A synopsis of the important ideas on this page:

Linear functions are functions with straight-line graphs.  The most basic linear function is y = f(x) = x; the generalized linear function is y = f(x) = mx + b.  Just as the parameters a, b and c of the  generalized quadratic function y = a t^2 + bt + c can be adapted to precisely model any parabolic graph, the parameters m and b of the generalized linear function can be adapted to precisely model any non-vertical linear graph.

Just as quadratic functions naturally model anything that changes at a constantly changing rate, linear functions naturally model anything that changes at a constant rate.   It is worth noting (and string away for a future calculus course) that since quadratic functions of time have rates that change at a constant rate, the rate of change of such a quadratic function is therefore a linear function of time.

Linear functions can be better understod by visualizing families consisting of those linear functions f(x) = mx + b with a fixed value of b, and those with a fixed value of m.   The former will appear as a sort of pincushion of straight lines through the cmmon point (0,b); the latter will appear as a stack of parallel lines, each with the given slope m.

The slope of a linear function y = mx + b between any two distinct graph points is always m.  This can be verified by calculating rise / run = (y2 - y1) / (x2 - x1) for arbitrary x1, x2.  Thus m represents the constant rate of change associated with the linear function.

Key points on the graph of a linear function include the y-intercept (substitute x = 0 into y = mx + b to get the point (0,b) ), the x-intercept (substitute y = 0 into the same equation to get point (-b/m,0)) and the point 1 unit to the right of the y intercept (since slope is m this point will be m units higher, at (1, b+m)).

Specific instances where linear functions are useful include the range of flow from a hole in the side of uniform cylinder, the force necessary to hold a pendulum or a spring at a given distance from equilibrium, demand vs. price, and the approximation of nonlinear functions over a restricted domain.

If we use `dy to stand for the difference in y values between two data points or graph points, and `dx for the x difference, we see that linear functions are characterized by the fact that `dy / `dx = constant.

When a sequence is generated by evaluating a linear function at constant intervals, the difference between any number in the sequence and the next is always the same.

The function a(n+1) = a(n) + c, with a given value of a(1), allows us to build up a sequence of a(n) values as long as we wish.  This sequence will be linear, in that the difference between one number and the next will always be c.

There are 17 exercises on this page.  They should be done as they are encountered, to provide a basic for understanding subsequent material.

The table is shown below, along with columns representing the function 3 f(x) = 3x and f(x+3) = x+3.

Note how the column for 3 f(x) has consecutive values that change by 3, as opposed to the columns for f(x) and f(x+3), which change by 1 between consecutive values. As a result, we conclude that the slope of every graph except 3 f(x) is 1. That is, f(x+3) has the same slope as f(x), while 3 f(x) has at every point 3 times the y coordinate of f(x) and therefore 3 times the slope.

Note also how every column except that for f(x)+3 has y = 0 when x = 0. At every point f(x) + 3 takes values which are 3 units greater than those for f(x). This has the effect of raising every graph point of y = f(x) by 3 units. The graph of y = f(x) + 3 is therefore at every point 3 units higher than that of y = f(x).

So we see that the graph of f(x) +3 has for every x the same slope as the graph of f(x), but lies 3 units higher. This will turn out to be a general principle: the graph of f(x)+c is identical to that of f(x), except that it is shifted in the y direction by c units. We call this a vertical shift.

So we see that the graph of f(x)+3 has the same slope at every x as the graph of f(x), but has been shifted 3 units higher, while the graph of 3 f(x) has been "stretched" away from the y axis by a factor of 3. These will turn out to be general principles for all functions:

The graph of f(x)+c is identical to that of f(x), except that it is shifted in the vertical or y direction by c units.

The graph of A*f(x), where A stands for a real number, is identical to that of f(x), except that it is stretched 3 times as far from the y axis at every point. As a consequence its slope is at every x three times as great as that of f(x).

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1. Make a table and sketch graphs of the following functions, where f(x) is the basic linear function y = f(x) = x:

y = -2 f(x)

y = .3 f(x)

y = f(x) -2

y = f(x) + 4

Label each graph.

2. Sketch graphs of the following, using common sense and patterns you can observe from the preceding exercise:

y = A f(x) for A = -.3 and then for A = 1.3.

y = f(x) + c for c = .3 and then for c = -2.7.

Label each graph

Just as the general quadratic function has a form, specifically y = at^2 + bt + c, which can be made to fit any parabola, the general linear function has a form which can be made to fit any straight line.

If we specify a y-intercept where the graph of a straight-line function crosses the y axis, and also a slope m, there is only one graph that will satisfy these conditions. Similarly any linear graph has one specific y intercept and slope m.

We can also say that any function of the form y = f(x) = mx + b has a graph with y intercept b and slope m.

We know the y intercept must be at y = b, since on the y axis x must be 0, which when substituted gives us y = b (i.e., y = f(0) = m * 0 + b = b).

We know that the slope must be m, because if x1 and x2 are two different values of x, we get corresponding y values y1 = mx1 + b and y2 = m x2 + b. To determine the corresponding slope we find that

rise = change in y value = second value - first value, so

rise = y2 - y1 = (m x2 + b) - (m x1 + b),

which when simplified gives us

rise = m (x2 - x1). <you should do the simplification yourself>

The run is

run = change in x value = second value - first value, so

run = x2 - x1.

The slope is therefore

slope = rise/run = m(x2-x1) / (x2-x1) = m.

The figure below illustrates the expression   m = slope = (y2 - y1) / (x2 - x1).

It logically follows that

(1)  For any straight-line graph there is a function y = f(x) = mx + b whose graph is identical, and

(2) Any function y = f(x) = mx + b has a graph through (0,b) such that the slope between any two of its points is m.

Find the coordinates of the graph points corresponding to x = -3 and to x = 1.

Determine the slope between these points and sketch the right triangle corresponding to your slope calculation, with the legs representing the rise and the run between the points.

Repeat the exercise for points corresponding to x = 1.5 and to x = 3.

4. For the linear function y = f(x) = -1.77 x - 3.87, what aspect of the graph is represented by the number -1.77, and what is the meaning of -3.87 for the graph?

Let x1 and x2 stand for the x coordinates of two points on the graph of the function. Find symbolic expressions, using x1 and x2, for the corresponding y coordinates y1 and y2.

Find a symbolic expression for `dy = difference in y = rise = y2 - y1.

Explain why the corresponding run is x2 - x1.

Write your expression for rise / run and simplify it.

What is the significance of your result?

Use appropriate substitutions and simplifications to obtain, in terms of x2 and x1, the simplest form of the expression

( f(x2) - f(x1) ) / (x2 - x1).

In terms of the graph, what is the meaning of f(x2) - f(x1)?

What is the meaning of (x2 - x1)?

What is the meaning of ( f(x2) - f(x1) ) / (x2 - x1)?

We already pretty much understand what sorts of graphs we can get from y = f(x) = mx + b form various values of the parameters m and b. However if we try to imagine all possible graphs of this form, we end up with an incoherent image of criss-crossing straight lines. It is therefore instructive to investigate in a systematic way the effects these parameters have on the graph.

For example we can agree to let b = 0 while we let m vary over a wide range of positive and negative values. We know that m is the slope, while b = 0 tells us that the graph passes through the y axis at y = 0. Thus the b = 0 family of linear functions consistent of all possible graphs, with all possible slopes, through y = 0.

We can imagine a such graph with slope 0, running along the x axis. We can imagine the graph with slope 1, which is our basic y = x graph, making a 45 degree angle with the x axis. We can imagine every graph between slope 0 and slope 1. Then we can imagine graphs with greater slopes, making smaller and smaller angles with the y axis. And we can imagine graphs with negative slopes, sloping down and to the right, each one like the reflection across the x axis of a corresponding positive-slope graph.

The graph below shows the b = 0 family for a few values of m.

:

If we let b = 1, we would obtain a similar picture except that the graphs would all pass through the y axis at y = b = 1. For any value of b, we obtain a picture which is identical to this, except that all graphs cross at the y axis point y = b.

As another example we could agree to let m = 1, and allow b to take a wide range of positive and negative values. All these graphs would have slope 1, and would differ only in that they cross the y axis at different points, corresponding to the different values of b.

We can imagine the graph with b = 0, which since m = 1 is just the y = x graph again, sloping up and to the right at a 45 degree angle. We can then imagine the graph with b = 1, parallel to the y = x graph but 1 unit higher. We can imagine all the graphs between b = 0 and b = 1, an infinite set of parallel graphs passing through the y axis between y=0 and y= 1. Then we can imagine such a graph passing through every point of the y axis. We end up with an infinite stack of parallel graphs.

The m=1 family is shown below with a few different values of b.

m = 2, b varies from -3 to 3 by step 1.

b = 2, m varies from 1 to 6 by step 1, and from 1/2 to 1/6 by analogous steps.

constant slope m

As shown previously, if x1 and x2 are distinct values of x, we obtain y1 = m x1 + b and y2 = m x2 + b. We know that rise / run = (y2 - y1) / (x2 - x1); for the above expressions this simplifies to m. So for any two points on the graph, the slope is m.

y-intercept b

As stated before, the y intercept occurs when x = 0. In this case we get y = f(0) = m * 0 + b = b.

Straight line, since slope is same between any two points.

Never vertical, since then x2 = x1 and we don't have two distinct points with which to calculate a slope.  Also would imply division by (x2 - x1) = 0 in order to calculate slope.

We have already seen that calculation of a slope involves two points (x1, y1) and (x2,y2) on the graph. Any two points on the graph will give us the same slope. The figure below depicts a variety of triangles on a single straight line, each triangle defined by two points on the line. The triangles are all geometrically similar, which means that the corresponding sides of any two triangles are all in the same ratio.

Pendulum: Force vs. displacement. Required force vs. displacement of pendulum.

A simple pendulum is pretty much just a small dense and symmetrical weight hanging freely by a relatively light string or rope from a fixed point. A pendulum tends to eventually come to rest, and this resting or equilibrium point is always the same as long as the pendulum itself isn't changed.

In order to move the pendulum from this equilibrium position, we have to push or pull it. The further we want to hold it from equilibrium the harder we have to push or pull.

It turns out that the force y required to hold the pendulum at a distance x from equilibrium is a linear function y = f(x) = mx + b, for appropriate values of the parameters m and b. (This is actually an approximation which holds with very good accuracy as long as x is small compared to the length of the string or rope).

If we observe force vs. distance from equilibrium, our data set will consist of points (x,y), with y = force vs. x = distance points. A graph of these points will form a straight line, with small discrepancies due to our inability to measure forces and positions with perfect accuracy.   We can therefore obtain a function

force = m * distance + b

which very accurately models the force vs. distance behavior of the pendulum.

If force is measured in Newtons (a Newton is the weight of a good mouthful of water) and distance in cm, then the slope m represents the number of extra Newtons required to hold the pendulum an additional cm away from its equilibrium position.

Since the force ideally should be zero at the equilibrium position, where distance = 0, the constant b should ideally be zero.  If our model is good, b will be very small.

Flow: Horizontal range of stream vs. time for flow from side of vertical uniform cylinder

If a uniform vertical cylinder with a hole in the side is positioned at some distance above a level horizontal surface such as a floor or level ground, then as the depth of water in the cylinder decreases the water stream flowing from the hole slows down and travels less and less distance before meeting the horizontal surface (e.g., the floor).

The distance from a point on the floor directly below the hole to the point where the water stream meets the floor is called the horizontal range of the water stream.

If we observe y = horizontal range vs. x = clock time we will obtain a data set consisting of points y = horizontal range vs. x = distance. A graph of these data points will form a straight line, with very small discrepancies. These discrepancies are due mostly to our inability to measure the precise point where the stream meets the floor with complete accuracy (we can measure time much more accurately). We can therefore obtain an extremely accurate linear model

horizontal range = m * t + b

of horizontal range vs. clock time t.

If horizontal range is measured in centimeters and clock time in seconds, the slope m of the graph measures the number of centimeters per second at which the point where the stream meets the floor moves. The slope is therefore the velocity of that point.

Since b is obtained by letting clock time t = 0, b will stand for the horizontal range at this time. That is, b is the horizontal range when clock time is 0.

Income: Money earned vs. hours worked

If you are earning money at a fixed hourly rate, then the total amount of money earned since clock time t = 0 is a linear function of time. For example if you earn 6 dollars per hour, and clock time is measured in hours, then since time t = 0, you will have earned 6t dollars.

In general then, your earnings function is therefore

earnings = m t

where m is the number of dollars earned per hour.

We could extend this to a total wealth function. If you start out with a certain total wealth at t = 0 and increase your wealth at a constant rate m (in dollars per hour; e.g., 6 dollars per hour), then at clock time t hours your total wealth will be m t + original wealth.

Thus your wealth function is

wealth = m t + b,

where b is your wealth at time t = 0.

Demand: Demand for a product vs. selling price (simplified economic model)

When deciding on a price for a commodity, it is natural for a manufacturer or seller to assume that the lower the price, the more people will want to buy it and the more units will be sold. The number of units to be sold is called the demand. So demand is a decreasing function of price.

The actual way demand changes with price is complex and somewhat unpredictable. However, in many situations, demand is pretty much a linear function of price. Many basic models then assume that the demand function of price is of the form

demand(price) = m * price + b.

The slope m of the demand vs. price function corresponds to the change in demand per unit of price. We note that m would therefore be negative, since an increase in price is expected to result in a decrease in demand.

Such a model is generally applicable only over a certain price range, consisting of realistic prices. It doesn't make much sense to ask what the demand would be if the price was 0, for example. The demand would go up very rapidly, in a way not typical in the domain of reasonable prices, as the price approached 0. So even though the model predicts a demand of b when price is 0 (since demand(0) = m * 0 + b = b), this prediction isn't realistic. Similarly even though the model predicts some demand even if the price is exorbitant, this is not realistic neither. In a sense the model works only for realistic prices, and we might define the function a little more precisely as

demand(price) = m * price + b, where price is in a realistic range.

Straight-line approximation to any continuously changing quantity over a short time interval

Just as the linear demand vs. price model works well only over a restricted price domain, just about any real-world phenomenon can be modeled reasonably well by a linear function of the form y = f(x) = m x + b, as long as x is restricted to a sufficiently narrow domain of values. Any smoothly varying quantity, no matter how much its actual graph curves, becomes pretty much straight if we look at a short enough segment. By analogy, we know that the Earth is pretty much round, but from where we stand it seems darned near flat. If we get close enough to a point on any smooth curve, it is similarly going to look straight.

If we are interested only in a restricted domain, it is often useful to approximate a function by a linear function.

If we know two points on the graph of a function, or two data points for a real-world phenomenon, we can easily find a corresponding linear function whose graph goes through those points. If those points are part of a domain over which the function or the data is very nearly linear, then the linear function provides a good model over that domain.

7. A simple pendulum with length 8 meters weighs 200 pounds. The following information is obtained from measurements of force and distance:

A force of .21 pounds is required to displace this pendulum 1.1 cm from equilibrium.

A force of .54 pounds is required to displace this pendulum 2.0 cm from equilibrium.

A force of .77 pounds is required to displace this pendulum 2.9 cm from equilibrium.

A force of 1.04 pounds is required to displace this pendulum 4.1 cm from equilibrium.

A force of 1.22 pounds is required to displace this pendulum 5.2 cm from equilibrium.

Make a table for this data, using a force vs. distance format.

Sketch the corresponding data points on the graph below. Use an appropriate scale for each axis.

Sketch the straight line you think best fits this data.

Creating the first model using two points and simultaneous equations:  Choose two appropriate points on the line and use them to obtain two simultaneous linear equations for the parameters of your linear model.  Solve for the parameters and get your linear model.

Creating the second model by measuring the slope and y-intercept:  Determine the slope and y intercept of the function that seems to best fit this data and give the corresponding linear function force(distance) that models the data.

Find the average deviation corresponding to the first model.

Find the square root of the average of the squared deviations for the second model.

Creating a third model using DERIVE or a graphing calculator:  When you have learned to use DERIVE or your graphing calculator:

Obtain the linear regression model for this data and compare it with your other models.

For the linear regression model, calculate the square root of the average of the squared deviations. Compare this result with the result you obtained for the second model. Why should this quantity be less for the linear regression model?

Use whichever model you consider best to answer the following questions:

What force would be required to hold the pendulum 47 centimeters from its equilibrium position?

Why would it not make sense to ask what force would be necessary to hold the pendulum 80 meters from its equilibrium position?

How far could you hold the pendulum from its equilibrium position using a string with a breaking strength of 25 pounds?

What is the average rate of change associated with this model? Explain this average rate in common-sense terms.

What is the average slope associated with this model? Explain this average slope in common-sense terms.

As you gradually pull the pendulum from a point 30 centimeters from its equilibrium position to a point 80 centimeters from its equilibrium position, what average force must you exert?

8. If the flow range from a uniform cylinder changes from 80 cm to 20 cm between clock times t = 34 sec and t = 97 sec, what is the corresponding linear function range(time)?

What is the significance of the average rate of change associated with this model? Explain this average rate in common-sense terms.

What is the average slope associated with this model? Explain this average slope in common-sense terms.

Use the procedures of the preceding exercise to determine and evaluate linear models for this situation, with the following additional information:

A range of 57.3 cm was observed at clock time t = 60 seconds.

A range of 35 cm was observed at clock time t = 80 seconds.

At clock time t = 50 seconds a range of 64 cm was observed.

Answer the following questions based on your model:

How long does it take for the range to change from 70 cm to 60 cm? How long from 30 cm to 20 cm? Why is there or is there not a difference between your answers?

By how much does the range change between clock times t = 30 sec and t = 40 sec? By how much between t = 70 sec and t = 80 sec? Why is there or is there not a difference between your answers?

What is the average rate of change for the time interval 45 sec < t < 55 sec?

9. If your total wealth at clock time t = 0 hours is $3956, and you earn $8/hour for the next 10 hours, then what is your total wealth function totalWealth( t ), where t is time in hours?

If the solution isn't obvious, follow this procedure:

Sketch a graph of total wealth vs. time. You should pick a y scale that will show a significant slope for a 10-hour-long x scale. And your y scale need not start at y = 0, especially since the x intercept isn't important to this problem. Suggestion: let the y scale start at $3950.

Determine the slope and y intercept of your graph.

Use this slope and intercept to write your function.

Use your function model to answer the following questions:

At what clock time will your total wealth reach $4000?

What is the meaning of the slope of your graph?

10. Experience shows that when a certain widget is sold for $30, a certain store can expect to sell 200 widgets per week, while a selling price of $28 increases the number sold to 300.

What linear function numberSold(price) describes this situation?

If the solution isn't obvious, note that you have two data points. Your linear function is numberSold(price) = m ( price ) + b, where m and b are your unknown parameters. Obtain two simultaneous linear equations and evaluate these parameters.

Use the resulting function model to answer the following questions:

If the store must meet a quota by selling 220 units per week, what price should they set?

If each widget costs the store $25, then how much total profit will be expected from selling prices of $28, $29 and $30?

It turns out that the total-profit function is quadratic. If you have learned to use DERIVE, put your three total-profit data points into a data set and perform a quadratic fit (syntax is fit([x, ax^2 + bx + c], &&), where && is the line number of the data set. If you have not yet learned to use DERIVE, defer this part of the problem until you have.

What total profit would be expected from a $28.50 selling price? If you have completed the quadratic model, check to see that it predicts the same total profit as the function numberSold(price).

11. Let the quadratic function depth(t) = .01 t^2 - 2t + 100 represent water depth vs. time for a uniform cylinder.

Find the equation of the straight line connecting the t = 20 point of the graph to the t = 60 point.

This straight line is more or less close to the graph of the depth function. So the linear function is more or less close to the quadratic function. In that sense, the linear function is a linear approximation of the quadratic function.

Based on the t = 0, 20, 40, 60 and 80 values of the depth function and of your linear approximation, what is the approximate average deviation of the straight line from the depth function?

Repeat this exercise for the linear approximation based on the t = 35 and t = 45 points of the depth function. Calculate your average deviation based on the t = 30, 35, 40, 45 and 50 values of each function.

Use the graph below to explain the meaning of your work:

Over which domain of x values is the depth function approximated reasonably well by the linear function?

If we repeated this exercise for the interval from t = 39.5 to t = 40.5, how well do you think the linear function would approximate the quadratic function?

A linear sequence f(n) has constant first difference.

If we have a linear function y = f(x) = mx + b, we say that its basic sequence {f(n), n = 1, 2, 3, ...} consists of the set of numbers {f(1), f(2), f(3), f(4), ... }.

This is really a simple idea. For example if f(x) = 3x + 2, the basic sequence is {f(1), f(2), f(3), ... } = {5, 8, 11, 14, ... }, since f(1) = 3 * 1 + 2 = 5, f(2) = 3 * 2 + 2 = 8, etc..

We see that for this example, the difference between any number and the next is 3.

In general, if f(x) = mx + b, then f(1) = m * 1 + b, f(2) = m * 2 + b, f(3) = m * 3 + b, etc.. Each result is m greater than the one before it (for example m * 3 is greater than m * 2 by m).

So we say that if y = f(x) = mx + b, the basic sequence consists of numbers with a constant difference of m from one number to the next.

a(n+1) = a(n) + c

A constantly changing sequence of numbers is naturally associated with linear behavior. (For example the numbers 2, 5, 8, 11, 14, ..., changing by 3 each time, would naturally be associated with a straight-line graph. The numbers 2, 3, 5, 9, 14, ..., changing by an increasing number every time, would not be associated with linear behavior.)

If we think of the numbers as the values of a function a(n), where for example a(1) = 2, a(2) = 5, a(3) = 8, etc., we can define a rule for finding a(2) from a(1), then using this to find a(3) from a(2), then using this result to find a(4) from a(3), etc.

We see that every number in this example is 3 greater than the one before. So we could write a(2) = a(1) + 3, a(3) = a(2) + 3, a(4) = a(3) + 3, etc.

If we note that the number following a general number n is n + 1, we can write the general pattern as

a(n+1) = a(n) + 3.

If we add to this the information that a(1) = 2, we have a complete rule for our linear function a(n).

In general if we write

a(1) = **, a(n+1) = a(n) + &&,

where ** and && standard for specific unchanging numbers, we have a 'stepping rule' for finding a(n) for any n we wish.

We simply start with a(1) = **; then we add && to get a(2), since our rule tells us that a(2) = a(1) + &&.

Then, since our rule tells us that a(3) = a(2) + &&, we add && to a(2), which we just found.

Then, since our rule tells us that a(4) = a(3) + &&, we add && to a(3), which we just found.

Then, since our rule tells us that a(5) = a(4) + &&, we add && to a(4), which we just found.

Then we continue in the same manner until we find the number we want.

It is of course clear that to get a(n), we must add && to a(1) a total of n-1 times. This gives us a formula for a(n):

a(n) = ** + (n-1) &&.

If we let c stand for && and use a(1) to stand for itself we have

a(n) = a(1) + (n-1) c.

12. Explain in your own words exactly how `dy / `dx represents the average rate of changes between two x values, and why it represents the average slope of the graph of y vs. x between two points.

13. For the quadratic depth function y = depth(t) = .01 t^2 - 2t + 100, determine `dy / `dt based on the two time values t = 30 sec and t = 40 sec. What does this `dy / `dt tell you about the depth vs. time situation?

Determine `dy / `dt based on t = 30 sec and t = 31 sec.

Determine `dy / `dt based on t = 30 sec and t = 30.1 sec.

What do these values of `dy / `dt tell you about the depth vs. time situation?

Use the graph below to illustrate what you have done in this situation, and what it means.

What do you think you would get if you continued this process?

Evaluate the linear function rate(t) = .02 t - 2 for t = 30.

Repeat the above process starting with the time interval t = 10 sec to t = 20 sec.

14. Use the linear function y = f(x) = .37 x + 8.09 to obtain the first five terms of the basic sequence {f(n), n = 1, 2, 3, ...} for this function.

What is the pattern of these numbers?

If you didn't know the equation for the function, how would you go about finding the 100th member of the sequence? How can you tell your method is valid?

Now use the quadratic function y = g(x) = .01 x^2 - 2x + 100 to obtain the first five terms of the basic sequence {g(n), n = 1, 2, 3, ...} for this function.

What is the pattern of these numbers?

If you didn't know the equation for the function, how would you go about finding the next three members of the sequence? How can you verify that your method is valid?

15. The difference equation a(n+1) = a(n) + .4, a(1) = 5 tells you two things. The first is how to find the next number when you have a member of the sequence. The other is the first member. This allows you to determine as many members of the sequence as you like.

If you substitute n = 1 into a(n+1) = a(n) + .4, what equation do you get? How can you use this equation to determine a(2) from the information you know to this point?

If you substitute n = 2 into a(n+1) = a(n) + .4, what equation do you get? How can you use this equation to determine a(3) from the information you know to this point?

If you substitute n = 3 into a(n+1) = a(n) + .4, what equation do you get? What can you determine from this equation and what you know to this point?

What is a(100)? Note: don't go through 97 more substitution steps to get your answer. If you can't see another way to do it, ask for a hint.

16. What difference equation would be associated with the linear function y = f(x) = -.7x + 4?

17. Determine the first five members of the sequence defined by the difference equation

a(n+1) = a(n) + 2 n, a(1) = 4.

What is the pattern of these numbers?