# Chapter 5

“Line art” made entirely out of segments of linear polynomials

Linear Polynomials In One Variable

So far, we have been exploring polynomial expressions, their graphs, and the connections between these two representations.

There is a third representation for polynomials as well, called xy tables.

Below is a snapshot of an applet which will allow you to discover what an xy table is.

Run the applet  “table.html”  , answer the given question, and follow the resulting directions.

After the word “Correct” appears twice, go on to the next page.

Polynomial expressions algebraically descibe the exact relationship, or pattern, between the input value and its resulting output value of any solution.  Graphs geometrically give a picture representation of this relationship, or pattern, between input values and output values for a range of solutions.  Tables are finite lists of input values and their resulting output values of actual solutions.  In other words, xy tables are solution tables.  This means, tables can be used to sketch the graphs of polynomials, when they are not given or available.  As you will see later, tables can also be used to determine the polynomials themselves, when they are not given or known.

To the right, are snapshots of a program which will randomly generate 4 linear polynomials, and a pdf for showing work.

The challenge is to fill in the tables, plot the solutions, and sketch the graphs.

Run  program 501  , until you correctly fill in the table and sketch the graph, for at least 3 of the 4 polynomials.

Use a copy of the pdf table to fill in the tables and sketch the graphs.

After correctly filling in the table and sketching the graphs for at least 3 of the 4 polynomials, go on to the next page.

Since the graphs of linear polynomials are lines, all we had to do was find and plot enough solutions to accurately sketch them.  However, it’s pretty time consuming to create a table, choose what input values to use, find the output values, plot the points, and then sketch the line.  All of this work might be necessary for higher degree polynomials, but for a linear one, only two points are necessary for graphing.  Since only two solutions are needed, it would be great if we could find a couple, without even having to create a table at all.

To the right are snapshots of an applet, which will allow you to discover a way to graph linear polynomials, without tables.

1)  Choose, or have someone else choose for you, a linear polynomial.

2)  Click and drag the two points to where you believe two solutions are located.

3) Use the line slider to see which two solutions the applet would choose, and repeat, until you discover a pattern which explains why.

Run the applet  “lines.html”  until you discover a way to graph linear polynomials without using tables.

Use a copy of the pdf  appletnotes2g  to record examples and your discoveries

After discovering a way to graph linear polynomials without using tables, go on to the next page.

When a linear polynomial is written in one variable, it’s in y = ax + b form. As you just discovered, the constant b, is the y-intercept (the point where the line crosses the y-axis). To be on the y-axis, the input value x has to be zero.  When you plug in zero for x into y = ax + b, you’re just left with y = b.  In other words, you can always use the solution ( 0 , b ) as one of your points to graph, without having to make a table.

We also discovered, to get a second solution to plot, all we do is look at the x-coefficient (the “a” value) in fraction form.  As we discovered, the fraction in front of the x, describes the steepness of the line.

The steepness of the line is referred to as the slope of the line,  and is usually represented by the letter m.

When linear polynomials are written in one variable, it’s referred to as y = mx + b form, or slope-intercept form.

For example, the linear polynomial y = x + 4 , is written in y = mx + b or slope-intercept form.

In the example, m =, is the slope (the steepness of the line), and b = 4, is the y-intercept (where the line intercepts the y-axis).

The slope is often described as rise over run, where the numerator describes vertical movement, and the denominator describes horizontal movement.

To see more examples, run the applet  “slope.html”  .

Below are snapshots of a program which will randomly generate 8 linear polynomials, in slope-intercept form, for you to graph (without having to make a table of solutions).  The challenge, is to plot the y-intercept and one other solution based on the slope, graph the line, and then give the coordinates of the second solution.  To make an accurate sketch of the line, you can use the slope to plot as many additional solutions as necessary.

Run  program 502  until you correctly graph and answer the questions, for at least 7 of the 8 probems.

Use a copy of the pdf  ymxb  to record the linear polynomials in slope-intercept form, plot your solutions, and sketch the lines.

After correctly graphing and answering the questions, for at least 7 of the 8 problems, go on to the next page.

Below are snapshots of an applet, which will allow you to discover how to go the other way, and determine the linear polynomial, based on the graph.

Run the applet  “linescomputer.html”  until you are able to determine the linear polynomial, based on the graph.

Use a copy of the pdf  appletnotes2g  to record at least 3 examples, and any discoveries.

After recording at least 3 examples (including sketches of the lines, and the linear polynomials they represent), go on to the next page.

Below are snapshots of a program, which will randomly generate the graphs of 8 linear polynomials for you. The challenge, is to determine the linear polynomials, based on their graphs.

Run  program 503  until you correctly determine the linear polynomials for at least 7 of the 8 graphs.

After correctly determining the linear polynomials for at least 7 of the 8 graphs, go on to the next page.

The pdf  slope-int challenge contains 10 challenges.  Below are snapshots of a program for finding and graphing solutions to each challenge.

Use  program 504  for graphing linear polynomials and solving the challenges.

Run  program 505  to check your solutions.

Use a copy of the pdf  slope-int challenge to record your solutions, in slope-intercept form.

After successfully solving at least 9 of the 10 challenges, go on to the next page.

So far, we’ve had the advantage of being able to use an xy grid to help solve the linear polynomial examples and challenges.  Rise over run works well for calculating slopes, when the values involved are integers.  This is because all you have to do is count.  This won’t always be the case however.  When there isn’t an xy grid available and/or the numbers involved aren’t integers, we’re going to need another way of determining slope.

To the right,  are snapshots of and applet, which will allow you to discover another way for determining the slope of a line.

Run the applet  “slopenogrid.html”  , until you discover a way to determine the slope of a line, without a grid.

Use a copy of the pdf appletnotes  to record your discoveries.

After you know how to determine the slope of a line, given two solutions, go on to the next page.

As we discovered, if we know the y-intercept and one other solution to a linear polynomial, we can determine the slope of its graph (the slope of the line), and the linear polynomial itself as well.  Knowing two solutions of a linear polynomial, means we know the coordinates of two points on a line.  Using the coordinates, we were able to determine how much the x values and y values changed, as we went from one point to the other.  From that, we were able to determine the slope, because the slope is just the comparison of those two changes, or the rate of change between the y values and the x values.

Mathematicians have an easier way to write this though.  They use the 4th letter of the Greek alphabet, delta, to represent the word change, which looks like a triangle.

As we discovered, to find the change in y and the change in x, we had to find the difference between the y values and the x values of the two points.  In other words, we had to subtract the y values and subtract the x values of the two points.

Using subscripts to show which point the y values and x values come from (point 1 or point 2), the definition for slope is as follows.

Below are snapshots of an applet, which will randomly generate two points and a line connecting them.  The challenge is to determine the slope of the line, and the linear polynomial it represents.

Run the applet  “deltayoverdeltax.html”  to try some challenges like the one above.

Use a copy of the pdf appletnotes to record at least two examples (one with a positive slope, and one with a negative slope).

After solving and recording at least two examples, go on to the next page.

In the examples and challenges so far, one of the solutions given, has been the y-intercept.  Below are snapshots of an applet, which will allow you to explore what happens to the slope, when you get to choose the two points.

Run the applet  “tangent.html”  to explore what happens to the slope when you change one or both of the points.

Use a copy of the pdf  appletnotes2g to record what each of the 4 sliders do, and your discoveries.

If you are wondering why the name of the applet is called “tangent”, open the pdf  “trig ratios”  .

After recording examples, sketches, and your discoveries, for all 4 sliders, go on to the next page.

As we discovered, nothing happens to the slope, when you get to choose the two points.  The steepness of the line is the same everywhere, so it doesn’t matter which two points you use to determine the slope.

As we discovered, any two points will do, to determine the slope of a line.  Below are snapshots of a program, which will randomly generate 7 pairs of points.  The challenge is to determine the slope of the line connecting the two given points.

Run  program 506  , until you correctly determine the slope for at least 6 out of the 7 problems.

Use a copy of the pdf  programnotes to record the given points, your work, and any discoveries.

After correctly determining the slope for at least 6 of the problems, and recording your work, go on the the next page.

So far, the change in y and the change in x for all of the slopes in the examples and challenges, have both been integers.  Below is a snapshot of an applet, which shows two views of the ramp in the school’s cafeteria.  The challenge is to estimate the slope of the ramp, and to determine if it meets the Americans with Disabilities Act’s specifications for ramps.

Run the applet  “ramp.html”  , and use the lines and line segments, to estimate the slope of the ramp.

In the pdf  ramp, you will find the A.D.A.’s specifications for ramps.

Use a copy of the pdf  appletnotes to record your sketches and work to defend your answer to the challenge.

After defending whether or not you believe the ramp meets the A.D.A.’s specifications, go on to the next page.

We started off this chapter by using xy tables, or solution tables, to graph linear polynomials.  Our next challenge is to use tables to determine the linear polynomials themselves, in slope-intercept form.  Below is a snapshot of an applet, which will randomly generate xy tables for linear polynomials.  The challenge, is to determine the linear relationship between the x and the y values of the solutions.  In other words, the challenge is to find a formula for y, in terms of x.

Run the applet  “linearrelations.html”  , until you discover how to determine the linear polynomials, without guessing.

Use a copy of the pdf  appletnotes, to record at least one example, and your discoveries.

After discovering and recording how to determine the linear polynomials, without guessing, go on to the next page.

In the last applet, all of the x coefficients were integers.  Below is a snapshot of an applet, which will randomly generate xy tables for linear polynomials, where the x coefficients will be written as rational numbers (fractions).  Again, the challenge, is to determine the linear relationship between the input and the output values, in slope-intercept form.

Run the applet  “linearrelations2.html”  , until you discover how to determine the linear polynomial, without guessing.

Use a copy of the pdf  appletnotes, to record at least one example, where the denominator is not a 1.

After discovering and recording how to determine the linear polynomials, without guessing, go on to the next page.

As we discovered, to write a linear polynomial in slope-intercept form, or y = mx + b form, we need to know the slope m and the y-intercept b.  The y-intercept is the solution where the line crosses the y-axis.  To be on the y-axis though, the x value must be a zero.  In other words, to find b, the y-intercept, we just have to find the solution with an x value of 0.  The slope is the steepness of the line, which is described by the change in y over the change in x, or ∆ y / ∆ x.  In other words, to find m, the slope, we just have to determine how the y values and the x values are changing from solution to solution.

Run the applet  “linearrelations3.html”  , until converting solution tables to slope-intercept form, makes sense to you.

After converting solution tables to slope-intercept form, makes sense to you, go on to the next page.

Below are snapshots of a program, which will randomly generate 12 xy tables for linear polynomials.  The challenge, is to determine the linear relationship between the y values and the x values (find a formula for y, in terms of x).

Run  program 507  , until you correctly determine the linear polynomials, for at least 10 of the 12 tables.

Use a copy of the pdf  programnotes, to record at least 3 of the solution tables, the linear polynomials to which they belong (written in slope-intercept form), and to explain how you found them.

After correctly determining the linear polynomials for at least 10 of the 12 solution tables, and recording and then defending your reasoning as well, for at least 3 of them, go on to the next page.

We can learn a lot about mathematical and natural occurances, by noticing linear relationships, and determining a way to represent them.  There is a linear relationship between temperature, and the height of the liquid in a thermometer.   In Celsius, the scale used by most countries, water freezes at around 0º and boils at around 100º.  In Fahrenheit, the scale still used in the U.S., water freezes at around 32º and boils at around 212º.  Since both scales represent linear relationships, there is a linear relationship between them as well (in chapter 3, we found out lines can be transformed into other lines, by scaling and translating vertically).

 C F 0º 32º 100º 212º

From the two solutions in the CF table, we can determine a linear polynomial, or formula for F in terms of C.

Since the F-intercept is 32º, ∆ F is 180, and ∆ C is 100, the linear relationship between C and F is F = (180/100) C + 32º .  After reducing to a simpler fraction, the linear relationship, or formula for F in terms of C, is as follows.

How many degrees Fahrenheit, are the following temperature readings in Celsius?

1)     20º C                         2)     37º C                         3)     -40º C

Use a copy of the pdf  booknotes to record the linear polynomial, the conversion steps, and any discoveries.

Determine the linear relationships for y, in terms of x, for each of the following 6 tables.

Determine what the x values and the y values represent in each table, or make up possible explanations.

Use a copy of the pdf  booknotes  to record and defend your work.

After determining the linear relationships of the 6 tables, and defending your work, go on to the next page.

In all of the examples and challenges so far, the linear relationship between the input values and the output values, have been perfect.  When data is collected which is close to having a linear relationship, valuable information can be gained from determining the linear relationship which best fits the data.

The x values and the y values in the table below, are close to having a perfect linear relationship. In other words, as the graph shows, the points almost line up perfectly.

The graph of the linear polynomial, which best describes the relationship between the input and the output vlaues, is called the best fit line, or more formally, the regression line.

As it turns out, the graph of y = 1.9x + .4 is the best fit line for the four data points.

This means it’s impossible to draw a line which comes closer to all four points, than the one shown below. In other words, the linear polynomial y = 1.9x + .4 is the best linear description for the relationship of the data in the table.

Now the question is, how do we measure how close the line comes to the points, in order to claim it’s the closest?

Before we can understand best fit lines though, we have to understand best fit points.  Below are snapshots of an applet, which will allow you to discover what the best fit point is, for a group of four points on a number line.

Run the applet  “bestfitpoint.html”  until you discover what the best fit point is.

Use a copy of the pdf  appletnotes  to record your answers to the two given questions, and your discoveries.

After recording and defending your answers to the two given questions, go on to the next page.

As we discovered, the least total error occurred, when the black point was anywhere between the two middle points.  In other words, for an even number of points, there is no unique point where the least total error occurs.  The total of the errors squared however, did have a unique point.  In other words, there was only one point where the total of the squares of the errors was the least.  As we discovered, and may have expected, that unique point was the mean, or average, of the four points.  So now we know the best fit point is the average, and it can be found by determining the least amount of total squared error between it and the other points.

Looking at our original example, shown again below, the best fit x value would be 4, and the best fit y value would be 8.

This means ( 4 , 8 ) is the best fit point for the data above, and the point we want our best fit line to go through. Unfortunately, there are an infinite number of lines which go through the point ( 4 , 8 ).

We want the line which has the least amount of squared error, where the errors are the distances between the y values of the line, and the actual y values of the points.  In other words, we want to minimize the total area of the squares, whose sides are the vertical distances between the points and the line.  This is why it’s called the method of least squares.

Run the applet  “leastsquares.html”  , to show that the linear polynomial y = 1.9x + .4 is the best fit line, becuase it minimizes the total squared error.

After showing the least amount of squared error occurs for the line y = 1.9x + .4, go on to the next page.

Now the question is, how do you find the best fit line, without using the applet?

The goal is to minimize the sum of the squares, of the errors between the actual y values of the points and the y values of a line.  In other words, we want the sum below to be as small as possible, where y1 throgh y4 are the y values of the four points, and mx1 + b through mx4 + b are the y values for the four input values x1 througho x4 .

[ y1 – ( mx1 + b )]2  +  [ y2 – ( mx2 + b )]+  [ y3 – ( mx3 + b )] +  [ y4 – ( mx4 + b )]2

To represent multiple additions, mathematicians use sigma, the 18th letter of the Greek alphabet.  This means, by using the summation symbol ∑, the above sum can be shortened to the following

Even though we now have an easier way to write the sum, it will still be too difficult to determine what “m”, the slope, and “b”, the y-intercept, of the best fit line should be.  In other words, we need an expression which doesn’t have, both an “m” and a “b”.  As you will see later, if we know what “m” is, we can figure out what “b” is.  This means, we would like an expression which doesn’t contain the “b”.  In other words, we would like the “b” value to be zero.  This can be done by using the best fit point, ( 4 , 8 ), as our origin.  If we do this, “b” will be zero, because we know the best fit line goes through it.

To use the best fit point as the origin, all we have to do is subtract 4 from each x value, and subtract 8 from each y value.  In other words, whether we use ( 1 , 2 ), ( 3 , 6 ), ( 5 , 11 ), ( 7 , 13 ) or ( -3 , -6 ), ( -1 , -2 ), ( 1 , 3 ), ( 3 , 5 ) as our 4 points, the relationship between the x values and the y values will stay the same.  This is because the slope of the best fit line for either group of 4 points will be the same, since all we are doing is translating to the left 4 and down 8 (as you discovered earlier, translations don’t change the steepness of lines).

By using the best fit point as the origin, we now have the following, which no longer involves a “b”.

If we look closely at the above trinomial, we will discover, it’s really just a quadratic (degree 2) polynomial in terms of m.  Since the m2 coefficient is a square, it has to be positive, so the graph of the quadratic is a parabola which opens upward.

Remember, our goal is to find an m value which will make the output value of the above trinomial, as small as possible.

In other words, we want to know the m value of the vertex (the bottom point of the parabola).  As we discovered in the last chapter, any quadratic can be vertically magnified and transformed into a monic quadratic, where the lead coefficient is a 1, without changing the input value of the vertex.  To transform the previous quadratic into a monic one, we geometrically magnify it by 1 over ∑( xi – 4 )2 , or algebraically divide each term by ∑( xi – 4 )2 .

The nice thing about this is, as we discoverd in the last chapter, the m value of the vertex is just the average value of the two roots.  So all we have to do is add the two roots and divide by 2.  As we also discovered in the last chapter, the m coefficient for a monic quadratic is always the opposite sign of the sum of the roots.  So we don’t need to find the roots, we just have to divide the opposite of the m coefficient by 2, which gives us the following.

Dividing by 2 is the same as dividing by 2 over 1, which gives us the following.

Dividing by 2 over 1 is the same as multiplying by 1 over 2, which gives us the following.

After reducing, we have the m value which will make the original sum of the squares of the errors, the smallest it can possibly be.  In other words, the following is the slope of the best fit line.

To determine the slope of the best fit line, all we have to do is calculate the sums, using the following values.

To determine the y-intercept, we just have to use the definition for slope, m = 1.9, and the two points ( 4 , 8 ) and ( 0 , b ).

$latex m=\frac{\left( y2-y1 \right)}{x2\; -\; x1}$

$latex 1.9=\frac{\left( 8-b \right)}{4-0}$

$latex 1.9=\frac{\left( 8-b \right)}{4}$

( 1.9 )( 4 ) = 8 – b

b + ( 1.9 )( 4 ) = 8

b = 8 – ( 1.9 )( 4 )

b = 8 – 7.6

b = .4

In slope-intercept form,  y = 1.9x + .4  , is the best fit line for ( 1 , 2 ), ( 3 , 6 ), ( 5 , 11 ) and ( 7 , 13 ).

At the Hatch Hill Solid Waste Facility in Augusta, Maine, you can drop off both garbage and recycling.  Each time you do, your vehicle is weighed when you enter, and then again when you leave.  This way the amount of garbage/recycling can be determined and recorded.  Customers who use the facility, receive tickets as they leave, with current and running totals.  Below are actual statisitics, collected by a couple from the Augusta area.

• Ticket # 458026 on September 25, 2007:  Load 108, 80 lbs (.04 tons), 6.67 total tons
• Ticket # 498536 on September 30, 2008:  Load 136, 60 lbs (.03 tons), 7.89 total tons
• Ticket # 537415 on September 19, 2009:  Load 161, 100 lbs (.05 tons), 8.95 total tons
• Ticket # 577442 on September 25, 2010:  Load 189, 80 lbs (.04 tons), 10.2 total tons

There is a close linear relationship between the year and the total tonnage, as the scatterplot to the right shows.

• ( 7 , 6.67 )
• ( 8 , 7.89 )
• ( 9 , 8.95 )
• ( 10 , 10.2 )

Run  program 508  to determine the best fit line for the above points.

Use a copy of the pdf programnotes to record your work and any discoveries.

After determining the best fit line for the scatterplot above, go on the next page.

Now that we know the linear polynomial for the garbage/recycling data, we can entertain the following questions.

1. What year did the couple start bringing garbage/recycling to the facility?
2. What is the expected total tonnage for today’s date?
3. What is the expected total tonnage for ten years from today?
4. What is the average total tonnage of garbage/recycling produced by the couple each year?
5. How much residential garbage/recycling would you expect to be produced by your town or city?
6. How much residential garbage/recycling would you expect to be produced by the State of Maine?
7. What assumptions did you use to answer questions 5 and 6?

Determine and defend an answer to each of the above 7 questions.

Use a copy of the pdf  booknotes to defend your answers to the questions below.

After you have defended your answers to the above questions, go on to the next page.

Now that we understand how to find the best fit line for a scatterplot, we can use various software packages to do the calculations for you.

To the right, is a snapshot of the garbage/recycling data, using the software  “Logger Pro”.

All we have to do, is enter the data in the spreadsheet on the left, and then click “Linear Fit” under the “Analyze” menu. The points and the best fit line will be graphed, and an information box will appear on the screen with the best fit line information.

After experimenting with “Logger Pro” , go on to the next page.

If you thought the “line art” picture at the beginning of the chapter was interesting, run  program 512  , and explore.

Below is a snapshot of the chapter 5 programs menu program.

Run the program  “5menu.bas”   , or its alias to access every program used in chapter 5.

“How can it be that mathematics, a product of human thought independent of experience, is so admirably adapted to the objects of reality?” — Albert Einstein —

“What sciences can there be more noble, more excellent, more useful for people, more admirably high and demonstrative, than mathematics?” — Benjamin Franklin —

“There is no branch of mathematics, however abstract, which may not someday be applied to phenomena of the real world” — Lobachevsky —

“Mathematics is the abstract key which turns the lock of the physical universe.” — John Polkinghorne —

“Mathematics is not only real, but it is the only reality.” — Martin Gardner —