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Section 7.6, Slide *

CHAPTER 7

Algebra: Graphs, Functions, and Linear Systems

Section 7.6, Slide *

7.6

Modeling Data: Exponential, Logarithmic, and Quadratic Functions

Section 7.6, Slide *

Objectives

Graph exponential functions.
Use exponential models.
Graph logarithmic functions.
Use logarithmic models.
Graph quadratic functions.
Use quadratic models.
Determine an appropriate function for modeling data.

Section 7.6, Slide *

Scatter Plots & Regression Lines

Data presented in a visual form as a set of points is called a scatter plot.

A line that best fits the data points in a scatter plot is called a regression line.

For example, the graph displays the relationship between literacy and child mortality. Each point represents a country.

Section 7.6, Slide *

Modeling with Exponential Functions

Section 7.6, Slide *

Example: Graphing an Exponential Function

Graph: f(x) = 2x.

Solution: We start by selecting numbers for x and finding ordered pairs.

We make a table:

x	f(x) = 2x	(x,y)
−3	f(-3) = 2-3 =	(−3, )
−2	f(-2) = 2-2 = ¼	(−2,¼)
−1	f(-1) = 2-1 = ½	(−1,½)
0	f(0) = 20 = 1	(0,1)
1	f(1) = 21 = 2	(1,2)
2	f(2) = 22 = 4	(2,4)
3	f(3) = 23 = 8	(3,8)

Section 7.6, Slide *

Example continued

Next, plot the ordered pairs and connect them with a smooth curve.

Section 7.6, Slide *

Example: Comparing Linear and Exponential Models

The graphs below show the world population for seven selected years from 1950 through 2010. One is a bar graph and the other is a scatter plot.

Section 7.6, Slide *

Example continued

After entering the data in a calculator, the graphing calculator displays the linear model, y = ax + b, and the exponential model, y = abx, that best fit the data.

Express each model in function notation, with numbers rounded to three decimal places.
How well do the functions model the world population in 2000?

Section 7.6, Slide *

Example continued

By one projection, world population is expected to reach 8 billion in the year 2026. Which function serves as a better model for this prediction?

Solution:

a. Using the figure from the graphing calculator, the functions f(x) = 0.074x + 2.294 and g(x) = 2.577(1.017)x

model world population, in billions, x years after 1949. The linear function is f and the exponential function g.

Section 7.6, Slide *

Example continued

The graph shows that the world population in 2000 was 6.1 billion. The year 2000 is 51 years after 1949. Hence, we substitute 51 for x in each function and then compare with the actual population in 2000.

f(x) = 0.074x + 2.294

f(51) = 0.074(51) + 2.294

f(51) ≈ 6.1

g(x) = 2.577(1.017)x

g(51) = 2.577(1.017)51

g(51) ≈ 6.1

Section 7.6, Slide *

Example continued

Since the world population was 6.1 billion, it seems both functions model world population well for 2000.

Now, we compare the models to a world population of 8 billion in the year 2026. We use 77 for x since 2026 is 77 years after 1949.

f(x) = 0.074x + 2.294

f(77) = 0.074(77) + 2.294

f(77) ≈ 8.0

Section 7.6, Slide *

Example continued

g(x) = 2.577(1.017)x

g(77) = 2.577(1.017)77

g(77) ≈ 9.4

It seems the linear function f(x) serves as a better model for a projected world population of 8 billion by 2026.

Section 7.6, Slide *

The Role of e in Applied Exponential Functions

An irrational number, symbolized by the letter e, appears as a base in many exponential functions.

This irrational number e ≈ 2.72 or more accurately

e ≈ 2.71828…

The number e is called the natural base.

The function f(x) = ex is called the natural exponential function.

Section 7.6, Slide *

Example: Alcohol and Risk of a Car Accident

Medical research indicates that the risk of having a car accident increases exponentially as the concentration of alcohol in the blood increases. The risk is modeled by

R = 6e12.77x,

where x is the blood alcohol concentration and R, given as a percent, is the risk of having a car accident. In many states, it is illegal to drive with a blood alcohol concentration at 0.08 or greater. What is the risk of a car accident with a blood alcohol concentration at 0.08?

Section 7.6, Slide *

Example continued

Solution: We substitute 0.08 for x in the function.

R = 6e12.77x

R = 6e12.77(0.08)

Putting this in the calculator, we get an approximation of 16.665813. Rounding to one decimal place, the risk of getting in a car accident is approximately 16.7% with a blood alcohol concentration at 0.08.

Section 7.6, Slide *

Modeling with Logarithmic Functions

Section 7.6, Slide *

Example: Graphing a Logarithmic Function

Graph: y = log2x.

Solution: Because y = log2x means 2y = x, we will use the exponential form of the equation to obtain the function’s graph.

x = 2y	y	(x,y)
2-2 = ¼	−2	(¼,−2)
2-1 = ½	−1	(½,−1)
20 = 1	0	(1,0)
21 = 2	1	(2,1)
22 = 4	2	(4,2)
23 = 8	3	(8,3)

Section 7.6, Slide *

Example: Dangerous Heat: Temperature in an Enclosed Vehicle

When the outside air temperature is anywhere from 72° to 96°F, the temperature in an enclosed vehicle climbs by 43°in the first hour. The bar graph and scatter plot are given below.

Section 7.6, Slide *

Example continued

After entering data in a graphing calculator, the calculator displays the logarithmic model y = a + b ln x, where ln x is called the natural logarithm.

Express the model in function

notation, with numbers rounded to

one decimal place.

Use the function to find the

temperature increase, to the nearest

degree, after 50 minutes. How well does the model resemble the actual increase?

Section 7.6, Slide *

Example continued

Solution:

Using the calculator figure and rounding to one decimal place, the function

f(x) = −11.6 + 13.4 ln x

models the temperature increase after x minutes..

Section 7.6, Slide *

Example continued

Solution:

We substitute 50 for x and evaluate the function.

f(x) = −11.6 + 13.4 ln x

f(50) = −11.6 + 13.4 ln 50

f(50) ≈ 41

Since the actual temperature increases 41° after 50 minutes, the function models the actual increase well.

Section 7.6, Slide *

Modeling with Quadratic Functions

A quadratic function is any function of the form

y = ax2 + bx + c or f(x) = ax2 + bx + c,

where a, b, and c are real numbers, with a ≠ 0.

The graph of any quadratic function is called a parabola.

The vertex of the parabola is the lowest point or the highest point on the graph.

Section 7.6, Slide *

Modeling with Quadratic Functions

Section 7.6, Slide *

Modeling with Quadratic Functions
Vertex of a Parabola

Section 7.6, Slide *

Graphing Quadratic Equations

Section 7.6, Slide *

Example: Graphing a Parabola

Graph the quadratic function: y = x2 – 2x – 3.

Solution: We follow the steps:

Determine how the parabola opens. Since a is the coefficient of x2 and a = 1 in this case, then the parabola opens upward.
Find the vertex.

Section 7.6, Slide *

Example continued

We use the formula to find the x-coordinate:

We plug x = 1 into the original function to find the y-coordinate:

y = x2 – 2x – 3

y = (1)2 – 2(1) – 3

y = −4

The vertex is (1, −4).

Section 7.6, Slide *

Example continued

3. Find the x-intercepts. Replace y with 0 in y = x2 – 2x – 3.

y = x2 – 2x – 3

0 = x2 – 2x – 3

0 = (x – 3)(x + 1)

x – 3 = 0 or x + 1 = 0

x = 3 or x = 1

The x-intercepts are 3 and −1. The parabola passes through (3,0) and (−1,0).

Section 7.6, Slide *

Example continued

4. Find the y-intercept. Replace x with 0 in y = x2 – 2x – 3.

y = x2 – 2x – 3

y = 02 – 2(0) – 3 = −3

The parabola passes through the point (0,−3).

Section 7.6, Slide *

Example continued

Steps 5. and 6. Plot the intercepts and vertex. Connect these points with a smooth curve.

Section 7.6, Slide *

Determine an Appropriate Function for Modeling Data

Description of Data Points in a Scatter Plot	Model
Lie on or near a line	Linear function y = mx + b or f(x) = mx + b
Increasing more and more rapidly	Exponential function y = bx, or f(x) = bx, b > 1
Increasing, although rate of increase is slowing down	Logarithmic function, y = logbx, b > 1 y = logbx means by = x.
Decreasing and then increasing	Quadratic Function y = ax2 + bx + c or f(x) = ax2 + bx + c, a > 0. The vertex is a minimum.
Increasing and then decreasing	Quadratic Function y = ax2 + bx + c or f(x) = ax2 + bx + c, a < 0. The vertex is a maximum.

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