Chapter 5 - Linear Functions
Algebra I
Table of Contents
•
•
•
•
•
5.6 - Direct Variation
5.7- Slope – Intercept Form
5.8 – Point – Slope Form
5.9 – Slopes of Parallel and Perpendicular Lines
5.10 – Transforming Linear Functions
5.6 - Direct Variation
Algebra I
5.6
Algebra 1 (bell work)
Just Read Top until Definitions (then copy)
A recipe for paella calls for 1 cup of rice to make 5 servings. In other words, a chef
needs 1 cup of rice for every 5 servings.
The equation y = 5x describes this relationship. In this relationship, the number of
servings varies directly with the number of cups of rice.
A direct variation is a special type of linear relationship that can be
written in the form y = kx,
where k is a nonzero constant called the constant of variation.
5.6
Example 1
Identifying Direct Variations from Equations
Tell whether the equation represents a direct variation. If so, identify the constant of
variation.
y = 3x
This equation
represents a direct
variation because it is in
the form of y = kx. The
constant of variation is
3.
3x + y = 8
–3x
–3x
y = –3x + 8
–4x + 3y = 0
+4x +4x
3y = 4x
This equation is
not a direct
variation because
it cannot be
written in the form
y = kx.
This equation represents a
direct variation because it
is in the form of y = kx.
The constant of variation
is
5.6
What happens if you solve y = kx for k?
y = kx
So, in a direct variation, the ratio
is equal to the constant of variation.
Another way to identify a direct variation is to check whether
is the same for
each ordered pair (except where x = 0).
5.6
Example 2
Identifying Direct Variations from Ordered Pairs
Tell whether the relationship is a direct variation.
Explain.
Method 1 Write an equation.
y = 3x
This is direct variation because it can
be written as y = kx, where k = 3.
Method 2 Find
for each ordered pair.
This is a direct variation because
is the same for each ordered pair.
5.6
Tell whether the relationship is a direct variation.
Explain.
Method 1 Write an equation.
y=x–3
This is not a direct variation because it cannot be written as y = kx.
5.6
Math Joke
• Parent: How is your math homework going?
• Child: Well, m is negative, so at least I’m on
the down slope
5.6
Example 3
Writing and Solving Direct Variation Equations
The value of y varies directly with x, and y = 3, when x = 9. Find y when x = 21.
Method 1 Find the value of k and
then write the equation.
Method 2 Use a proportion.
y = kx
3 = k(9)
9y = 63
y=7
The equation is y =
x = 21, y =
x. When
(21) = 7.
5.6
The value of y varies directly with x, and y = 4.5 when x = 0.5. Find y when x = 10.
Method 1 Find the value of k and
then write the equation.
Method 2 Use a proportion.
y = kx
4.5 = k(0.5)
9=k
The equation is y = 9x.
When x = 10, y = 9(10) = 90.
0.5y = 45
y = 90
HW pg. 339
• 5.6– 2-9 (9 No Graph), 36, 39, 47-55 (Odd)
– Ch: 17-20, 37, 38, 40
5.7 - Slope – Intercept Form
Algebra I
5.7
Example 1
Graph the line given the
slope and y-intercept.
y intercept = 4
Graphing by Using Slope and y-intercept
5.7
Graph the line given the
slope and y-intercept.
slope = 4;
y-intercept =
5.7
Graph the line given the
slope and y-intercept.
slope = 2,
y-intercept = –3
5.7
Graph the line given the
slope and y-intercept.
slope =
y-intercept = 1
5.7
Example 2
Writing Linear Equations in Slope-Intercept Form
Write the equation that describes the line in slope-intercept form.
slope =
; y-intercept = 4
y = mx + b
y= x
+4
slope =
; y-intercept = –6
y = mx + b
5.7
Write the equation that describes the line in slope-intercept form.
slope = –9; y-intercept =
slope = 3; y-intercept =
y = mx + b
y = mx + b
y = –9x +
5.7
Write the equation that describes the line in slope-intercept form.
slope = 2; (3, 4) is on the line
A line has a slope of 8 and (3, –1) is on the line
Step 1 Find the y-intercept.
Step 1 Find the y-intercept.
y = mx + b
y = mx + b
4 = 2(3) + b
–1 = 8(3) + b
4=6+b
–6 –6
–1 = 24 + b
–24 –24
–2 = b
–25 = b
Step 2 Write the equation.
Step 2 Write the equation.
y = mx + b
y = mx + b
y = 2x + (–2)
y = 8x + (–25)
y = 2x – 2
y = 8x – 25
5.7
Math Joke
• Q: Why was the scientist smiling as he
worked?
• A: His graph had a great plot!
5.7
Example 3
Using Slope-Intercept From to Graph
Write the equation in slope -intercept form. Then graph the line described by
the equation.
2y + 3x = 6
Step 1 : Solve for Y=
2y + 3x = 6
–3x –3x
2y = –3x + 6
Day 2
5.7
Write the equation in slope-intercept form. Then graph the line described by the
equation.
5.7
Write the equation in slope-intercept form. Then graph the line described by the
equation.
6x + 2y = 10
Step 1 – Solve for Y =
6x + 2y = 10
–6x –6x
2y = –6x + 10
5.7
Example 4
Application
A closet organizer charges a $100 initial consultation fee plus $30 per hour.
The cost as a function of the number of hours worked is graphed below.
a. Write an equation that represents the cost as a function of the number of hours.
for each hour
Cost
is
$30
y
=
30
•x
An equation is y = 30x + 100.
plus
$100
+
100
5.7
A closet organizer charges a $100 initial consultation fee plus $30 per hour.
The cost as a function of the number of hours worked is graphed below.
An equation is y = 30x + 100.
HW pg.348
• 5.7– Day 1: 5-9, 14-22, 45-50
– Day 2: 10-12, 23-31, 35
– Ch: 36
5.8 – Point-Slope Form
Algebra I
5.8
Algebra 1 (bell work)
5.8
Example 1
Writing Linear Equations in Point-Slope Form
Write an equation in point-slope form for the line with the given slope that contains the
given point.
A.
B.
Then graph A/B on a Graph
C.
5.8
A.
B.
Example 2
Using Point-Slope Form to Graph
5.8
Optional
Write an equation in point-slope form for the line with the given slope that contains the
given point.
a.
b. slope = 0; (3, –4)
y – (–4) = 0(x – 3)
y + 4 = 0(x – 3)
Example 3
Writing Linear Equations in Slope-Intercept Form
Write an equation in slopeWrite an equation in slopeintercept form for the line with
intercept form for the line with
slope 3 that contains (–1, 4).
slope that contains (–3, 1).
Step 1 Write the equation in point-slope form: Step 1 Write the equation in point-slope
form:
y – y1 = m(x – x1)
y – y1 = m(x – x1)
5.8
y – 4 = 3[x – (–1)]
Step 2 Write the equation in slopeintercept form by solving
for y.
Step 2 Write the equation in slopeintercept form by solving for y.
y – 4 = 3(x + 1)
y – 4 = 3x + 3
+4
+4
y = 3x + 7
+1
+1
5.8
Example 3
Writing Linear Equations in Slope-Intercept Form
Write an equation in slope-intercept form for the line through the two points.
(2, –3) and (4, 1)
Step 1 Find the slope.
Step 2 Substitute the slope and one of the points into the point-slope form.
y – y1 = m(x – x1)
y – (–3) = 2(x – 2)
Step 3 Write the equation in slope-intercept form.
y + 3 = 2(x – 2)
y + 3 = 2x – 4
–3
–3
y = 2x – 7
5.8 Write an equation in slope-intercept form for the line through the two points.
(0, 1) and (–2, 9)
Step 1 Find the slope.
Step 2 Substitute the slope and one of the points into the point-slope form.
y – y1 = m(x – x1)
y – 1 = –4(x – 0)
Step 3 Write the equation in slope-intercept form.
y – 1 = –4(x – 0)
y – 1 = –4x
+1
+1
y = –4x + 1
Optional
5.9
Math Joke
• Q: What did the horizontal line say to the
vertical line?
• A: It was nice running into you
5.8
Example 5
Application
The cost to stain a deck is a linear function of the deck’s area.
The cost to stain 100, 250, 400 square feet are shown in the table.
Write an equation in slope-intercept form that represents the function.
Then find the cost to stain a deck whose area is 75 square feet.
Step 1 Choose any two ordered pairs from the table to find the slope.
Skip 15_16
5.8
Step 2 Substitute the slope and any ordered pair from the table into the pointslope form.
y – y1 = m(x – x1)
y – 150 = 1.25(x – 100)
Step 3 Write the equation in slope-intercept form by solving for y.
y – 150 = 1.25(x – 100)
y – 150 = 1.25x – 125
y = 1.25x + 25
5.8
Step 4 Find the cost to stain an area of 75 sq. ft.
y = 1.25x + 25
y = 1.25(75) + 25 = 118.75
The cost of staining 75 sq. ft. is $118.75.
5.8
What if…? At a newspaper the costs to place an ad for
one week are shown.
Write an equation in slope-intercept form that represents
this linear function.
Then find the cost of an ad that is 21 lines long.
Step 1 Choose any two ordered pairs from the table to find the slope.
Step 2 Substitute the slope and any ordered pair from the table into the pointslope form.
y – y1 = m(x – x1)
y – 17.25 = 2.25(x – 5)
5.8
Step 3 Write the equation in slope-intercept form by solving for y.
y – 17.25 = 2.25(x – 5)
y – 17.25 = 2.25x – 11.25
y = 2.25x + 6
Step 4 Find the cost for an ad that is 21 lines long.
y = 2.25x + 6
y = 2.25(21) + 6 = 53.25
The cost of the ad 21 lines long is $53.25.
HW pg. 355
• 5.8– 1-15, 50, 59, 60
– Ch: 34, 35, 36, 40-43, 51, 53
5.9 – Slopes of Parallel and
Perpendicular Lines
Algebra I
5.9
Just Read
To sell at a particular farmers’ market for a year, there is a $100 membership fee.
Then you pay $3 for each hour that you sell at the market.
However, if you were a member the previous year, the membership fee is reduced to
$50.
• The red line shows the total cost if you
are a new member.
• The blue line shows the total cost if you
are a returning member.
5.9
Just Read
These two lines are parallel.
Parallel lines are lines in the same plane that
have no points in common.
In other words, they do not intersect.
5.9
5.9
Identify which lines are parallel.
The lines described by
and
both have slope
.
These lines are parallel. The lines described by y = x and
y = x + 1 both have slope 1. These lines are parallel.
5.9
Identify which lines are parallel.
Write all equations in slope-intercept form to determine the slope.
2x + 3y = 8
–2x
y + 1 = 3(x – 3)
– 2x
y + 1 = 3x – 9
3y = –2x + 8
–1
–1
y = 3x – 10
y = 2x – 3 slope-intercept form
slope-intercept form


5.9
Identify which lines are parallel.
Write all equations in slope-intercept form to determine the slope.
y = 2x – 3
y + 1 = 3(x – 3)
5.9
Identify which lines are parallel.
Write all equations in slope-intercept form to determine the slope.
Slope-intercept form

y = 3x
Slope-intercept form
–3x + 4y = 32
y – 1 = 3(x + 2)
+3x
y – 1 = 3x + 6
+3x
4y = 3x + 32
+1
+1
y = 3x + 7

5.9
–3x + 4y = 32
The lines described by
–3x + 4y = 32 and y =
+ 8 have the same
y = 3x
slope, but they are not parallel lines. They are the
same line.
The lines described by
y = 3x and y – 1 = 3(x + 2) represent parallel lines.
They each have slope 3.
y – 1 = 3(x + 2)
5.9
Math Joke
• Q: What did the lines do at the gym to avoid
each other?
• A: They used the parallel bars.
5.9
Show that JKLM is a parallelogram.
Use the ordered pairs and the slope formula to
find the slopes of MJ and KL.
MJ is parallel to KL because they have the same slope.
JK is parallel to ML because they are both horizontal.
Since opposite sides are parallel, JKLM is a parallelogram.
5.9
Perpendicular lines are lines that intersect to form right angles (90°).
5.9
Identify which lines are perpendicular: y = 3; x = –2; y = 3x;
The graph described by y = 3 is a horizontal line,
and the graph described by x = –2 is a vertical
line.
x = –2
y=3
These lines are perpendicular.
The slope of the line described by y = 3x is 3. The
slope of the line described by
y =3x
is
.
These lines are perpendicular because
the product of their slopes is –1.
.
5.9
Identify which lines are perpendicular: y = –4; y – 6 = 5(x + 4); x = 3; y =
The graph described by x = 3 is a vertical line, and
the graph described by y = –4 is a horizontal line.
x=3
These lines are perpendicular.
The slope of the line described by y – 6 = 5(x + 4) is 5.
The slope of the line described by
y=
is
y = –4
These lines are perpendicular because the
product of their slopes is –1.
y – 6 = 5(x + 4)
5.9
Show that ABC is a right triangle.
If ABC is a right triangle, AB will be perpendicular
to AC.
slope of
slope of
AB is perpendicular to AC because
Therefore, ABC is a right triangle because it contains a right angle.
5.9
Write an equation in slope-intercept form for the line that passes through (4, 10)
and is parallel to the line described by y = 3x + 8.
Step 1 Find the slope of the line.
y = 3x + 8
The parallel line also has a slope of 3.
Step 2 Write the equation in point-slope form.
y – y1 = m(x – x1)
y – 10 = 3(x – 4)
Step 3 Write the equation in slope-intercept form.
y – 10 = 3(x – 4)
y – 10 = 3x – 12)
y = 3x – 2
5.9
Write an equation in slope-intercept form for the line that passes through (2, –1) and is
perpendicular to the line described by y = 2x – 5.
Step 1 Find the slope of the line.
y = 2x – 5
The perpendicular line has a slope of
because
Step 2 Write the equation in point-slope form.
y – y1 = m(x – x1)
Step 3 Write the equation in slope-intercept form.
5.9
Helpful Hint
If you know the slope of a line, the slope of a perpendicular
line will be the "opposite reciprocal.”
5.9
Write an equation in slope-intercept form for the line that passes through (5, 7) and
is parallel to the line described by y =
x – 6.
Step 1 Find the slope of the line.
y=
x –6
The parallel line also has a slope of .
Step 2 Write the equation in point-slope form.
y – y1 = m(x – x1)
Step 3 Write the equation in slope-intercept form.
5.9
Write an equation in slope-intercept form for the line that passes through (–5, 3) and
is perpendicular to the line described by y = 5x.
Step 1 Find the slope of the line.
y = 5x
The perpendicular line has a slope of
because
.
Step 2 Write the equation in point-slope form.
y – y1 = m(x – x1)
Step 3 Write in slope-intercept form.
HW pg. 365
• 5.9– 2-8, 9 – 17 (Odd), 18, 20, 46 – 48, 51,
– B: 26, 30, 40
5.10 - Transforming Linear
Functions
Algebra I
5.10
A family of functions is a set of functions whose graphs have basic characteristics in
common.
For example, all linear functions form a family because all of their graphs are the same
basic shape.
A parent function is the most basic function in a family.
For linear functions, the parent function is f(x) = x.
The graphs of all other linear functions are transformations of the graph of the
parent function, f(x) = x.
A transformation is a change in position or size of a figure.
5.10
The graphs of g(x) = x + 3, h(x) = x – 2, and k(x) = x – 4, are vertical translations of the
graph of the parent function, f(x) = x.
A translation is a type of transformation that moves every point the same distance in
the same direction. You can think of a translation as a “slide.”
5.10
Graph f(x) = 2x and g(x) = 2x – 6. Then describe the transformation from the graph of
f(x) to the graph of g(x).
f(x) = 2x
f(x) = 2x
g(x) = 2x – 6
g(x) = 2x −6
The graph of g(x) = 2x – 6 is the result of translating the graph of f(x) = 2x 6 units down.
5.10
Graph f(x) = x + 4 and g(x) = x – 2. Then describe the transformation from the graph of
f(x) to the graph of g(x).
f(x) = x + 4
f(x) = x + 4
g(x) = x −2
g(x) = x – 2
The graph of g(x) = x – 2 is the result of translating the graph of f(x) = x + 4 6 units down.
5.10
Math Joke
• Q: Why did the student need directions during
math class?
• A: He was lost in translation
5.10
The graphs of g(x) = 3x, h(x) = 5x,
and k(x) =
are rotations of the graph
f(x) = x.
A rotation is a transformation about a
point.
You can think of a rotation as a “turn.” The
y-intercepts are the same, but the slopes
are different.
5.10
Graph f(x) = x and g(x) = 5x.
Then describe the transformation from the graph of f(x) to the graph of g(x).
f(x) = x
g(x) = 5x
f(x) = x
g(x) = 5x
The graph of g(x) = 5x is the result of rotating the graph of f(x) = x about (0, 0). The
graph of g(x) is steeper than the graph of f(x).
5.10
Graph f(x) = 3x – 1 and g(x) =
x – 1. Then describe the transformation from the
graph of f(x) to the graph of g(x).
f(x) = 3x – 1
f(x) = 3x – 1
g(x) = x – 1
g(x) = x – 1
The graph of g(x) is the result of rotating the graph of f(x) about (0, –1). The graph of
g(x) is less steep than the graph of f(x).
5.10
The diagram shows the reflection of the
graph of f(x) = 2x across the y-axis, producing
the graph of g(x) = –2x.
A reflection is a transformation across a line
that produces a mirror image. You can think
of a reflection as a “flip” over a line.
5.10
Graph f(x) = 2x + 2. Then reflect the graph of f(x) across the y-axis. Write a function g(x)
to describe the new graph.
f(x) = 2x + 2
g(x)
f(x)
To find g(x), multiply the value of m by –1.
f(x) = 2x + 2, m = 2.
2(–1) = –2
g(x) = –2x + 2
f(x)
g(x)
5.10
Graph
. Then reflect the graph of f(x) across the y-axis. Write a function
g(x) to describe the new graph.
f(x)
g(x)
f(x)
g(x)
To find g(x), multiply the value of m by –1.
f(x) =
x + 2, m = .
(–1) = –
g(x) = –
x+2
HW pg. 373
• 5.10– 3-19 (Odd), 28, 35 - 37, 45
– B: 38, 48,