Download section 1.4

CHAPTER 1:
Graphs, Functions,
and Models
1.1
1.2
1.3
1.4
1.5
1.6
Introduction to Graphing
Functions and Graphs
Linear Functions, Slope, and Applications
Equations of Lines and Modeling
Linear Equations, Functions, Zeros and Applications
Solving Linear Inequalities
Copyright © 2009 Pearson Education, Inc.
1.4
Equations of Lines and Modeling




Determine equations of lines.
Given the equations of two lines, determine
whether their graphs are parallel or perpendicular.
Model a set of data with a linear function.
Fit a regression line to a set of data; then use the
linear model to make predictions.
Copyright © 2009 Pearson Education, Inc.
Slope-Intercept Equation
Recall the slope-intercept equation y = mx + b or
f (x) = mx + b.
If we know the slope and the y-intercept of a line,
we can find an equation of the line using the slopeintercept equation.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 4
Example
7
A line has slope  and y-intercept (0, 16). Find an
9
equation of the line.
Solution:
7
We use the slope-intercept equation and substitute 
9
for m and 16 for b:
y  mx  b
7
y   x  16
9
or
7
f x    x  16
9
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 5
Example
2
A line has slope  and contains the point (–3, 6).
3
Find an equation of the line.
Solution:
2
We use the slope-intercept equation and substitute 
3
for m:
2
y   xb
3
Using the point (3, 6), we substitute –3 for x and 6 for
y, then solve for b.
2
6  2b
4b
6   3  b
3
2
2
Equation of the line is y   x  4, or f x    x  4.
3
3
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 6
Point-Slope Equation
The point-slope equation of the line with slope m
passing through (x1, y1) is
y  y1 = m(x  x1).
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 7
Example
Find the equation of the line containing the points
(2, 3) and (1, 4).
Solution: First determine the slope
4  3 7
m

7
1 2
1
Using the point-slope
equation,
substitute 7 for m and
either of the
points for (x1, y1):
Copyright © 2009 Pearson Education, Inc.
y  y1  m(x  x1 )
y  3  7(x  2)
y  3  7x 14
y  7x 11
or f x   7x  11
Slide 1.4 - 8
Parallel Lines
Vertical lines are parallel. Nonvertical lines are
parallel if and only if they have the same slope and
different y-intercepts.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 9
Perpendicular Lines
Two lines with slopes m1 and m2 are perpendicular if
and only if the product of their slopes is 1:
m1m2 = 1.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 10
Perpendicular Lines
Lines are also perpendicular if one is vertical (x = a)
and the other is horizontal (y = b).
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 11
Example
Determine whether each of the following pairs of lines
is parallel, perpendicular, or neither.
a) y + 2 = 5x,
5y + x =  15
Solve each equation for y.
y  5x  2
5y  x  15
1
y x3
5
The slopes are negative reciprocals.
The lines are perpendicular.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 12
Example (continued)
Determine whether each of the following pairs of lines
is parallel, perpendicular, or neither.
b) 2y + 4x = 8,
5 + 2x = –y
Solve each equation for y.
2y  4x  8
y  2x  4
y  2x  5
y  2x  5
The slopes are the same.
The lines are parallel.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 13
Example (continued)
Determine whether each of the following pairs of lines
is parallel, perpendicular, or neither.
c) 2x +1 = y,
y + 3x = 4
Solve each equation for y.
y  2x  1
y  3x  4
The slopes are not the same and their product is not –1.
They are neither parallel nor perpendicular.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 14
Example
Write equations of the lines (a) parallel and
(b) perpendicular to the graph of the line 4y – x = 20
and containing the point (2,  3).
Solution: Solve the equation for y:
4y  x  20
1
y x5
4
1
So the slope of this line is .
4
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 15
Example (continued)
(a) The line parallel to the given line will have the same
1
slope, . We use either the slope-intercept or point4
1
slope equation for the line. Substitute for m and
4
use the point (2,  3) and solve the equation for y.
y  y1  m(x  x1 )
1
y  3  x  2 
4
1
1
y3 x
4
2
1
7
y x
4
2
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 16
Example (continued)
(b) The slope of the perpendicular line is the
1
negative reciprocal of , or – 4. Use the point4
slope equation, substitute – 4 for m and use the
point (2, –3) and solve the equation.
y  y1  m(x  x1 )
y  3  4 x  2 
y  3  4 x  8
y  4x  5
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 17
Mathematical Modeling (optional)
When a real-world problem can be described in a
mathematical language, we have a mathematical
model. The mathematical model gives results that
allow one to predict what will happen in that realworld situation. If the predictions are inaccurate or the
results of experimentation do not conform to the
model, the model must be changed or discarded.
Mathematical modeling can be an ongoing process.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 18
Curve Fitting b (optional)
In general, we try to find a function that fits, as well as
possible, observations (data), theoretical reasoning, and
common sense. We call this curve fitting, it is one
aspect of mathematical modeling.
In this chapter, we will explore linear relationships.
Let’s examine some data and related graphs, or scatter
plots and determine whether a linear function seems to
fit the data.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 19
Example (optional)
Model the data in the table on the number of U.S.
households with cable television with a linear function.
Then predict the number of cable subscribers in 2010.
Scatterplot
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 20
Example (continued) (optional)
Choose any two data points to determine the equation.
We’ll use (2, 81.5) and (6, 94.0). First, find the slope.
94.0  81.5
m
 3.125
Here’s the graph of
62
this line.
Substitute 3.125 for m and
use either point, we’ll use
(6, 94.0) in the point-slope
equation.
y  94.0  3.125 x  6 
y  3.125x  75.25
x is number of years after
1999, y is in millions.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 21
Example (continued) (optional)
Now we can estimate the number of cable subscribers in
2010 by substituting 11 for x in the model
(2010 – 1999 = 11).
y  3.125x  75.25
y  3.125 11  75.25
y  109.6
We predict there will be about 109.6 million U.S.
households with cable television in 2010.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 22
Linear Regression (optional)
Linear regression is a procedure that can be used
to model a set of data using a linear function.
We use the data on the number of U.S. households
with cable television.
We can fit a regression line of the form y = mx + b
to the data using the LINEAR REGRESSION
feature on a graphing calculator.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 23
Example (optional)
Fit a regression line to the data given in the table.
Use the function to predict the number of cable
subscribers in 2010.
Solution:
Enter the data in lists on the
calculator. The independent
variables or x values are
entered into List 1, the
dependent variables or y
values into List 2. The
calculator can then create a
scatterplot.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 24
Example (continued) (optional)
Here are screen shots of the calculator showing
Lists 1 and 2 and the scatterplot of the data.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 25
Example (continued) (optional)
Here are screen shots of selecting the LINEAR
REGRESSION feature from the STAT CALC menu.
From the screen on the right, we find the linear
equation that best models the data is
y  2.863095238x  76.74166667.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 26
Example (continued) (optional)
The calculator can also graph the regression line on
the same graph as the scatterplot.
Substitute 11 into the regression equation.
It predicts 108.2 million cable subscribers in 2010.
Copyright © 2009 Pearson Education, Inc.
Slide 1.4 - 27