Polynomial Functions and Models

You will in this section learn how to fit functions that have the familiar shape of a parabola or a cubic to data. Using your calculator to find these equations involves the same procedure as when using it to fit linear, exponential, log, or logistic functions.

FINDING SECOND DIFFERENCES When the input values are evenly spaced, you can

use program DIFF to quickly compute second differences in the output values. If the data are perfectly quadratic (that is, every data point falls on a quadratic function), the second differ- ences in the output values are constant. When the second differences are close to constant, a quadratic function may be appropriate for the data.

We illustrate these ideas with the roofing jobs data given in Table 1.26 of Section 1.5 in

Calculus Concepts. We align the input data so that 1 = January, 2 = February, etc. Clear any

old data and enter these data in lists L1 and L2:

Month of the year 1 2 3 4 5 6

Number of roofing jobs 90 91 101 120 148 185

The input values are evenly spaced, so we can see what information is given by viewing the second differences.

Run program DIFF and observe the first differences in list L3, the second differences in L4, and the percentage differences in list L5.

The second differences are close to constant, so a quadratic function may give a good fit for these data.

Construct a scatter plot of the data. (Don’t forget to clear the Y= list of previously-used equations and to turn on Plot 1.) The shape of the data confirms the numerical investigation result that a quadratic function is appropriate.

FINDING A QUADRATIC FUNCTION TO MODEL DATA Use your TI-83 to find a

quadratic function of the form y = ax2 + bx + c to model the roofing jobs data.

Find the quadratic function and paste the equation into the Y1 location of the Y= list by pressing STAT ► [CALC] 5 [QuadReg] VARS ► [Y−VARS] 1 [Function] 1 [Y1] ENTER .

Press GRAPH to overdraw the graph of the function on the scatter plot. This function gives a very good fit to the data.

How many jobs do we predict the company will have in August? Because August corresponds to x = 8, evaluate Y1 at x = 8. We

predict that there will be 1559 jobs in August.

FINDING A CUBIC FUNCTION TO MODEL DATA Whenever a scatter plot of data

shows a single change in concavity, we are limited to fitting either a cubic or logistic function. If one or two limiting values are apparent, use the logistic equation. Otherwise, a cubic func- tion should be considered. When appropriate, use your calculator to obtain the cubic function of the form y = ax3 + bx2 + cx + d that best fits a set of data.

We illustrate finding a cubic function with the data in Table 1.30 in Example 3 of Section 1.5 in Calculus Concepts. The data give the average price in dollars per 1000 cubic feet of

natural gas for residential use in the U.S. for selected years between 1980 and 2005.

Year 1980 1982 1985 1990 1995 1998 2000 2003 2004 2005

Price

First, clear your lists, and then enter the data. Next, align the input data so that x represents the number of years since 1980.

(We do not have to align here, but we do so in order to have smaller coefficients in the cubic function.)

Clear any functions in the Y= list, and draw a scatter plot of these data.

Notice that a concavity change is evident, but there do not appear to be any limiting values. Thus, a cubic model is appropriate to fit the data.

Find the cubic function and paste the equation in Y1 with STAT ► [CALC] 6 [CubicReg] VARS ► [Y−VARS] 1 [Function] 1 [Y1] ENTER .

Overdraw the graph of the function on the scatter plot with GRAPH or ZOOM 9 [ZoomStat].

What was the price of 1000 cubic feet of natural gas in 1993? Because x represents the number of years after 1980, evaluate Y1 at x = 13 to find that the answer is approximately $5.89.

Part c of Example 3 asks you to find when, according to the model, the price of 1000 cubic

feet of natural gas first exceed $6. You can find the answer by using the SOLVER to find the solution to Y1 – 6 = 0. However, because we already have a graph of this function, the following may be easier. Note that if you do use the SOLVER, you should first TRACE the graph to estimate a guess for x that you need to enter in the SOLVER.

Press Y= and enter 6 in Y2. Press GRAPH . If you wish, reset the window for a better view of the intersections. The graph to the right was drawn by changing the vertical view to Ymin = 4and Ymax = 8.

Note that there are 3 solutions to the equation Y1= 6. We are asked to find the one that occurs first. Press 2nd TRACE [CALC] 5 [intersect], press ENTER ENTER to choose the 2 functions, and use ◄ to move the cursor near the first intersection point.

Press ENTER to find the input at the point of intersection. Because x is the number of years after 1980, the answer to the

question posed in part c of Example 3 is either 1984 or 1985.

The price had not exceeded $6 in 1984, so the answer to the

question is 1985. Carefully read the question you are asked.

Notice that the natural gas prices could have also been modeled with a cubic equation by renumbering the input data so that x is the number of years after 1900.

Return to the data and add 80 to each input value so that L1 represents the number of years after 1900 rather than the number of years after 1980. Return to the home screen. Press

2nd ENTER (ENTRY) to return the instruction “Cubic Y1” to the screen. Press ENTER . Press Y= and have Y2 = 6.

Because the data are realigned, draw a graph of the equation with ZOOM 9 [ZoomStat] rather than GRAPH . Graphically finding where Y1 first exceeds 6 is done exactly as was previously discussed, but you need to carefully interpret the solution.

Because x is now the number of years

after 1900, the answer to the question posed in part c of Example 3 is still