x^3+y^3

Ans: -350

Notes: 

(1) Before giving students this question you may wish to scaffold with finding xy first.
Ans: 45

(2)  To promote connection-making and to deepen their thought processes, give them the answer -350 and ask:
(a)  Without graphing. explain why the graphs of the 2 given eqns DO NOT INTERSECT!
(b) Then how can there be a solution!

Sent from my Verizon Wireless 4GLTE Phone

Algebra 2/Precalculus "Extended" Activity Based on an SAT-Type Question

Consider the following problem:

If -5 ≤ x ≤ 4, and f(x) = 2x² - 3, how many integer values are possible for f(x)? 


One can simply view this as a more challenging question to pose to your honors/accelerated students, but, for me, it's an opportunity for all your students to think more deeply about important concepts. I feel strongly that our role here is to ask the key questions which will guide them toward understanding the "big ideas" underlying this problem. In fact, we can turn this question into an extended activity: 15-20 minutes).

Here is one idea for creating the environment currently being recommended. Please keep an open mind before concluding that there is simply not enough time for these explorations...


WITH YOUR LEARNING PARTNER(S):

1.  Sketch the graph of the function on the given domain from recognition of quadratic functions and by making an x-y table with 4-5 points. WRITE YOUR INFERENCES FROM THIS. For example, from the sketch we believe that the greatest y-value on this domain is ___.

WRITE your conjecture for the answer to the problem: ____

2.  Using the TABLE feature of your graphing calculator, with TblStart = -5 and ΔTbl = 1, display the Table.  Now turn TRACE on and analyze the graph on this domain. Does this alter or confirm your conjecture from Step 1?  YES   NO

3.  The following statement is plausible but FALSE.

The domain consists of 10 integer values. Therefore there are also 10 integer values for f(x), so the answer is 10.

Explain why this is wrong. There is more than one error! 

4.  The correct answer is 51. Depending on the class, a few, if not several,  students should be able to come up with the correct answer and provide a thorough explanation.

5.  Group Discussion:

• Ask students how they might have approached this question if it appeared on a standardized test? Plug in x-values? Use the graphing calculator? Guess? Skip it?
• Ask the group what made this questionable formidable for some students? How important was understanding what was asked for?
• Review one successful approach to solving the problem by calling on individual students to give the "next" step.

NOTE: This  problem also presents a highly teachable moment for students to see an application of the Intermediate Value Theorem in Precalculus (or more intuitively in Algebra 2).  Help them make the connection! Is this easy for us to do?

 Your thoughts?



"All Truth passes through Three Stages: First, it is Ridiculed... Second, it is Violently Opposed... Third, it is Accepted as being Self-Evident." - Arthur Schopenhauer (1778-1860)

 You've got to be taught To hate and fear, You've got to be taught From year to year, It's got to be drummed In your dear little ear You've got to be carefully taught. --from South Pacific

Video Solutions to Two Twitter SAT Problems of the Day

Please note correction to 2nd problem in the video. The correct answer is 4096 "real" values. The original answer, 13, applies to rational solutions only. Thanks to Nick Hobson for pointing out my careless error. Haste makes waste!!


Please vote in the poll at the right. Be candid in your opinion of these videos. It will guide me in the future to improve. Don't hesitate to share your opinions on MathNotationsVids and rate each video there as well. If you subscribe to my feed, please vote directly on the site. Only a few days left...


The title says it all so here is the video as promised:

Note: See above correction to 2nd problem! The video has not been corrected so beware!




Comments on 2nd problem:


If x is greater than or equal to 0 and less than or equal to 3, for how many values of x will 16^x be an integer?

As mentioned above, Nick pointed out my error. I should have restricted x to be of the form a/b, where a and b are integers, b ≠ 0. Normally, SAT questions avoid use of the term rational so they would spell it out. This problem however is very questionable for SATs. If real solutions were sought, this question would be more appropriate for a math contest. Here's one way of explaining why the answer is 4096 for real solutions:

16^x = k, k an integer → 2^(4x) = k
3 ≥ x ≥ 0 → 12 ≥ 4x ≥ 0 →  4096 ≥ 2^(4x) ≥ 1 since the exponential function 2^(4x) is increasing. This argument is reversible, so there are 4096 solutions for x, one of each integer value of k from 1 to 4096 inclusive. This solution could be written more concisely using log base 16 or log base 2 as Nick did, but I wanted to show a method without the log symbol.

Again, the video solution is WRONG as it shows only rational solutions! Well, at least i was thinking "rationally!"

I fully realize that the school year is over for some and about to end for others but these SAT Problems will be around for you or your students in perpetuity! Let me know if you like the questions. They are now appearing in the right sidebar of my blog so you will need to visit the page to see them.
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"All Truth passes through Three Stages: First, it is Ridiculed... Second, it is Violently Opposed... Third, it is Accepted as being Self-Evident." - Arthur Schopenhauer (1778-1860) You've got to be taught To hate and fear, You've got to be taught From year to year, It's got to be drummed In your dear little ear You've got to be carefully taught. --from South Pacific

An Equation Which May Be More 'Complex' Than It Appears!

Maybe I should rename this blog to Saturday 'Morning' Post. After all, no one reads that either anymore!

As the school year comes to a close (and I'm assuming it's already over for some), here's an innocent-looking equation which might be worth discussing with your advanced algebra/precalculus students now or next year. I might have considered saving this for our next online math contest but it's complex nature makes it more suitable for discussion in the classroom than on a test. Have you seen exercises like this in your Algebra or Precalculus texts? Do students often delve beneath the surface of these? It's kind of like a black box. We often feel we simply cannot reveal too much of the mystery here or we will not finish required content. Well, you know my philosophy of 'less is more' and I don't even live in Westport, CT. (Ok, that's a post for another day!).

SOLVE (by at least two different methods):

2a^-3/2 - a^-1/2 - a^1/2 = 0

Preliminary Comments/Questions/Issues

• Is the term solve ambiguous here, i.e., should we always specify the domain to be over the reals or over the complex numbers or is that understood in the context of the problems? I'm guessing that most advanced algebra students learn that the domain of the variable or solve instructions may impact on the result, but, that is precisely one of the objectives of this problem.
  
• Should students immediately change all fractional exponents to radical form? OR use the gcf approach (which requires strong skill)?
  
• It's not hard to guess that 1 is a solution but is it the only solution? Can we make a case for -2 being the other solution? The graph doesn't reveal this and surely, -2 doesn't make sense or does it....
  
• Is there ambiguity in raising a negative real number to a fractional exponent (never mind raising i to the i)? Why? Isn't there a principal value for such an expression? How is it defined? This problem raises fundamental and sophisticated issues about numbers which can be taken as far as one chooses to go Just how complex can complex numbers get?
• What is the role of the graphing calculator here? Mathematica? Wolfram Alpha? In addition to verifying solutions or determining answers, can these tools also be useful in clarifying ideas or raising new questions?
  
• Students (and the rest of us) are now capable of quickly filling in the gaps in their knowledge base by visiting Wolfram's MathWorld or Wikipedia for more background. Should this impact on how we present material? Typically, in the pre-web days teachers would avoid opening up a can of worms like complex solutions here, but, with your more capable groups, the sky's the limit now IMO...
  

Classic Exponent Challenge for SATs, Algebra 2, Math Contests...

Don't forget to register for the upcoming 2nd MathNotations Free Online Contest for secondary students. Click here for more info.



The first 4 terms of a sequence are 2, 6, 18, and 54.
Each term after that is three times the preceding term.
If the sum of the 49th, 50th and 51st terms of this sequence is expressed as k⋅3⁴⁹, then k = ?

Click Read more to see the answer, solution, discussion...


Answer: 26/3

Suggested Solution
The first three terms can be written as
2(3⁰), 2(3¹) and 2(3²). (***)
In general, the nth term is 2(3^n-1).
The sum of the 3 desired terms would then be 2(3⁴⁸) + 2(3⁴⁹) +2(3⁵⁰). Factoring out 3⁴⁹, we obtain 3⁴⁹(2/3 + 2 + 6) = (26/3)(3⁴⁹), so k = 26/3.

Comments
(1) Too hard for SATs? Similar (but slightly easier) problems have appeared on the test.
(2) Could students use the "Make it simpler strategy" here to reduce the problem to the sum of just the first three terms? But this is the essence of geometric sequences (or exponential functions):
From (***) above, this sum would be
2(3⁰) + 2(3¹) + 2(3²) = 2 + 6 + 18 = 3(2/3 + 2 +6) = 3(26/3). The coefficient 26/3 would be the same for any three consecutive terms! Is this concept/technique worth developing?

...Read more

Algebra For All - Another Report...

A recent article in Education Week, entitled Algebra-for-All Policy Found to Raise Rates Of Failure in Chicago, is generating some provocative online comments. Although access is usually restricted to subscribers to Education Week, there may be limited access to this article.

Meanwhile, California currently is mired in legal action regarding implementation of Algebra for all eighth graders.

What does all of this mean? IMO, mandating that all students take Algebra at the same point in time, ready or not, reflects a lack of understanding of the prerequisites for student success in learning algebra.

I've been advocating for some time that the content of an algebra course be standardized. As long as the course contains a common body of knowledge, I would argue that the name of the text, the approach, the instructional strategies, and the extent of integration of technology are of less importance.

Teachers should also be provided with samples of the kinds of assessment questions students should be able to handle. If these exemplars reflect a variety of question-types, balancing skill, conceptual understanding and problem-solving we can be reasonably certain that students are getting an authentic algebra course.

After reading some of the excellent comments on this article, I decided to share my own thoughts. Click on Read more if you would like to see these comments...


First Comment
Considering that this debate has been ongoing for years, one would hope that our current administration will listen to the voices of reason, many of whom have already submitted excellent comments to this article. If you bring together 100 parents and educators and sit them down in a room to discuss this issue, a consensus could be reached that would probably be far more reasonable and helpful to our children than all the research studies and commission reports that have been published.

Here's my best guess of what this group would recommend (much of which was stated above by some of the commenters:

(1) Algebra for All makes sense only if we have Arithmetic for All, i.e., a STANDARDIZED body of content/core knowledge of skills AND concepts, K-7 or K-8. Yes, it is possible to balance UNDERSTANDING AND SKILL and, yes, the preparation of our K-8 teachers must similarly be upgraded to deliver this!

(2) Students must be expected to demonstrate conceptual understanding of and proficiency in basic arithmetic skills including fractions, decimals, percents and ratios. Does this sound impossible given the current levels of student performance? What seems far more absurd to me is expecting proficiency in algebraic reasoning and skill UNLESS this foundation is in place.

(3) We also know that mandating ALL children to demonstrate proficiency AT THE SAME TIME time is unreasonable, however, there must still be strong EXPECTATIONS THAT ALL WILL GET THERE if we provide enough support and demand the needed effort and commitment from each child. Why should parents (and most do not have the means) have to pay thousands of dollars to private after-school companies to supplement their children's learning? This administration should provide whatever funds are needed to provide extra tutorial time in mathematics DURING THE SCHOOL DAY or before or afterwards or on Saturdays or during the summer or whatever is needed to bring children up to level. In return, students must be expected to work hard - NO EXCUSES. Coming for this extra help should not be optional! IMO, that would truly be a 'stimulus' for success.

Other countries have shown that most children can be ready for more algebra at an earlier age provided the necessary foundation is laid.

Prof. Escalante demonstrated that, through superhuman effort, it may even be possible to have CALCULUS FOR ALL! I'm not advocating this nor am I convinced that this is feasible unless one can clone this remarkable educator, however, the main lesson to take from him is the POWER OF HIGH EXPECTATIONS.

Until our society believes that EDUCATION OF OUR CHILDREN IS AN INVESTMENT NOT AN EXPENSE, all the recommendations from all the experts will fail. Listen to the voices of reason, please, before another generation is lost...
Dave Marain

Second Comment

How about some real specifics of what it means to develop algebra sense...

Here is one of many possible ways of developing the distributive property (the basis of 'combining like terms').

Visual:
[{&&&&}{&&&&}{&&&&}] combined with [{&&&&}{&&&&}] =
[{&&&&}{&&&&}{&&&&}{&&&&}{&&&&}]

Verbal:
Three groups of 4 added to two groups of 4 equals how many groups of 4?

Numerical:
(3x4) + (2x4) = 5x4

Symbolic:
3a + 2a = 5a

The language of algebra is the generalization of the language of numbers and arithmetic.

Other countries introduce the symbolic form early on as children are learning their arithmetic facts. Do we? In fact, children can develop both number sense and symbol sense if these are presented in a systematic organized manner. BTW, the use of multiple representations I've shown above is not just to get at different learning styles; it also deepens the child's understanding of numbers and relationships.

BUT, in the end, children also need to KNOW that 3x4 = 12 without hesitation. Knowledge of fact and skills can only be achieved through repetition and practice. Educators know this self-evident truth and they also know that one can accomplish this while students are gaining insight from solving problems and communicating their thoughts. Until we all make a commitment to this BALANCED VIEW of learning, it won't make any difference what curriculum a district purchases.
Dave Marain
MathNotations


Your thoughts...

...Read more

Using Algebra to Enhance Numerical Accuracy On Your Calculator: When is ZERO really ZERO!

When a calculator displays zero as a result should students assume that is exact or only accurate to the precision the machine can store and/or display?

The next time students ask you why we use the conjugate method to rationalize denominators, here's an example of why we sometimes use the method in "reverse". This happens more frequently in calculus but the following is an apparently trivial numerical computation your students can try on their graphing calculators. The results of this computation depend heavily on the specific technology used (e.g., expect different results between the TI-89 and the TI-84), but hopefully they will get the idea. This numerical issue came up as I was solving an applied problem which required finding the difference between two very large numbers (the difference between distances from the center of the earth to a point slightly above its surface and the radius of the earth). This numerical issue has come up more before on this blog. Look here if you want to see another application.

Here's the computation:
Let R = 2.0916 x 10^7
We need to compute the following expression (denoted by **)



For the Student
(a) Do the calculation directly on your calculator. You will want to store this value of R as a variable for later use:
2.0916x10^7 STO> ALPHA R
Does your calculator display zero? If so, explain this "error."
Note: This display depends on the calculator being used. I experimented with the -84 and -83. Let me know how the display appears on other machines. Of course, one would expect a very different outcome if using Mathematica!

(b) Rewrite the above expression ** by multiplying the numerator and denominator by the conjugate of the expression. (Hint: Put the original expression "over 1").

(c) Recalculate the value of ** using the modified but equivalent form from part (b).
What result do you see this time? Can you explain what may be going on?

(d) Find other numerical expressions that produce an incorrectly displayed result on your calculator! Post these in the comments section pls!

Solve x^2 - 10000x - 10000 = 0 without a calculator! A Precalculus Investigation

BACKGROUND/OVERVIEW
No, there's no mistake in the constant term in that equation. Imagine giving this to your Precalculus/Math Analysis/Adv Math students! Actually, 'solving' it with the TI-84 requires some effort using Solver since one needs to make an approximate guess or adjust the lower bound so that the positive root is obtained. Using the graph is no 'walk in the park' either! The TI-89 or Mathematica would have much less difficulty in displaying the exact radical form or a suitable decimal approximation but they may not be within reach. Perhaps an important issue here is that sometimes technology gives us unexpected or even inaccurate results. That's when students need some understanding of theory to recognize the limitations of the technology and adjust accordingly.

Here's the point of all this. The given quadratic is not factorable over the integers, however we can replace it with a 'nicer' quadratic that is. The roots of the desired quadratic can be shown to be approximately the same as the 'nice' quadratic and we can show that the absolute error is less than two ten-thousanths (and a much much smaller relative or % error)! Does this 'numerical analysis' have any practical value? Why approximate roots when powerful technology can produce exact answers? Do professionals who need to apply mathematics to the solution of 'real' problems ever use such approximation techniques? Could it be that theory actually provides practical application!

THE INVESTIGATION
(1) Show that the roots of the x²-10000x-10001 = 0 are 10001 and -1 by factoring.

(2) Show that the roots of x²-10000x-10000=0 can be approximated by 10001 and -1 with an error of less than 0.0001.
(a) By direct calculation: Using the quadratic formula and, yes, you may use the calculator!

(b) (Challenging) By comparing, in general,
(*) the roots of x²-bx-(b+1)=0 and
(**) the roots of x²-bx-b=0.Here we are assuming that b > 0.
(i) First show by factoring that the roots of equation (*) are b+1 and -1.
(ii) Then use the quadratic formula to express the roots of (**) in terms of b.
(iii) Compare the positive roots of these equations by subtracting them and (after algebraic manipulation and simplication), show that the absolute value of the difference is less than 1/(b+1).
Note: For b=10000, this error is therefore less than 0.0001.

(c) Explain intuitively why the roots of the original equation and the 'approximating' equation are virtually the 'same' for 'large' values of b. One possibility here is to consider how the graphs of the associated quadratic functions are related. What do they have in common? How are they different?
Note: Subtle point here for students. Even though the difference of the function values (i.e., y-values) is always 1, this is not true of the difference between their zeros! This may be the essence of the numerical analysis in this investigation.

EXTENSION/PRACTICE
Ok, now "solve" x² - (googol)x - googol = 0 without a calculator.

RELATED PROBLEM:
Without your calculator show that √(10001) - √(10000) is less than 0.005.
Does this provide us with an effective method of approximating the square root of some large numbers or is it limited and impractical?

For Calculus students: How does this compare to using linearization to approximate the square root?

For more advanced calculus students: Newton's Method? The Binomial Formula (using fractional exponents)? A Taylor Polynomial approximation? All equivalent?

When Curves Collide Part II - Quadratic Systems Re-Explored!

One of MathNotations more popular posts (hundreds of views) was published one year ago this week: When Curves Collide.

Here's a variation to review the essential ideas or to use as an assessment problem or just to challenge yourself. Parts (a) thru (d) require some theoretical analysis and algebraic skill. Part (e) is the main challenge...

An Investigation for Algebra 2/Precalculus
Consider the quadratic-quadratic system:

x² + y² = 1
y = ax² -1, a>0

(a) Show that (0,-1) is always a solution to this system.

(b) For what values of the parameter 'a' will there be 3 distinct solutions to the system?
Coordinate Interpretation: For what values of 'a' will the parabola and circle intersect in 3 distinct points?

(c) For what value(s) of the parameter 'a' will two of the points of intersection be above the x-axis? Below the x-axis (in addition to (0,-1))? On the x-axis?

(d) For the case that there are 3 distinct solutions, determine the two solutions, other than (0,-1), in terms of 'a'.

(e) Now for the main problem:

Assume the graph of our system has three points of intersection: P, Q and R(0,-1). If the area of ΔPQR is 32/25, determine the coordinates of P and Q and the value of 'a'.

(f) Can you think of an even more clever variation!

Monthly Math League Challenge: List all ordered triples of positive integers (x,y,z) whose product is 4 times their sum and x>y>z.

The problem in the title is another wonderful challenge for our readers or for students. Questions like these are powerful tools to develop student reasoning and problem-solving prowess. Once again I have received permission from the directors of the Math League to publish this question on MathNotations. This was the last question from the first contest this year.

Please cite the question in the title as:
Copyright Mathematics Leagues Inc 2007. May not be reproduced without permission of the copyright holder.

You can learn more about Math League Contests at the Math League website.

Now you know that I'm not going to simply copy a problem and just leave it at that!
How can we make this more of an enrichment experience for our algebra students who may not quite be ready for the contest level. Can you guess? Scroll down...


Well, what makes the contest problem particularly formidable is the use of three variables. In fact, two numbers is already a difficult problem for most! So we use the "Let's Make It Simpler" strategy:

Find all ordered pairs (x,y) of positive integers whose product is four times their sum and x>y.

I'll begin some analysis, starting with our basic equation:
xy = 4(x+y).

Before one starts the traditional solving for y in terms of x ritual (then let's go to the graphing calculator), we need to remind our students that this is a positive integer problem, which allows for a somewhat different kind of approach.
I also frequently suggest to students to consider the case that x = y even though the restrictions do not allow this.
If x=y, then x² = 8x leading to x=8, y = 8 (or ____?). Even though this is not allowed, it could suggest other solutions. Those who enjoy graphical solutions will also appreciate that this solution is one of the two points of intersection of the graph of our basic equation with the line y = x. I'll leave it to our readers to find the other.

Note that from xy = 4x + 4y we can see that neither x nor y can equal 4. For example, if x=4, we'd obtain 4y = 16+4y, which is impossible.
Similarly, neither x nor y could be less than 4. This is more readily proved algebraically. I'll omit the details.

Note: Another important tool for students in solving these kinds of problems is to consider symmetry. The basic equation is symmetrical in x and y. Symmetry can be very useful.

Ok, while most students are using guess-test methods (they would call it 'plugging in'), we will solve for y in terms of x (skipping a few steps):
y = 4x/(x-4).
There's a well-known algebraic device students need to see here when finding integer solutions, which is equivalent to long division. Rewrite the previous equation as:
y = (4x-16)/(x-4) + 16/(x-4). I'll let you guess why I subtracted 16 in the first fraction then added it back in the second. This is equivalent to:
y = 4 + 16/(x-4).

Now we use the integer condition (and the fact that x > 4 and x > y) to find solutions:
x-4 has to be a factor of 16 which implies that x-4 could equal 16, 8, 4, 2, or 1. In fact we can show that only 16 and 8 are possible. I'll leave the rest to you...

At r% interest compounded annually, $400 earns $63.05 interest in 3 years. r = ?? Developing Greater "Interest" in Algebra...

SILLY RIDDLE OF THE WEEK
Why were the Romans so good at algebra?
You have to think outside the box and be in the mood for this groaner! Of course you've probably seen this elsewhere on the web...

It's been awhile since we've worked on financial math applications. Anyone recall those 3 mortgage investigations from last year? [Note: To see other mortgage/finance posts, click on the mortgage label/tag in the sidebar].
Considering the current economic situation, perhaps we should devote more attention in our math classes to the subtle trap of running up credit card debt. I'm working on that. There are strong mathematical similarities between loans, mortgages and investments and in this investigation students will focus on the investment problem in the title of this post.

The Problem in the Title of this Post:
At r% compounded annually, $400 earns $63.05 interest over 3 years. What is the value of r?
Let's agree, that r% has already been converted to a decimal so that we do not have to work with r/100 in the formulas below. That is, if r = 10% for example, we will work with r = 0.1.


OVERVIEW OF ACTIVITY
We will first consider a quick estimate of the interest rate by using simple interest to approximate compound interest. This develops sense about the formulas and could be helpful if a question like this appears as a multiple choice question on the SATs or other standardized tests. We will then apply standard compound interest formulas to validate our estimate. Students will be asked to use more than one method for this. Finally, there will be an extension for your students to try.
----------------------------------------------------------------------------------------------------
KEY for this activity (not necessarily standard notation)
[Assume one interest period per year; no additional money deposited or withdrawn]

P = original amount invested (principal)
r = annual rate of interest (decimal form)
n = number of years
A_n = Amount original money is worth after n years
I_n = Interest earned during the nth year
T_n = Total Interest earned over n years
-----------------------------------------------------------------------------------------------------
Background for Simple vs. Compound Interest

Simple: Interest each year is constantly Pr so total interest for n years is T_n = Prn.
Example: If $400 is invested at 10% annually simple interest, then over 3 years one would earn (400)(0.1)(3) = $120 in interest.

Compound Interest
Example: Suppose $400 is compounded annually at 10%.
1st year: Interest earned = I₁ = (400)(0.1) = $40; money is now worth A₁ = $440.
2nd year: Interest earned = I₂ = (440)(0.1) = $44; A₂ = $484
In general:
A₁ = P + Pr = P(1+r)
A₂ = P(1+r) + rP(1+r) = P(1+r) (1+r) = P(1+r)²
(*) A_n = P(1+r)ⁿ

Beginning of Activity
I. Approximating the Rate using Simple Interest:
If the total interest over 3 years is about $63, show that r = 0.05 is a reasonable estimate for our problem using the simple interest formula above.

II. Using Compound Interest Formula
There are several approaches to solving the title problem:

Method I: Use the above compound interest formula (*) directly to solve for r.
Remember: The formula expresses A_n but it's the total interest, T_n that's given.

Method II: Derivation of Related Formulas

(a) Show that or explain why the total interest earned after n years can be expressed as
T_n = P[(1+r)ⁿ - 1].
(b) Use the formula in (a) to solve for r in our problem. Here you will be substituting the values for n, P and I_n first, then solve for r.
(c) Alternate Approach: Use the formula in (a) to derive a general formula for r in terms of n, P and I_n. Then use this formula to find the value for r in our problem. When do you think it makes more sense to use (b)? (c)?

Extension
In the above problem, we knew what the total interest was after 3 years and we needed to manipulate a formula to determine the rate. In other applications, we might want to determine the interest earned each year. This is usually done for us by our bank -- we certainly need this amount for federal and state income taxes. We will now derive the formula for I_n by two different methods:

(a) Derive a formula for I_n using the fact that I_n = A_n - A_n-1, for n = 1,2,3,...

(b) Derive a formula for I_n using the following pattern:
I₁ = rA₀ = rP = rP(1+r)⁰
I₂ = rA₁ = rP(1+r)¹
....
In general: I_n = ____________.
Note: This formula makes sense. Why? Can you show that the results in (a) and (b) are equivalent?

(c) For the original problem in the title of this post, complete the following table:

n................A_n....................I_n
0...............$400...............

1...............$400...............$40

2

3
.
.
.
10


Comments:

• The instructor may choose to use this activity to develop recursive functions. For example, A_n = (1+r)⋅A_n-1
  
• The chart above can be generated using the graphing calculator of course. More importantly, ask students to discover relationships among the columns.
• Much of the above is standard 'stuff' and not very challenging. However, the goal here is to help our students develop a feel for these formulas, rather than mechanically 'plugging in.' Considering that this topic is related to exponential functions, recursive thinking, and geometric sequences, there is unlimited potential for bringing more financial math into the algebra or precalculus classroom. And, yes, it's all standards-based...
  

A 5-7-8 Triangle for Starters -- An In-Depth Exploration in Algebra, Geometry and Trigonometry

As promised, here is an investigation/activity/challenge/.../ (Anyone remember a certain star college and pro football player, with initials K.S., who was nicknamed 'Slash' because he could play several positions?).

The following extensive activity is designed as a long-term assignment, however, modify it as you wish. Don't forget to forgive proper attribution as indicated in the sidebar.


A Possible Instructional Scenario...
"Girls and boys, what do you think of when you're told that a triangle has a 60° angle? What if you're given that all the sides have integer lengths? Why can't it be 30-60-90 in that case? Of course, it could still be equilateral, but for this exploration, we are looking for scalene triangles with integer sides and a 60° angle. By the way, if it's not equilateral, how do we know it must be scalene? Why couldn't it be isosceles?"
------------------------------------------------------------------------------------------------------------
The Exploration

Let's agree on the following labeling for our triangle:
ΔABC with ∠C = 60° with opposite side c. Other angles are labeled A and B with opposite sides a and b, respectively. For this investigation, b > a.

(a) Show that the angle opposite the '7' side in the 5-7-8 triangle in the title of this post is 60°. Using the labeling above: If a=5, b=8, c=7, show that ∠C = 60°.

Note: This can be done with or without the Law of Cosines. As Joshua pointed out in a previous post, the student can work with the altitude on the '5' side and use the Pythagorean Theorem and algebra to show that a 30-60-90 triangle is formed. This method is instructive (and constructive too!).

For parts (b) - (e), we will no longer be focusing only on the 5-7-8 triangle. Consider any integer-sided ΔABC with ∠C = 60° and with opposite side of length 7.
(b) Show that a² + b² - ab = 49. [*]
See note after part (a) for two possible methods.

(c) We are looking for integer solutions to [*]. There are several methods including a Pell equation approach (we will not go in that direction). Students can certainly try a guess-test strategy, however we can initially restrict the possible values of b, can't we?
As an initial boundary, show that 13 >b ≥ 8 using basic geometry.

(d) Show that a better restriction for b is 14/√3 ≥ b ≥ 8 by:

(i) [Trig] Using the Law of Sines
(ii) [Advanced Algebra] Use the quadratic formula in [*] to solve for a in terms of b. Using the discriminant, show that b ≤ 14/√3.

From this result, explain why it follows that b must equal 8.

(e) Solve for a.
Suggestions: From [*] OR use your result from the quadratic formula in (d)(ii) to show that there are exactly 2 scalene triangles satisfying the above conditions.
The answers for this are:
a=3, b=8, c=7;
a=5, b=8, c=7
BTW, is it a coincidence that the two values of a happen to add up to b?

(f) REPEAT PARTS (b) - (e) for an integer-sided scalene triangle with a 60° angle and an opposite side of length 13. There will be some slight modifications needed such as formula [*]. State and use a similar inequality from (d) to show that there are two possible values for b, namely 14 and 15. Go further and show b=14 is not possible (e.g., using the discriminant).
Then show that b = 15 leads to two solutions (triangles) - sorry, I'm not giving these away yet!

(g) REPEAT PARTS (b) - (e) for an integer-sided scalene triangle with a 60° angle and an opposite side of length 19. Again, you should find that the inequality from (d) leads to two possible values for b, only one of which works. Give the two solutions (triangles).

(h) Some of you will no doubt wonder why we did 3 separate analyses, when a slightly more general approach could have been used, specifically a more general inequality in part (d). Ok, so do that (keep all conditions, except side 'c' will now be a parameter).
Show that, in general, 2c/√3 ≥ b ≥ c.

(j) Surely, we can't go further other than searching for a general solution for the 60° problem. Of course, we can: Come up with similar questions and solutions to all parts above if ∠C = 120°. I'll start you off: a=3, b=5, c=7. Show that ∠C = 120°, etc...

I'm sure our astute and talented readers will pick up on my usual errors or omissions or make suggestions to improve the flow of the activity. I'm counting on you! Enjoy...

Beyond Mixed Nuts --A More Challenging Ratio Problem

Readers are strongly encouraged to read the extensive discussion of methods in the comments.


In Virtual HS, the ratio of the number of juniors to seniors is 7:5.
The ratio of (the number of) junior males to junior females is 3:2.
The ratio of senior males to senior females is 4:3.
What is the ratio of junior males to senior females?

Comments:
(1) Isn't it a shame that the Jn:Sn ratio isn't 5:7!
(2) Algebraic methods seem to be the most reasonable here, but would some students attempt straight numerical methods using common multiples? Can you find such a way?
(3) Does anyone ever use or teach tree models for ratio problems? We know they are useful for probability problems, but isn't there a similarity here?
(4) Would a Singapore bar model approach work here? I'm counting on those who are proficient with this method! This is not my area of expertise...
(5) Is this question too difficult for the SATs? More appropriate for a math contest question?

16/64 = 1/4...How to Reduce Fractions the 'Easy Way'!

Totally Clueless sent me an email reminding me of some famous 'fractured fraction' examples like the one in the title. Can you think of a couple of other two-digit examples of the same type that 'reduce' this way? Note that 10/30 = 1/3 doesn't qualify (the digits have to 'cancel' diagonally!).

Here is TC's version:
Note that the product 16 x 4 can be obtained by deleting the '1' and the 'x'!

READER/STUDENT CHALLENGE

(a) Find the other two instances of this 'weird' multiplication. The two factors have to be of the same type as in the example, i.e., a 2-digit number by a 1-digit number and the tens' digit of the 2-digit number must be 1.

(b) Most would find the other instances by guess-test. Here's a more significant challenge. Verify algebraically that there are exactly three such solutions.

(c) Is this problem equivalent to the 'easy way' to reduce fractions mentioned in the title of this post? Why or why not?

An Exceptional Math Contest Problem

Update: The solution I emailed to the director of the Math League is at the bottom after you scroll down a ways. Note how this ties into the difference of squares problem I posted the other day!

A very good friend of mine who runs one of the most successful math league contests in the US has given me permission to reprint the question below with the following attribution:
“Copyright Mathematics Leagues Inc 2007. May not be reproduced without permission of the copyright holder.”

This question was the last question on a recent contest. It certainly made me think! I felt the need to be rigorous about justifying one of the steps and that was the reason for the
M² - N² = 12 post from the other day. There are at least 3 methods that have been found thus far, according to my friend. Teachers often send in their own solutions and some even get included with the official solutions. My feeling is that questions like these are vital for our students intending to pursue higher mathematics, not just for those who happen to participate in contests. I also believe that variations on great questions like this will have an eternal life...

THE QUESTION

There's exactly one real number a for which ax² + (a+3)x + (a-3) = 0 has two positive integer solutions for x. What are these values of x?

Have fun with this!
For more information about the Math League, visit here.

Scroll down to see the solution I sent to the directors of the Math League...











First of all, I'll state the answer I obtained before giving the details:

A real number a which produces two positive integer solutions is a = -3/7 which does lead to the solutions x = 4 and x = 2. Proving that this is the only real value of a is far more challenging and I suspect that there's a much easier solution than the one I found. Anyway, here's my approach:

In order to simplify the original equation, divide both sides by a to make the leading coefficient equal to 1:

x^2 + (1+(3/a))x + (1 - (3/a)) = 0
To simplify further, let b = 3/a:
x^2 + (1+b)x + (1-b) = 0
From a well-known rule about the roots of a quadratic equation, the sum of the roots in this case equals -1-b and the product of the roots is 1-b. Since the roots must be positive integers, it follows that b is an integer less than -1. Remember for later:
b must be negative!

From the quadratic formula, we obtain the roots to be:
x = ((-1-b) +- SQRT(b^2+6b-3))/2.
In order for the roots to be integers, the expression b^2+6b-3 must be a perfect square.
I will rewrite this expression by completing the square:
(b+3)^2 - 12 and this must equal a perfect square , call it N^2.
Let b+3 = M, then we must have M^2 - 12 = N^2 or
M^2 - N^2 = 12. It's easy to guess a difference of perfect squares equal to 12, namely 16-4, but I will now prove this is the only solution:

Factoring, we have (M+N)(M-N) = 12 where M,N are integers.
The only pairs of factors to consider are: 12,1; 6,2; 4,3 as well as their negatives.
In order for M,N to be integers, the factors of 12 must BOTH be EVEN so the only possibilities are 6 and 2 or -6 and -2..
If M+N = 6, M-N = 2, then, by adding, M=4 and N=2. From b+3 = M, we would have b = 1, but this doesn't work since b must be negative (see above!)
Therefore M+N = -6 and M-N = -2. From this we have M=-4 and N=-2.
Then b = M-3 = -4 - 3 = -7.
b =-7 leads to the quadratic equation: x^2 + (1+(-7))x = (1-(-7)) = 0 or
x^2 - 6x + 8 = 0. Factoring we obtain the roots to be x = 4 and x = 2.

Since b = 3/a, it follows that -7 = 3/a or a = -3/7.

And vs. Or: Developing Deeper Understanding in Algebra

While some are awaiting the remaining chapters in Alec Klein's interview, I know the rest of my readers are wondering why I've gone into early hibernation. I'm actually doing some independent mathematical research apart from anything on this blog but I will not bore you with the details at this time. That may change (i.e., I may bore you later!).

In the meantime, since the central theme of this blog has always been developing student conceptual understanding, here are a couple of problems for you to consider giving to your Algebra 2 students (or beyond). Conjunction vs. Disjunction is often misunderstood by students and these ideas appear in so many contexts in mathematics, from absolute values to inequalities and beyond. Consider giving these as warm-ups, for review, practice for SAT's, etc. I'm not suggesting these are difficult or challenging problems. Their purpose is to promote deeper reflection on the part of the student. Students who have strong background and understanding will simply solve these quickly and not see why anyone would make a big deal over them. However, you may find other students who don't grasp the ideas as readily or have forgotten. Comparing/contrasting is a powerful heuristic when trying to develop a more profound understanding of mathematics...

1. If (a -4)² + (b+4)² = 0, what is the least possible value of a² + b²?

2. If (a-4)(b+4) = 0 , what is the least possible value of a² + b²?
(A) 0 (B) 8 (C) 16 (D) 32 (E) 64

Ask students to explain to each other, why the word 'least' is irrelevant in Question 1 but not in Question 2. Also, how does question 1 relate to circles?

But it's not in the Standards: Finding imaginary roots, completing the square, factoring and other 'obsolete' topics...

Remember the good old days when students solved 'quadratic-type' equations? Of course, many are still doing this but it is fast becoming a lost art (and some of you may feel it should be!). It is not required in any state math standards or Achieve's, so there's no reason to mention it, right?

Below you will find a 4th-degree (quartic) polynomial equation. The rational root theorem won't help because there are no rational roots. The graphing calculator won't help because there are no real roots! Ok, maybe Mathematica and other Symbolic algebra software could do this, but who exactly programmed this?

Using substitution to rewrite certain 4th degree equations as quadratics (so-called 'biquadratic' equations) used to be covered in some Algebra 2 or advanced classes. Some of you may feel nostalgic about this. However, our challenge today is to solve this by at least TWO 'radically' different methods and then show the solutions are equivalent!

Here's your equation:
x⁴ + 3x² + 4 = 0

(a) Explain, without solving, why this equation has no real roots. Should ALL students in Algebra 2 and beyond be able to answer this one?

(b) Solve, by substituting y for x² and using the quadratic formula. You should eventually arrive at 4 imaginary solutions. This is the way I was taught to solve it, eons ago.

(c) Solve by completing the square and factoring. [Definitely not the first method I would have thought of way back when...]

(d) Show your results are equivalent. This may be annoying! So, which method is easier in your opinion?

(e) Any other method for finding imaginary solutions?

QUICK OPINION POLL
(1) Completing the square (not to mention factoring) is no longer an important topic and should be deemphasized in our curriculum (or omitted).
By the way, is it explicitly mentioned in your state's math standards for Gr 8-12?
(YES NO)

(2) The equation in this post has little relevance to the 21st century and Dave should be ashamed for publishing such trash. Besides, this topic is not included in the Algebra 2 Standards developed by Achieve and ADP.
(YES NO)

You've perhaps assumed that since I've been discussing and complimenting Achieve's standards and the new Algebra 2 End of Course Exam, that I would no longer advocate exposing students to this kind of traditional mechanical 'exercise.'

Well, I taught from the AP Calculus syllabus and I still made time to discuss some ideas and methods that were not 'required'! Further, who exactly will be the ones left on this planet who know how to find imaginary roots for this type of polynomial equation that has no real roots! In case you're wondering, this kind of question has traditionally been taught in Asian countries and still is! (Dave, can you document that? Sure...)

An In(de)scribable Rectangle in a Rectangle - A Geometry Challenge

Now that the summer months have arrived, I thought it was time for a geometry challenge problem to chew on. Although this is a departure from the lesson plans I have been writing, it's still an enrichment experience. I'm often asked by students and parents how one becomes better at solving 'hard' math problems. My response is: "Keep trying hard problems!" One can only improve at problem-solving by challenging one's mind. Also, learn from others - we all learn from good models. There are no shortcuts here. Some frustration is healthy and if you want more cliches, let me know!

OVERVIEW
This question should definitely challenge your geometry students. It was brought to my attention by a teacher via a student who was given this by his honors geometry teacher. I'd provide proper attribution if I knew the original source. However, it is possible to go beyond this question and generalize. There are endless problems one could generate from inscribing Figure A in Figure B. Rectangles in rectangles, other than special cases (square in a square) are not often seen by students.

In addition, strong algebra skill and a graphing calculator would be useful. Use of Geometer's Sketchpad (or traditional drawing tools) would also make sense here as a fairly accurate construction of the diagram (better than my crude attempt) would be highly instructive and students enjoy 'solving' the problem this way. Of course they need to understand that such a solution is not mathematically valid!

THE PROBLEM
In the diagram below, ABCD is a rectangle with AB = 8 and BC = 6. Rectangle PQRS is inscribed in ABCD, i.e., the vertices of PQRS lie on the sides of ABCD. If PQ = 8, what is the length of QR?

Notes:
(a) Figure not drawn to scale! Drawing this was not fun!
(b) Someone out there will argue that side PQ could coincide with side AB by my definition of inscribed. After all, the diagram is not drawn accurately! I should have added that the rectangles share only those vertices in common!
(c) Students often begin by assuming that PQ is parallel to the diagonal AC. Careless use of similar triangles could lead to an answer of 2. The only problem is that the actual answer is 2.2085 rounded! Does PQ have to be parallel to AC? In fact, is it even possible here? The instructor might begin with assuming parallelism and asking students to see where that leads and if the conclusion makes sense.
(d) What might a mathematician do to extend this numerical problem? Would they consider the issue of a unique solution here, i.e., is the given length of PQ enough to produce only one such inscribed rectangle? What is the range of possible values for PQ (assuming that PQ represents the longer dimension)? Could PQ be 10 or more? Explain. Could PQ be 6 or less? How could we generalize this result further?
(e) As always the disclaimer: My results need independent verification - I depend on my astute readers to check them and correct any careless errors. You are always my best editors!

Take any number, Add Three, Divide the Result by -1. Now Repeat this! Recursive Sequences and Functions Part I: Grades 7-12

Here is the link to the Carnival of Math Edition X.

The following is the first in a series of investigations in recursive sequences and functions for middle school and secondary students. This apparently advanced topic is accessible to prealgebra students at an introductory level. The first few parts of the investigation below are appropriate for the younger students. The remaining parts require more algebraic facility and reasoning. The problem in the title of this post doesn't begin until more than halfway down the page (after some background is developed). Do not skip the background below since it's referred to frequently in the activity. My personal experience is that this topic is highly engaging to students. Considering the connection between recursion and fractals, this topic is certainly part of most standards-based curricula. The terminology of recursion (recursively-defined sequences, recursive description, recursive function, recurrence relations, etc.) is quite confusing at first. Many confuse these ideas with iteration, a general term for describing repetitive algorithms.
Finally, from a pedagogical point of view, please note how the Rule of Four is implemented in the activity below: We start with a verbal description of the rule of formation of a sequence (in natural language), followed by a concrete numerical representation of the terms, followed by symbolic representation. One could also depict the terms graphically on a number line or in the coordinate plane if the function model is used for the sequence.
----------------------------------------------------------
I should probably save this for the new school year but it's hard for me to suppress ideas when they begin to crystallize. I've been thinking for some time about how we can introduce recursive functions in prealgebra through advanced algebra and beyond. I enjoy taking sophisticated ideas and reducing them to basic principles, then developing lessons that explore the topic in some depth. Moreover, this particular topic reveals the interconnectedness of mathematics in a particularly elegant and beautiful way.

Background (Needed for the Investigation Below!)
Consider the sequence 1,2,4,8,...
Elementary students can generally guess the most likely value for the next term, 16. They also are expected to identify the 'rule' of forming the 'next' term, namely doubling or multiplying by 2. This is an important stage in their development of algebraic reasoning - abstraction or generalization. In addition, they should begin to recognize that the terms of the sequence can be described generally as powers of 2, even though a formal introduction to exponents normally begins in 7th grade.
Middle school students should progress to the function table format of a sequence:
n.....a_n
1.....1
2....2
3....4
4....8
5....16
...

Elementary and middle school students should be able to verbalize in natural language that 'you double the terms'. As educators, we need to lead them to a more formal relationship by a line of Socratic questioning like: "Double what? To get what?" Students should then be able to express the idea that each term is twice the previous term. We can ask, "Which term doesn't follow that rule?"
To symbolically describe this sequence, we can write:
a₁ = 1
a_n+1 = 2 ⋅ a_n, n = 1,2,3,...
This is known as a recursive description of the sequence. Try it - replace n by 1,2, and 3 and see if it produces the terms above.
[Note: Later on, in more advanced algebra, students should be able to express this as a recursive function: f(1) = 1; f(n+1) = 2f(n), n = 1,2,3,...]

The closed or general form requires a knowledge of exponents but is accessible to 7th graders
a_n = 2^n-1, n = 1,2,3,... Try it!
(If you're questioning my sanity (you wouldn't be the first!) about introducing such sophisticated mathematics to general 7th graders, well, I do have a legitimate basis for this curricular decision - more later...).

Powers lend themselves naturally to a recursive description and this is why I begin with the above example. Recursive thinking develops when we ask questions like:
If we know what 2⁵ is, how would we obtain 2⁶?
To deepen this understanding further:
If we know what 2⁹⁸ is, how would we obtain 2¹⁰⁰?
Does the exponent key on a calculator help students see these relationships? Not really! The calculator is useful to demonstrate powers and exponents but not for this discussion. Later on, the graphing calculator can be used to enter recursively-defined functions (after they've learned the ideas!).

If you're very familiar with recursively defined sequences and functions, you've probably left this page already! However, the idea of an operation or function being defined in terms of itself is a beautiful and very important notion in mathematics. This type of thinking was necessary for Mandelbrot to develop the notion of fractals, which defines a process in which each stage is defined in terms of the preceding stage or stages - that is recursive thinking!

Ok, by now you're wondering what happened to the title of this blog!
THE PROBLEM:
TAKE ANY NUMBER, ADD THREE, DIVIDE (OR MULTIPLY) THE RESULT BY -1. NOW REPEAT THIS SEQUENCE OF OPERATIONS ON THE RESULT YOU OBTAINED.

Student Activity:
1. Start with the number 6 and follow the instructions above. Repeat this 2 more times. List the first 4 terms of the sequence obtained. Write a brief description of what you observe about this sequence.
2. This time start with a different integer. Again, list the first 4 terms of the sequence obtained and your observations.
3. By now you've concluded that the sequence obtained will alternate in the form a,b,a,b,...
Which one of the original operations (add 3, multiply result by -1, etc.) do you believe is causing the sequence to repeat like this?

The remaining parts require algebra background.

4. If the first term is x, verify algebraically that the sequence will alternate.
5. You've now determined that the sequence appears to be repeating but not constant like a,a,a,a,... For what value of x, the first term, will the sequence be constant, i.e., all terms will have the same value?
6. Write a recursive definition (refer to how we did this for powers) for our sequence whose first term is x. We'll start you off:
a₁ = ____
a_n = __________, n = _______
7. BACKGROUND
The algebraic formulation of the recursive description in #6 was fairly straightforward, since it is just a symbolic representation of the verbal "Take any number, add three, then divide the result by -1." As useful as this may be, we often a need a general formula for the nth term as a function of n, rather than in terms of preceding terms of the sequence. That last sentence was fairly complex, so here's an illustration:
Let's assume the first term is 6. Then the nth term can be described as:
a_n = 6 if n = 1,3,5,7,... (i.e., n is odd)
a_n = -9 if n = 2,4,6,... (i.e., n is even).
This would allow us to find any particular term, say the 100th term, without knowing the values of preceding terms. Such a description is known as a general description or the closed form of the sequence. Such a formula is often very hard to determine, whereas the recursive form is easier to formulate. In our problem, the general formula for the nth term had to be given in two cases or piecewise as mathematicians term it.
It is possible to give a single formula for all of the terms of our sequence as a function of n:
a_n = -7.5(-1)ⁿ - 1.5, n = 1,2,3,...
Verify this formula for our sequence above: 6,-9,6,-9,6,-9,...
8. Change the original problem to:
Take any number, add 2 to it and multiply the result by -1. Repeat.
(a) Starting with an original value of 6 (as the first term), list the first 5 terms of the sequence.
(b) Write a recursive description for this sequence.
(c) Write a piecewise formula for the nth term as in the background example above.
(d) Write a single formula (closed form) for the nth term in terms of n for this sequence.

9. (More Challenging)
A sequence is defined verbally by:
Take any number, add k to it and multiply the result by -1.
(a) If the first term is x, write a recursive description for this sequence.
(b) Write a piecewise formula for the nth term.
(c) (Super Challenge) Write a single formula (closed form) for the nth term in terms of n for this sequence.