Lesson 4-8

Example 1
Determine the type of polynomial function that could be used to represent the data in each scatter plot.

a.		b.		
	The scatter plot seems to change direction one time, so a quadratic function would best fit the plot.		The scatter plot seems to change direction two times, so a cubic function would best fit the plot.	

Example 2
Use a graphing calculator to write a polynomial function to model the set of data.

x	-3	-2.5	-2	-1	0	1	2	2.5	3	
f(x)	-4	-3	-1	0.5	-1	-0.5	1.5	2	4	

Clear the statistical memory and input the data.	

Using the standard viewing window, graph the	
data. The scatter plot seems to change direction
two times, so a cubic function would best fit the
scatter plot.

Press  and highlight CALC. Since the scatter	
plot seems to be a cubic function, choose
6:CubicReg.

Press  [L1]   [L2] .

Rounding the coefficients to the nearest hundredth,
f(x) = 0.18x3 + 0.03x2 - 0.20x - 0.29.

To check the polynomial function, enter the function	
in the Y= list. Then use the TABLE feature for x values
from -3 with an interval of 0.5. Compare the values of
y with those given in the data. The values are similar,
and the polynomial function seems to be a good fit.

	

Example 3
AMUSEMENT PARK  The annual attendance at a local amusement park for the past ten years is shown in the table below.

Year	1993	1994	1995	1996	1997	1998	1999	2000	
Annual Attendance
(in hundred thousands)	15.7	17.8	18.4	23.1	19.6	18.8	18.3	16.9	

a.	What polynomial function could be used to model these data?

b.	Use the model to predict the annual attendance in 2003.

c.	Use the model to predict when the annual attendance will drop to 200,000.

a.	Let L1 be the number of years since 1993	
and L2 be the annual attendance
(in hundred thousands). Enter this data 
into the calculator.

	Adjust the window to an appropriate	
setting and graph the statistical data.
The data seems to change direction
one time, so a quadratic function will
fit the scatter plot.

	Press , highlight CALC, and	
choose 5:QuadReg.

	Press  [L1]   [L2] 

	Rounding the coefficients to the
nearest hundredth,
f(x) = -0.36x2 + 2.64x + 15.68 models
the data.

b.	Since 2003 is 10 years after 1993, find f(10).

	f(x)	= -0.36x2 + 2.64x + 15.68
	f(10)	= -0.36(102) + 2.64(10) + 15.68
	= 6.08

	According to the model, the annual attendance in 2003 will be about 608,000.

c.	To find when the annual attendance drops to 200,000, solve the equation
2 = -0.36x2 + 2.64x + 15.68.

	2 = -0.36x2 + 2.64x + 15.68
	0 = -0.36x2 + 2.64x + 13.68	Add -2 to each side.

	Use the quadratic formula to solve for x.

	x = 
	x = 
	x = -3.51 or 10.84

	According to the model, an annual attendance of 200,000 occurs about 4 years before 1993 (in 1989) or about 11 years after 1993 (in 2004). Since you want to know when the annual attendance will drop to 200,000, the answer is in about 2004.

	To check the answer, graph the related	
function f(x) = -0.36x2 + 2.64x + 15.68.
Use the CALC menu to determine the
zero at the right. The answer of about
11 years checks.