Module 2
Linear Functions
What this module is about
This module is about linear function of the form f(x) = mx + b. As you go
over this material, you will develop the skill in determining different aspects of
linear function such as slope, trend, intercepts and some points that belong to the
graph of the linear function. It is also expected that you will develop the skill in
forming linear functions of the form f(x) = mx + b, given certain conditions.
What you are expected to learn
This module is designed for you to:
1. determine the following:
a. slope
b. trend (increasing or decreasing)
c. x- and y-intercepts; and
d. some points given f(x) = mx + b
2. determine f(x) = mx + b given:
a. slope and y-intercept
b. x- and y-intercept
c. slope and a point;
d. any two points
How much do you know
1. What is the slope of the f(x) = 2x – 3?
2. True or false? The trend of a linear function with negative slope is
increasing.
3. Solve for the y-intercept of 3x – 2y = 6.
4. Which of the following step will solve for the x-intercept of the linear
function y = 4x + 16?
2
a. Substitute zero to x and solve for y.
b. Substitute zero to y and solve for x.
c. Take the square root of the constant term.
d. Take the negative reciprocal of the coefficient of x.
5. What is the x-intercept of y = 4x + 16?
6. Name three points on the graph of y – 3x + 2 = 0.
7. Determine the linear function with slope 5 and y-intercept -2.
8. The x- and y-intercepts of a linear function are both 3. Find the linear
function.
9. The graph of a linear function passes through the point (-4, 0), and its
slope is -2. Express the linear function in the form f(x) = mx + b.
10.What linear function has (2, 4) and (-3, 1) as points on its graph?
What you will do
Lesson 1
Determining the Slope, Given f(x) = mx + b
In the linear function f(x) = mx + b, m is the slope and b is the y-intercept.
Examples:
Determine the slope and the y-intercept of each linear function.
1. f(x) = 3x + 8
The slope is 3 and the y-intercept is 8.
2. y = 2x – 5
The slope is 2 and the y-intercept is -5.
3. g(x) = -4x + 7
The slope is -4 and the y-intercept is 7.
4.
2
1
y = x
The slope is
2
1
and the y-intercept is 0.
5. 3x + 2y = 12
First, transform the equation 3x + 2y = 12 in of form f(x) = mx + b
3
3x + 2y = 12
2y = -3x + 12 -3x is added to both sides
2y 3x 12
2 2 2
−
= + Divide both sides by the coefficient of y
3
y x 6
2
= − + The fractions are simplified.
The slope is
3
2
− and the y-intercept is 6.
Try this out
Determine the slope and the y-intercept of each linear function.
Set A
1. f(x) = 4x + 5
2. g(x) = -7x + 1
3. h(x) = 4x
4. F(x) = 3 – 4x
5. 2x + 7y = 14
Set B
1. F(x) = 8x – 9
2. G(x) = -4x – 6
3. H(x) = x
4. 3x + 8y = 24
5. 51x – 17y – 4 = 1
Set C
1.
1
f (x) x
2
= +
2.
9 3
g(x) x
4 2
= −
3. h(x) = -0.3x + 0.1
4. f(x) = -5.2x – 4.4
5. -9x + 8y – 12 = 0
Lesson 2
Determining the Trend, Given f(x) = mx + b
The trend of a linear function is said to be increasing if the slope is
positive.
4
The trend of a linear function is said to be decreasing if the slope is
negative.
Examples:
Determine the trend of each linear function.
1. f(x) = 8x + 3
The slope is 8; hence, the trend of the function is increasing.
2. y = 5x – 2
The slope is 5; hence, the trend of the function is increasing.
3. g(x) = -7x + 4
The slope is -7; hence, the trend of the function is decreasing.
4.
1
y x
2
=
The slope is
1
2
; hence, the trend of the function is decreasing.
5. 4x + 3y = 12
First, transform the equation 4x + 3y = 12 in of form f(x) = mx + b.
4x + 3y = 12
3y = -4x + 12 -4x is added to both sides
3y 4x 12
3 3 3
−
= + Divide both sides by the coefficient of y
4
y x 4
3
= − + The fractions are simplified
The slope is
4
3
− , hence, the trend of the function is decreasing.
Try this out
Determine the trend of each linear function.
Set A
1. f(x) = 5x + 4
2. g(x) = -x + 7
3. h(x) = -4x – 5
4. F(x) = 4 – 3x
5. -5x + 3y + 15 = 0
Set B
1. F(x) = 9x – 8
2. G(x) = -5x + 1
5
3. H(x) = -5 – x
4. 8x + 3y = 24
5. 17x – 51y – 4 = -1
Set C
1. h(x) = -0.1x + 0.3
2. 5x – 6 = y
3. 3x + y = 11
4. -8x + 12y – 9 = 0
5. 25x – 15y – 3 = 2
Lesson 3
Determining the X-and Y-intercepts, Given f(x) = mx + b
To determine the y-intercept of a linear function, substitute x = 0 and solve
for the value of y. This value is the y-intercept.
To determine the x-intercept of a linear function, substitute y = 0 and solve
for the value of y. This value is the x-intercept.
Examples:
Determine the x- and the y-intercepts of each linear function.
1. f(x) = 3x + 8
Let y = f(x) = 3x + 8.
0 = 3x + 8 0 is substituted to y
-8 = 3x -8 is added to both sides
3x = -8 Apply symmetric property of equality
8
x
3
= − Both sides are divided by 3
The x-intercept is
8
3
− .
y = 3(0) + 8 0 is substituted to x
y = 8 The numerical expression is simplified
The y-intercept is 8.
2. 3x + 2y = 12
3x + 2(0) = 12 0 is substituted to y
3x = 12
3x 12
3 3
= Divide both sides by the coefficient of x
x = 4
6
The x-intercept is 4.
3(0) + 2y = 12 0 is substituted to x
2y = 12
2y 12
2 2
= Divide both sides by the coefficient of y
y = 6
The y-intercept is 6.
Try this out
Determine the x- and the y-intercepts of each linear function.
Set A
1. f(x) = 8x + 16
2. g(x) = 4x – 2
3. y = 3 – 4x
4. 2x – 5 = y
5. -7x + 7y = 14
Set B
1. F(x) = -14x + 7
2. 2y = -4x – 6
3. 4x + 6 = 2y
4. 16x – 12 = 4y
5. -12x + 22y – 14 = 0
Set C
1. 6x – 30 = 18y
2. -32x + 8y – 12 = 0
3. 28x – 12y – 3 = 1
4. G(x) = -0.3x + 0.1
5. y = -5.2x – 4.4
Lesson 4
Determining Some Points of a Given Linear Function
To determine a point on the graph of a given linear function y = f(x):
A. Get any element from the domain of the linear function. Recall that the
domain of any linear function is the set of real numbers. This element is
the abscissa of the point.
B. Substitute this abscissa to x in the linear function.
7
C. Solve for the corresponding value of y. This value is the ordinate of the
point.
D. Repeat steps A to C using other element from the domain of the linear
function to determine other points on the graph of the linear function.
Examples:
Determine three points on the graph of each linear function.
1. y = x + 4
If x = 0, then Step A
y = 0 + 4 Step B
y = 4 Step C
The point (0, 4) is on the graph of y = x + 4.
If x = 2, then
y = 2 + 4
y = 6
The point (2, 6) is also on the graph of y = x + 4.
If x = -5, then
y = -5 + 4
y = -1
The point (-5, -1) is also on the graph of y = x + 4.
2. 3x – 4y = 36
If x = 0, then Step A
3(0) – 4y = 36 Step B
0 – 4y = 36 Step C
-4y = 36
4y 36
4 4
−
=
− −
y = -9
The point (0, -9) is on the graph of 3x – 4y = 36.
If x = 2, then
3(2) – 4y = 36
6 – 4y = 36
-4y = 30
8
4y 30
4 4
−
=
− −
15
y
2
= −
The point
15
2,
2
 − 
 
is also on the graph of 3x – 4y = 36.
If x = -4, then
3(4) – 4y = 36
12 – 4y = 36
-4y = 24
4y 24
4 4
−
=
− −
y = -6
The point (-4, -6) is also on the graph of 3x – 4y = 36.
Try this out
Determine three points on the graph of each linear function.
Set A
1. f(x) = x + 6
2. g(x) = -2x + 7
3. h(x) = 2x
4. y = 4 – 3x
5. -2x + y = 4
Set B
1. g(x) = 5x – 4
2. 2y = -2x – 6
3. 4x + 6 = 2y
4. -12x + 22y – 14 = 0
5. 14x – 21y – 4 = 3
Set C
1. y = -7x
2.
x
y 2
4
= −
3. 28x – 12y – 3 = 1
4.
9 3
f(x) x
4 2
= −
5. h(x) = -0.3x + 0.1
9
Lesson 5
Determining f(x) = mx + b, Given the Slope and Y-intercept
To determine f(x) = mx + b, given the slope and y-intercept of the linear
function, substitute the given slope to m and the given y-intercept to b.
Examples:
Determine f(x) = mx + b, given the slope and y-intercept of the linear
function.
1. slope = 2, y-intercept 3
Since the slope is 3, m = 2 and the y-intercept is 3, b = 3. Hence,
f(x) = mx + b
f(x) = 2x + 3
2. slope = -3, y-intercept 1
Since the slope is -3, m = -3 and the y-intercept is 1, b = 1. Hence,
f(x) = mx + b
f(x) = -3x + 1
3. slope = 1, y-intercept
3
2
−
Since the slope is 1, m = 1 and the y-intercept is
3
2
− , b =
3
2
− . Hence,
f(x) = mx + b
3
f(x) (1)x
2
 = + − 
 
3
f(x) x
2
= −
4. slope =
4
3
, y-intercept 0
Since the slope is
4
3
, m =
4
3
and the y-intercept is 0, b = 0. Hence,
f(x) = mx + b
4
f(x) x 0
3
= +
4
f(x) x
3
=
10
Try this out
Determine f(x) = mx + b, given the slope and y-intercept of the linear
function.
Set A
1. slope 3, y-intercept 2
2. slope -2, y-intercept 4
3. slope 5, y-intercept -7
4. slope -8, y-intercept -1
5. slope 6, y-intercept 0
Set B
1. slope 1, y-intercept -1
2. slope -9, y-intercept 0
3. slope
4
3
, y-intercept -2
4. slope
1
3
− , y-intercept
1
3
−
5. slope 7, y-intercept
4
3
−
Set C
1. slope
2
3
, y-intercept
1
4
−
2. slope
2
5
− , y-intercept
1
3
−
3. slope 0.2, y-intercept 1
4. slope 1, y-intercept -0.1
5. slope 2, y-intercept 2
Lesson 6
Determining f(x) = mx + b, Given the X - and Y- intercepts
Let y = f(x). To determine f(x) = mx + b, given the x- and y-intercept of the
linear function, substitute the given x-intercept to a and the given y-intercept to b
in the form of the equation
yx
1
a b
+ =
and then, solve for y in terms of x.
11
Examples:
Determine f(x) = mx + b, given the x- and y-intercept of the linear function.
1. x-intercept 2, y-intercept 3
Since the x-intercept is 2, a = 2 and the y-intercept is 3, b = 3. Hence,
yx
1
a b
+ =
yx
1
2 3
+ =
( )
yx
6 6 1
2 3
 
+ = 
 
Multiply both sides by the LCD, 6
3x + 2y = 6
2y = -3x + 6 -3x is added to both sides
2y 3x 6
2 2 2
−
= + Both sides are divided by 2
3
y 3
2
= − +
3
f (x) 3
2
= − +
2. x-intercept = -3, y-intercept = 1, hence,
yx
1
a b
+ =
yx
1
3 1
+ =
−
( )
yx
3 3 1
3 1
 
− + = − − 
Multiply both sides by the LCD, –3
x – 3y = -3
-3y = -x – 3 x is subtracted from both sides
x 3
y
3 3
−
= −
− −
Both sides are divided by –3
1
y x 1
3
= +
1
f(x) x 1
3
= +
3. x-intercept 1, y-intercept
3
2
−
Since the x-intercept is 1, a = 1 and the y-intercept is
3
2
− , b =
3
2
− . Hence,
12
yx
1
a b
+ =
yx
1
1 3
2
+ =
−
( )
y3 x 3
1
2 1 3 2
2
 
 
− + = − 
− 
 
Multiply both sides by the LCD,
3
2
−
3 3
x y
2 2
− + = −
3 3
y x
2 2
= −
3
x
2
is added to both sides
3 3
f(x) x
2 2
= −
4. x-intercept
4
3
, y-intercept
6
5
−
Since the x-intercept is
4
3
, a =
4
3
and the y-intercept is
6
5
− , b =
6
5
− .
Hence,
yx
1
a b
+ =
yx
1
4 6
3 5
+ =
−
5y3x
1
4 6
− =
( )
5y3x
12 12 1
4 6
 
− = 
 
Multiply both sides by the LCD 12.
9x – 10y = 12
-10y = -9x + 12 9x is subtracted from both sides
9x 12
y
10 10
−
= +
− −
Both sides are divided by –10
9 6
y x
10 5
= −
9 6
f(x) x
10 5
= −
Try this out
Determine f(x) = mx + b, given the slope and y-intercept of the linear
function.
13
Set A
1. x-intercept 3, y-intercept 2
2. x-intercept -2, y-intercept 4
3. x-intercept -1, y-intercept -3
4. x-intercept -8, y-intercept -1
5. x-intercept -4, y-intercept -8
Set B
1. x-intercept 5, y-intercept 6
2. x-intercept -4, y-intercept 12
3. x-intercept -12, y-intercept -16
4. x-intercept -15, y-intercept -18
5. x-intercept
1
3
− , y-intercept
1
7
Set C
1. x-intercept 13, y-intercept
1
5
2. x-intercept
1
9
− , y-intercept 3
3. x-intercept
8
3
− , y-intercept -3
4. x-intercept
4
7
− , y-intercept
1
2
5. x-intercept
2
9
, y-intercept
13
12
Lesson 7
Determining f(x) = mx + b, Given the Slope and One Point
Let y = f(x). To determine f(x) = mx + b, given the slope and one point of
the linear function, substitute the given slope to m and the coordinates of the
given point to x1 and y1 in the form of the equation y – y1 = m(x – x1) and then,
solve for y in terms of x.
Examples:
Determine f(x) = mx + b, given the slope and one point of the linear
function.
1. slope = 2, (1, 3)
Since the slope is 2, m = 2 and the given point is (1, 3), x1 = 1 and y1 = 3.
14
Hence,
y – y1 = m(x – x1)
y – 3 = 2(x – 1)
y – 3 = 2x – 2 Distributive property is applied
y = 2x – 2 + 3 3 is added to both sides
y = 2x + 1
2. slope = -3, (2, -1)
Since the slope is -3, m = -3 and the given point is (2, -1), x1 = 2 and y1 = -1.
Hence,
y – y1 = m(x – x1)
y – (-1) = -3(x – 2)
y + 1 = -3x + 6 Distributive property is applied
y = -3x + 6 – 1 1 is subtracted from both sides
y = -3x + 5
3. slope
2
1
, (-1, -4)
Since the slope is
2
1
, m =
2
1
and the given point is (-1, -4), x1 = -1 and y1 =
-4. Hence,
y – y1 = m(x – x1)
[ ])1(x
2
1
)4(y −−=−−
( )1x
2
1
4y +=+
( ) ( )



+=+ 1x
2
1
24y2 Both sides are multiplied by 2
2y + 8 = x + 1 Distributive property is applied
2y = x + 1 – 8 8 is subtracted from both sides
2y = x – 7
2
7x
2
y2 −
= Both sides are divided by 2
2
7
x
2
1
y −=
4. )3,2(,
3
4
slope −−
15
Since the slope is
3
4
− , m =
3
4
− and the given point is (-2, 3), x1 = -2 and
y1 = 3. Hence,
y – y1 = m(x – x1)
[ ])2(x
3
4
3y −−−=−
)2x(
3
4
3y +−=−
3
8
x
3
4
3y −−=− (Distributive property is applied)
3
3
8
x
3
4
y +−−= (3 is added to both sides)
3
9
3
8
x
3
4
y +−−= (Renaming 3)
3
1
x
3
4
y +−=
Try this out
Determine f(x) = mx + b, given the slope and a point of the linear function.
Set A
1. slope 3, (2, 5)
2. slope 4, (3, -4)
3. slope 2, (-5, 1)
4. slope -2, (-4, -2)
5. slope 5, (-7, 0)
Set B
1. slope 5, (-2, 2)
2. slope -3, (3, 1)
3. slope 1, (6, -1)
4. slope -9, (-5, 0)
5. slope )3,2(,
3
1
−
Set C
1. slope )1,4(,
3
2
−
2. slope 





−
2
1
,4,
2
1
16
3. slope 





−
2
1
,
3
2
,
3
2
4. slope ( )3.0,1,2.0
5. slope ( )2.0,1.0,3.0−
Lesson 8
Determining f(x) = mx + b, Given Any Two Points
Let y = f(x). To determine f(x) = mx + b, given any two points of the linear
function, substitute the coordinates of the first point to x1 and y1, and the
coordinates of the second point to x2 and y2 in the form of the equation
( )1
12
12
1 xx
xx
yy
yy −
−
−
=−
and then, solve for y in terms of x.
Examples:
Determine f(x) = mx + b, given two points of the linear function.
1. (2, 1), (3, 4)
Since the first point is (2, 1), x1 = 2 and y1 = 1 and the second point is (3,
4), x2 = 3 and y2 = 4. Hence,
( )1
12
12
1 xx
xx
yy
yy −
−
−
=−
( )2x
23
14
1y −
−
−
=−
( )2x
1
3
1y −=−
y – 1 = 3(x – 2)
y – 1 = 3x – 6
y = 3x – 6 + 1
y = 2x – 5
2. (-2, 5), (0, -3)
Since the first point is (-2, 5), x1 = -2 and y1 = 5 and the second point is (0,
-3), x2 = 0 and y2 = -3. Hence,
17
( )1
12
12
1 xx
xx
yy
yy −
−
−
=−
( )[ ]2x
)2(0
53
5y −−
−−
−−
=−
( )2x
2
8
5y +
−
=−
y – 5 = -4(x + 2)
y – 5 = -4x – 8
y = -4x – 8 + 5
y = -4x – 3
3. (-2, -7), (-3, -1)
Since the first point is (-2, -7), x1 = -2 and y1 = -7 and the second point is (-
3, -1), x2 = -3 and y2 = -1. Hence,
( )1
12
12
1 xx
xx
yy
yy −
−
−
=−
( )[ ]2x
)2(3
)7(1
)7(y −−
−−−
−−−
=−−
( )2x
1
6
7y +
−
=−
y – 7 = -6(x + 2)
y – 7 = -6x – 12
y = -6x – 12 + 7
y = -6x – 5
Try this out
Determine f(x) = mx + b, given two points of the linear function.
Set A
1. (1, 3), (-2, 6)
2. (3, 0), (0, -2)
3. (-5, 0), (-4, 7)
4. (0, 4), (4, 0)
5. (0, 1), (1, -2)
Set B
1. (1, 4), (5, 6)
2. (8, -2), (6, -4)
3. (9, 0), (5, -2)
4. (5, -6), (6, -5)
5. (9, 3), (5, -6)
18
Set C
1. (-3, -2), (2, 1)
2. (-1, 3), (1, 0)
3. (-2, -3), (3, -1)
4. (-1, 3), (3, -3)
5. (-1, -1), (-3, -3)
Let’s summarize
1. In the linear function f(x) = mx + b, m is the slope and b is the y-intercept.
2. The trend of a linear function is said to be increasing if the slope is
positive. The trend of a linear function is said to be decreasing if the slope
is negative.
3. To determine the y-intercept of a linear function f(x) = y, substitute x = 0
and solve for the value of y. This value is the y-intercept. To determine the
x-intercept of the linear function, substitute y = 0 and solve for the value of
y. This value is the x-intercept.
4. To determine a point on the graph of a given linear function y = f(x):
a. Get any element from the domain of the linear function. Recall that the
domain of any linear function is the set of real numbers. This element
is the abscissa of the point.
b. Substitute this abscissa to x in the linear function.
c. Solve for the corresponding value of y. This value is the ordinate of the
point.
d. Repeat steps a to c using other element from the domain of the linear
function to determine other points on the graph of the linear function.
5. To determine f(x) = mx + b, given the slope and y-intercept of the linear
function, substitute the given slope to m and the given y-intercept to b.
6. To determine f(x) = y = mx + b, given the x- and y-intercept of the linear
function, substitute the given x-intercept to a and the given y-intercept to b
in the form of the equation
1
b
y
a
x
=+
and then, solve for y in terms of x.
7. To determine f(x) = y = mx + b, given the slope and one point of the linear
function, substitute the given slope to m and the coordinates of the given
19
point to x1 and y1 in the form of the equation y – y1 = m(x – x1)and then,
solve for y in terms of x.
8. To determine f(x) = y = mx + b, given any two points of the linear function,
substitute the coordinates of the first point to x1 and y1, and the
coordinates of the second point to x2 and y2 in the form of the equation
( )1
12
12
1 xx
xx
yy
yy −
−
−
=−
and then, solve for y in terms of x.
What have you learned
1. What is the slope of the f(x) = -5x + 4?
2. True or false? The trend of a linear function with positive slope is
increasing.
3. Solve for the y-intercept of 6x – 3y = 4.
4. Which of the following step will solve for the y-intercept of the linear
function y = 3x + 9?
a. Substitute zero to x and solve for y.
b. Substitute zero to y and solve for x.
c. Take the square root of the constant term.
d. Take the negative reciprocal of the coefficient of x.
5. What is the x-intercept of 12 = 9y + 18x?
6. Name three points on the graph of y – 2x + 4 = 0.
7. Determine the linear function with slope -3 and y-intercept -4.
8. The x- and y-intercepts of a linear function are both -2. Find the linear
function.
9. The graph of a linear function passes through the point (-3, 7), and its
slope is -4. Express the linear function in the form f(x) = mx + b.
10.What linear function has (-2, -5) and (-3, 1) as points on its graph?
20
Answer Key
How much do you know
1. 2
2. False
3. -3
4. b
5. -4
6. (-1, -5), (0, -2), (1, 1)
7. y = 5x – 2
8. y = -x + 3
9. y = -2x – 8
10.
5
14
x
5
3
y +=
Try this out
Lesson 1
Set A
1. The slope is 4 and the y-intercept is 5.
2. The slope is -7 and the y-intercept is 1.
3. The slope is 4 and the y-intercept is 0.
4. The slope is -4 and the y-intercept is 3.
5. The slope is
7
2
− and the y-intercept is 2.
Set B
1. The slope is 8 and the y-intercept is -9.
2. The slope is -4 and the y-intercept is -6.
3. The slope is 1 and the y-intercept is 0.
4. The slope is
8
3
− and the y-intercept is 3.
5. The slope is 3 and the y-intercept is
17
5
− .
Set C
1. The slope is 1 and the y-intercept is
2
1
.
2. The slope is
4
9
and the y-intercept is
2
3
− .
3. The slope is -0.3 and the y-intercept is 0.1.
4. The slope is -5.2 and the y-intercept is -4.4.
5. The slope is
8
9
and the y-intercept is
2
3
.
21
Lesson 2
Set A
1. increasing
2. decreasing
3. decreasing
4. decreasing
5. increasing
Set B
1. increasing
2. decreasing
3. decreasing
4. decreasing
5. increasing
Set C
1. decreasing
2. increasing
3. decreasing
4. increasing
5. increasing
Lesson 3
Set A
1. The x-intercept is -2 and the y-intercept is 16.
2. The x-intercept is
2
1
and the y-intercept is -2.
3. The x-intercept is
4
3
and the y-intercept is 3.
4. The x-intercept is
2
5
and the y-intercept is -5.
5. The x-intercept is -2 and the y-intercept is 2.
Set B
1. The x-intercept is
2
1
and the y-intercept is 7.
2. The x-intercept is
2
3
− and the y-intercept is -3.
3. The x-intercept is
2
3
− and the y-intercept is 3.
4. The x-intercept is
4
3
and the y-intercept is -3.
22
5. The x-intercept is 7/6 and the y-intercept is -
11
7
.
Set C
1. The x-intercept is 5 and the y-intercept is
5
3
− .
2. The x-intercept is
8
3
− and the y-intercept is
2
3
.
3. The x-intercept is
7
1
and the y-intercept is
1
3
− .
4. The x-intercept is
3
1
and the y-intercept is 0.1.
5. The x-intercept is
13
11
− and the y-intercept is -4.4.
Lesson 4
Set A (Answers may vary)
1. (-1, 5), (0, 6), (1, 7)
2. (-1, 9), (0, 7), (1, 5)
3. (-1, -2), (0, 0), (1, 2)
4. (-1, 7), (0, 4), (1, 1)
5. (-1, 2), (0, 4), (1, 6)
Set B (Answers may vary)
1. (-1, -9), (0, -4), (1, 2)
2. (-1, -2), (0, -6), (1, -8)
3. (-1, 1), (0, 3), (1, 1)
4. 





−
11
1
,1 , 





11
7
,0 , 





−
11
13
,1
5. ( )1, 1− − ,
1
0,
3
 − 
 
,
1
1,
3
 
 
 
Set C (Answers may vary)
1. (-1, 7), (0, 0), (1, -7)
2. (-4, 3), (0, 2), (4, 1)
3. ( )1, 3− − ,
2
0,
3
 − 
 
,
5
1,
3
 
 
 
4. 





−−
4
15
,1 , 





−
2
3
,0 , 





4
3
,1
5. (-1, 0.4), (0, 0.1), (1, -0.2)
Lesson 5
Set A
1. f(x) = 3x + 2
23
2. f(x) = -2x + 4
3. f(x) = 5x – 7
4. f(x) = -8x – 1
5. f(x) = 6x
Set B
1. f(x) = x – 1
2. f(x) = -9x
3. ( ) 2x
3
4
xf −=
4. ( ) 1 1
f x x
3 3
= − −
5. ( )
3
4
x7xf −=
Set C
1. ( )
4
1
x
3
2
xf −=
2. ( ) 2 1
f x x
5 3
= − −
3. f(x) = 0.2x + 1
4. f(x) = x – 0.1
5. ( )f x 2x 2= +
Lesson 6
Set A
1. ( ) 2x
3
2
xf +−=
2. ( ) 4x2xf +=
3. ( ) 3x3xf −=
4. ( ) 1x
8
1
xf −−=
5. ( ) 8x2xf −−=
Set B
1. ( ) 6x
5
6
xf +−=
2. ( ) 12x3xf +=
3. ( ) 16x
3
4
xf −−=
4. ( ) 18x
5
6
xf −−=
5. ( )
7
1
x
7
3
xf +=
24
Set C
1. ( )
5
1
x
65
1
xf +−=
2. ( ) 3x27xf +=
3. ( ) 3x
8
9
xf +−=
4. ( )
2
1
x
8
7
xf +=
5. ( )
12
13
x
8
39
xf +=
Lesson 7
Set A
1. ( ) 1x3xf −=
2. ( ) 16x4xf −=
3. ( ) 11x2xf +=
4. ( ) 10x2xf −−=
5. ( ) 35x5xf +=
Set B
1. ( ) 12x5xf +=
2. ( ) 10x3xf +−=
3. ( ) 7xxf −=
4. ( ) 45x9xf −−=
5. ( )
3
11
x
3
1
xf −=
Set C
1. ( )
3
11
x
3
2
xf −=
2. ( )
2
5
x
2
1
xf +−=
3. ( )
18
17
x
3
2
xf +−=
4. ( ) 1.0x2.0xf +=
5. ( ) 23.0x3.0xf +−=
Lesson 8
1. y = -x + 4
2. y =
2
3
x - 2
25
3. y = 7x + 35
4. y = -x + 4
5. y = -3x + 1
Set B
1. y = x + 7
2 2
2. y = x - 10
3. y = x – 9
2 2
4. y = x - 11
5. y = 9x - 69
4 4
Set C
1. y = 3x – 1
5 5
2. y = -
2
3
x +
2
3
3. y = 2x – 11
5 5
4. y = -
2
3
x +
2
3
5. y = x
What have you learned
1. -5
2. true
3.
3
4
−
4. a
5.
2
3
6. (0, -4), (1, -2), (2, 0)
7. y = -3x - 4
8. y = -x - 2
9. y = -4x - 5
10.y = -6x - 17

More Related Content

DOCX
co (addition of polynomials).docx
PDF
Math10 q2 mod3of8_theorems on chords, arcs, central angles and inscribed angl...
PPTX
Law of sine and cosines
PDF
Module 7 geometry of shape and size
PPTX
Graph of functions
PPTX
5.2 Solving Quadratic Equations by Factoring
PPT
Rational Root Theorem.ppt
PPTX
Lesson 1.9 the set of rational numbers
co (addition of polynomials).docx
Math10 q2 mod3of8_theorems on chords, arcs, central angles and inscribed angl...
Law of sine and cosines
Module 7 geometry of shape and size
Graph of functions
5.2 Solving Quadratic Equations by Factoring
Rational Root Theorem.ppt
Lesson 1.9 the set of rational numbers

What's hot (20)

PPTX
GRADE 7 INTERVENTION CAMP LESSON 1..pptx
PPT
Distance between two points
PPTX
Distance formula
PPTX
ppt for Properties of the Operations on Integers
PDF
Module 2 quadratic functions
PDF
Factoring Difference of Two Squares
PPT
Properties of operations
PPTX
Addition and subtraction of polynomial functions
PPTX
Graph of linear equations
PDF
3.2 factoring polynomials
PPTX
Math12 lesson9
PPT
Completing the square
PPTX
Cube of binomial
PPTX
Variation
PPTX
Circular permutation
PDF
Math 8 Curriculum Guide rev.2016
PPT
Equation and inequalities
PPT
Hyperbola
PPTX
Repeating decimal to fraction 1
PPT
Percent Increase And Decrease
GRADE 7 INTERVENTION CAMP LESSON 1..pptx
Distance between two points
Distance formula
ppt for Properties of the Operations on Integers
Module 2 quadratic functions
Factoring Difference of Two Squares
Properties of operations
Addition and subtraction of polynomial functions
Graph of linear equations
3.2 factoring polynomials
Math12 lesson9
Completing the square
Cube of binomial
Variation
Circular permutation
Math 8 Curriculum Guide rev.2016
Equation and inequalities
Hyperbola
Repeating decimal to fraction 1
Percent Increase And Decrease
Ad

Similar to Module 2 linear functions (20)

PDF
Module 1 linear functions
DOC
Mathematics 8 Linear Functions
DOCX
Sim (Linear Function)
PDF
Form 4 add maths note
PDF
Module 3 exponential and logarithmic functions
PPT
Grph quad fncts
PDF
Module 1 quadratic functions
PDF
Module 3 quadratic functions
DOCX
Function
PDF
Form 4 Add Maths Note
PDF
Form 4-add-maths-note
PPT
Solution 3
DOC
Mathematics 9 Quadratic Functions (Module 1)
PPT
Solution 3
PPTX
Linear functons
PPTX
Straight-Line-Graphs-Final -2.pptx
PDF
Quadratic Function Presentation
PPT
Ml lesson 4 7
PDF
Module 2 exponential functions
PPTX
20200830230859_PPT4-Lines, Parabolas and Systems.pptx
Module 1 linear functions
Mathematics 8 Linear Functions
Sim (Linear Function)
Form 4 add maths note
Module 3 exponential and logarithmic functions
Grph quad fncts
Module 1 quadratic functions
Module 3 quadratic functions
Function
Form 4 Add Maths Note
Form 4-add-maths-note
Solution 3
Mathematics 9 Quadratic Functions (Module 1)
Solution 3
Linear functons
Straight-Line-Graphs-Final -2.pptx
Quadratic Function Presentation
Ml lesson 4 7
Module 2 exponential functions
20200830230859_PPT4-Lines, Parabolas and Systems.pptx
Ad

More from dionesioable (20)

PDF
Squad drill
PDF
PDF
Modyul 01 hegrapiya ng daigdig
PPSX
Innovation presentation
PPTX
Results based performance management system (rpms) for dep ed
PDF
Unit 1, mod 3 Sulyap ng Buhay Panlipunan sa Sinaunang Panahon
PDF
Unit 1, mod 4 Pagtatag ng kolonyang Espanyol at mga patakarang kolonyal
PDF
Unit 1, mod 2 Ang bangang Manunggul at mga sinaunang paniniwala
PDF
1 1a modyul final ok
PDF
1 1c modyul final ok
PDF
1 1b modyul final ok
PPTX
Deped Sch calendar 2014 -15
PDF
Biology m13 human reproductive system
PDF
Biology m8 integumentary & excretory systems
PDF
Biology m6 the levels of biological organization
PDF
Biology m3 movement of matls thru the cell membrane
PDF
Biology m1 nature of biology
PDF
Biology m18 animals with backbones
PDF
Biology m16 diversity of plants
PDF
Biology m1 nature of biology
Squad drill
Modyul 01 hegrapiya ng daigdig
Innovation presentation
Results based performance management system (rpms) for dep ed
Unit 1, mod 3 Sulyap ng Buhay Panlipunan sa Sinaunang Panahon
Unit 1, mod 4 Pagtatag ng kolonyang Espanyol at mga patakarang kolonyal
Unit 1, mod 2 Ang bangang Manunggul at mga sinaunang paniniwala
1 1a modyul final ok
1 1c modyul final ok
1 1b modyul final ok
Deped Sch calendar 2014 -15
Biology m13 human reproductive system
Biology m8 integumentary & excretory systems
Biology m6 the levels of biological organization
Biology m3 movement of matls thru the cell membrane
Biology m1 nature of biology
Biology m18 animals with backbones
Biology m16 diversity of plants
Biology m1 nature of biology

Recently uploaded (20)

PPTX
Diploma pharmaceutics notes..helps diploma students
PDF
African Communication Research: A review
PDF
Fun with Grammar (Communicative Activities for the Azar Grammar Series)
PDF
anganwadi services for the b.sc nursing and GNM
PDF
M.Tech in Aerospace Engineering | BIT Mesra
PDF
Hospital Case Study .architecture design
PDF
Disorder of Endocrine system (1).pdfyyhyyyy
PDF
MICROENCAPSULATION_NDDS_BPHARMACY__SEM VII_PCI Syllabus.pdf
PPTX
Reproductive system-Human anatomy and physiology
PPTX
BSCE 2 NIGHT (CHAPTER 2) just cases.pptx
PDF
Chevening Scholarship Application and Interview Preparation Guide
PPTX
pharmaceutics-1unit-1-221214121936-550b56aa.pptx
PDF
Lecture on Viruses: Structure, Classification, Replication, Effects on Cells,...
PDF
PUBH1000 - Module 6: Global Health Tute Slides
PDF
Everyday Spelling and Grammar by Kathi Wyldeck
PPTX
PLASMA AND ITS CONSTITUENTS 123.pptx
PDF
Skin Care and Cosmetic Ingredients Dictionary ( PDFDrive ).pdf
PDF
Myanmar Dental Journal, The Journal of the Myanmar Dental Association (2013).pdf
PDF
LIFE & LIVING TRILOGY - PART (3) REALITY & MYSTERY.pdf
PDF
Farming Based Livelihood Systems English Notes
Diploma pharmaceutics notes..helps diploma students
African Communication Research: A review
Fun with Grammar (Communicative Activities for the Azar Grammar Series)
anganwadi services for the b.sc nursing and GNM
M.Tech in Aerospace Engineering | BIT Mesra
Hospital Case Study .architecture design
Disorder of Endocrine system (1).pdfyyhyyyy
MICROENCAPSULATION_NDDS_BPHARMACY__SEM VII_PCI Syllabus.pdf
Reproductive system-Human anatomy and physiology
BSCE 2 NIGHT (CHAPTER 2) just cases.pptx
Chevening Scholarship Application and Interview Preparation Guide
pharmaceutics-1unit-1-221214121936-550b56aa.pptx
Lecture on Viruses: Structure, Classification, Replication, Effects on Cells,...
PUBH1000 - Module 6: Global Health Tute Slides
Everyday Spelling and Grammar by Kathi Wyldeck
PLASMA AND ITS CONSTITUENTS 123.pptx
Skin Care and Cosmetic Ingredients Dictionary ( PDFDrive ).pdf
Myanmar Dental Journal, The Journal of the Myanmar Dental Association (2013).pdf
LIFE & LIVING TRILOGY - PART (3) REALITY & MYSTERY.pdf
Farming Based Livelihood Systems English Notes

Module 2 linear functions

  • 1. Module 2 Linear Functions What this module is about This module is about linear function of the form f(x) = mx + b. As you go over this material, you will develop the skill in determining different aspects of linear function such as slope, trend, intercepts and some points that belong to the graph of the linear function. It is also expected that you will develop the skill in forming linear functions of the form f(x) = mx + b, given certain conditions. What you are expected to learn This module is designed for you to: 1. determine the following: a. slope b. trend (increasing or decreasing) c. x- and y-intercepts; and d. some points given f(x) = mx + b 2. determine f(x) = mx + b given: a. slope and y-intercept b. x- and y-intercept c. slope and a point; d. any two points How much do you know 1. What is the slope of the f(x) = 2x – 3? 2. True or false? The trend of a linear function with negative slope is increasing. 3. Solve for the y-intercept of 3x – 2y = 6. 4. Which of the following step will solve for the x-intercept of the linear function y = 4x + 16?
  • 2. 2 a. Substitute zero to x and solve for y. b. Substitute zero to y and solve for x. c. Take the square root of the constant term. d. Take the negative reciprocal of the coefficient of x. 5. What is the x-intercept of y = 4x + 16? 6. Name three points on the graph of y – 3x + 2 = 0. 7. Determine the linear function with slope 5 and y-intercept -2. 8. The x- and y-intercepts of a linear function are both 3. Find the linear function. 9. The graph of a linear function passes through the point (-4, 0), and its slope is -2. Express the linear function in the form f(x) = mx + b. 10.What linear function has (2, 4) and (-3, 1) as points on its graph? What you will do Lesson 1 Determining the Slope, Given f(x) = mx + b In the linear function f(x) = mx + b, m is the slope and b is the y-intercept. Examples: Determine the slope and the y-intercept of each linear function. 1. f(x) = 3x + 8 The slope is 3 and the y-intercept is 8. 2. y = 2x – 5 The slope is 2 and the y-intercept is -5. 3. g(x) = -4x + 7 The slope is -4 and the y-intercept is 7. 4. 2 1 y = x The slope is 2 1 and the y-intercept is 0. 5. 3x + 2y = 12 First, transform the equation 3x + 2y = 12 in of form f(x) = mx + b
  • 3. 3 3x + 2y = 12 2y = -3x + 12 -3x is added to both sides 2y 3x 12 2 2 2 − = + Divide both sides by the coefficient of y 3 y x 6 2 = − + The fractions are simplified. The slope is 3 2 − and the y-intercept is 6. Try this out Determine the slope and the y-intercept of each linear function. Set A 1. f(x) = 4x + 5 2. g(x) = -7x + 1 3. h(x) = 4x 4. F(x) = 3 – 4x 5. 2x + 7y = 14 Set B 1. F(x) = 8x – 9 2. G(x) = -4x – 6 3. H(x) = x 4. 3x + 8y = 24 5. 51x – 17y – 4 = 1 Set C 1. 1 f (x) x 2 = + 2. 9 3 g(x) x 4 2 = − 3. h(x) = -0.3x + 0.1 4. f(x) = -5.2x – 4.4 5. -9x + 8y – 12 = 0 Lesson 2 Determining the Trend, Given f(x) = mx + b The trend of a linear function is said to be increasing if the slope is positive.
  • 4. 4 The trend of a linear function is said to be decreasing if the slope is negative. Examples: Determine the trend of each linear function. 1. f(x) = 8x + 3 The slope is 8; hence, the trend of the function is increasing. 2. y = 5x – 2 The slope is 5; hence, the trend of the function is increasing. 3. g(x) = -7x + 4 The slope is -7; hence, the trend of the function is decreasing. 4. 1 y x 2 = The slope is 1 2 ; hence, the trend of the function is decreasing. 5. 4x + 3y = 12 First, transform the equation 4x + 3y = 12 in of form f(x) = mx + b. 4x + 3y = 12 3y = -4x + 12 -4x is added to both sides 3y 4x 12 3 3 3 − = + Divide both sides by the coefficient of y 4 y x 4 3 = − + The fractions are simplified The slope is 4 3 − , hence, the trend of the function is decreasing. Try this out Determine the trend of each linear function. Set A 1. f(x) = 5x + 4 2. g(x) = -x + 7 3. h(x) = -4x – 5 4. F(x) = 4 – 3x 5. -5x + 3y + 15 = 0 Set B 1. F(x) = 9x – 8 2. G(x) = -5x + 1
  • 5. 5 3. H(x) = -5 – x 4. 8x + 3y = 24 5. 17x – 51y – 4 = -1 Set C 1. h(x) = -0.1x + 0.3 2. 5x – 6 = y 3. 3x + y = 11 4. -8x + 12y – 9 = 0 5. 25x – 15y – 3 = 2 Lesson 3 Determining the X-and Y-intercepts, Given f(x) = mx + b To determine the y-intercept of a linear function, substitute x = 0 and solve for the value of y. This value is the y-intercept. To determine the x-intercept of a linear function, substitute y = 0 and solve for the value of y. This value is the x-intercept. Examples: Determine the x- and the y-intercepts of each linear function. 1. f(x) = 3x + 8 Let y = f(x) = 3x + 8. 0 = 3x + 8 0 is substituted to y -8 = 3x -8 is added to both sides 3x = -8 Apply symmetric property of equality 8 x 3 = − Both sides are divided by 3 The x-intercept is 8 3 − . y = 3(0) + 8 0 is substituted to x y = 8 The numerical expression is simplified The y-intercept is 8. 2. 3x + 2y = 12 3x + 2(0) = 12 0 is substituted to y 3x = 12 3x 12 3 3 = Divide both sides by the coefficient of x x = 4
  • 6. 6 The x-intercept is 4. 3(0) + 2y = 12 0 is substituted to x 2y = 12 2y 12 2 2 = Divide both sides by the coefficient of y y = 6 The y-intercept is 6. Try this out Determine the x- and the y-intercepts of each linear function. Set A 1. f(x) = 8x + 16 2. g(x) = 4x – 2 3. y = 3 – 4x 4. 2x – 5 = y 5. -7x + 7y = 14 Set B 1. F(x) = -14x + 7 2. 2y = -4x – 6 3. 4x + 6 = 2y 4. 16x – 12 = 4y 5. -12x + 22y – 14 = 0 Set C 1. 6x – 30 = 18y 2. -32x + 8y – 12 = 0 3. 28x – 12y – 3 = 1 4. G(x) = -0.3x + 0.1 5. y = -5.2x – 4.4 Lesson 4 Determining Some Points of a Given Linear Function To determine a point on the graph of a given linear function y = f(x): A. Get any element from the domain of the linear function. Recall that the domain of any linear function is the set of real numbers. This element is the abscissa of the point. B. Substitute this abscissa to x in the linear function.
  • 7. 7 C. Solve for the corresponding value of y. This value is the ordinate of the point. D. Repeat steps A to C using other element from the domain of the linear function to determine other points on the graph of the linear function. Examples: Determine three points on the graph of each linear function. 1. y = x + 4 If x = 0, then Step A y = 0 + 4 Step B y = 4 Step C The point (0, 4) is on the graph of y = x + 4. If x = 2, then y = 2 + 4 y = 6 The point (2, 6) is also on the graph of y = x + 4. If x = -5, then y = -5 + 4 y = -1 The point (-5, -1) is also on the graph of y = x + 4. 2. 3x – 4y = 36 If x = 0, then Step A 3(0) – 4y = 36 Step B 0 – 4y = 36 Step C -4y = 36 4y 36 4 4 − = − − y = -9 The point (0, -9) is on the graph of 3x – 4y = 36. If x = 2, then 3(2) – 4y = 36 6 – 4y = 36 -4y = 30
  • 8. 8 4y 30 4 4 − = − − 15 y 2 = − The point 15 2, 2  −    is also on the graph of 3x – 4y = 36. If x = -4, then 3(4) – 4y = 36 12 – 4y = 36 -4y = 24 4y 24 4 4 − = − − y = -6 The point (-4, -6) is also on the graph of 3x – 4y = 36. Try this out Determine three points on the graph of each linear function. Set A 1. f(x) = x + 6 2. g(x) = -2x + 7 3. h(x) = 2x 4. y = 4 – 3x 5. -2x + y = 4 Set B 1. g(x) = 5x – 4 2. 2y = -2x – 6 3. 4x + 6 = 2y 4. -12x + 22y – 14 = 0 5. 14x – 21y – 4 = 3 Set C 1. y = -7x 2. x y 2 4 = − 3. 28x – 12y – 3 = 1 4. 9 3 f(x) x 4 2 = − 5. h(x) = -0.3x + 0.1
  • 9. 9 Lesson 5 Determining f(x) = mx + b, Given the Slope and Y-intercept To determine f(x) = mx + b, given the slope and y-intercept of the linear function, substitute the given slope to m and the given y-intercept to b. Examples: Determine f(x) = mx + b, given the slope and y-intercept of the linear function. 1. slope = 2, y-intercept 3 Since the slope is 3, m = 2 and the y-intercept is 3, b = 3. Hence, f(x) = mx + b f(x) = 2x + 3 2. slope = -3, y-intercept 1 Since the slope is -3, m = -3 and the y-intercept is 1, b = 1. Hence, f(x) = mx + b f(x) = -3x + 1 3. slope = 1, y-intercept 3 2 − Since the slope is 1, m = 1 and the y-intercept is 3 2 − , b = 3 2 − . Hence, f(x) = mx + b 3 f(x) (1)x 2  = + −    3 f(x) x 2 = − 4. slope = 4 3 , y-intercept 0 Since the slope is 4 3 , m = 4 3 and the y-intercept is 0, b = 0. Hence, f(x) = mx + b 4 f(x) x 0 3 = + 4 f(x) x 3 =
  • 10. 10 Try this out Determine f(x) = mx + b, given the slope and y-intercept of the linear function. Set A 1. slope 3, y-intercept 2 2. slope -2, y-intercept 4 3. slope 5, y-intercept -7 4. slope -8, y-intercept -1 5. slope 6, y-intercept 0 Set B 1. slope 1, y-intercept -1 2. slope -9, y-intercept 0 3. slope 4 3 , y-intercept -2 4. slope 1 3 − , y-intercept 1 3 − 5. slope 7, y-intercept 4 3 − Set C 1. slope 2 3 , y-intercept 1 4 − 2. slope 2 5 − , y-intercept 1 3 − 3. slope 0.2, y-intercept 1 4. slope 1, y-intercept -0.1 5. slope 2, y-intercept 2 Lesson 6 Determining f(x) = mx + b, Given the X - and Y- intercepts Let y = f(x). To determine f(x) = mx + b, given the x- and y-intercept of the linear function, substitute the given x-intercept to a and the given y-intercept to b in the form of the equation yx 1 a b + = and then, solve for y in terms of x.
  • 11. 11 Examples: Determine f(x) = mx + b, given the x- and y-intercept of the linear function. 1. x-intercept 2, y-intercept 3 Since the x-intercept is 2, a = 2 and the y-intercept is 3, b = 3. Hence, yx 1 a b + = yx 1 2 3 + = ( ) yx 6 6 1 2 3   + =    Multiply both sides by the LCD, 6 3x + 2y = 6 2y = -3x + 6 -3x is added to both sides 2y 3x 6 2 2 2 − = + Both sides are divided by 2 3 y 3 2 = − + 3 f (x) 3 2 = − + 2. x-intercept = -3, y-intercept = 1, hence, yx 1 a b + = yx 1 3 1 + = − ( ) yx 3 3 1 3 1   − + = − −  Multiply both sides by the LCD, –3 x – 3y = -3 -3y = -x – 3 x is subtracted from both sides x 3 y 3 3 − = − − − Both sides are divided by –3 1 y x 1 3 = + 1 f(x) x 1 3 = + 3. x-intercept 1, y-intercept 3 2 − Since the x-intercept is 1, a = 1 and the y-intercept is 3 2 − , b = 3 2 − . Hence,
  • 12. 12 yx 1 a b + = yx 1 1 3 2 + = − ( ) y3 x 3 1 2 1 3 2 2     − + = −  −    Multiply both sides by the LCD, 3 2 − 3 3 x y 2 2 − + = − 3 3 y x 2 2 = − 3 x 2 is added to both sides 3 3 f(x) x 2 2 = − 4. x-intercept 4 3 , y-intercept 6 5 − Since the x-intercept is 4 3 , a = 4 3 and the y-intercept is 6 5 − , b = 6 5 − . Hence, yx 1 a b + = yx 1 4 6 3 5 + = − 5y3x 1 4 6 − = ( ) 5y3x 12 12 1 4 6   − =    Multiply both sides by the LCD 12. 9x – 10y = 12 -10y = -9x + 12 9x is subtracted from both sides 9x 12 y 10 10 − = + − − Both sides are divided by –10 9 6 y x 10 5 = − 9 6 f(x) x 10 5 = − Try this out Determine f(x) = mx + b, given the slope and y-intercept of the linear function.
  • 13. 13 Set A 1. x-intercept 3, y-intercept 2 2. x-intercept -2, y-intercept 4 3. x-intercept -1, y-intercept -3 4. x-intercept -8, y-intercept -1 5. x-intercept -4, y-intercept -8 Set B 1. x-intercept 5, y-intercept 6 2. x-intercept -4, y-intercept 12 3. x-intercept -12, y-intercept -16 4. x-intercept -15, y-intercept -18 5. x-intercept 1 3 − , y-intercept 1 7 Set C 1. x-intercept 13, y-intercept 1 5 2. x-intercept 1 9 − , y-intercept 3 3. x-intercept 8 3 − , y-intercept -3 4. x-intercept 4 7 − , y-intercept 1 2 5. x-intercept 2 9 , y-intercept 13 12 Lesson 7 Determining f(x) = mx + b, Given the Slope and One Point Let y = f(x). To determine f(x) = mx + b, given the slope and one point of the linear function, substitute the given slope to m and the coordinates of the given point to x1 and y1 in the form of the equation y – y1 = m(x – x1) and then, solve for y in terms of x. Examples: Determine f(x) = mx + b, given the slope and one point of the linear function. 1. slope = 2, (1, 3) Since the slope is 2, m = 2 and the given point is (1, 3), x1 = 1 and y1 = 3.
  • 14. 14 Hence, y – y1 = m(x – x1) y – 3 = 2(x – 1) y – 3 = 2x – 2 Distributive property is applied y = 2x – 2 + 3 3 is added to both sides y = 2x + 1 2. slope = -3, (2, -1) Since the slope is -3, m = -3 and the given point is (2, -1), x1 = 2 and y1 = -1. Hence, y – y1 = m(x – x1) y – (-1) = -3(x – 2) y + 1 = -3x + 6 Distributive property is applied y = -3x + 6 – 1 1 is subtracted from both sides y = -3x + 5 3. slope 2 1 , (-1, -4) Since the slope is 2 1 , m = 2 1 and the given point is (-1, -4), x1 = -1 and y1 = -4. Hence, y – y1 = m(x – x1) [ ])1(x 2 1 )4(y −−=−− ( )1x 2 1 4y +=+ ( ) ( )    +=+ 1x 2 1 24y2 Both sides are multiplied by 2 2y + 8 = x + 1 Distributive property is applied 2y = x + 1 – 8 8 is subtracted from both sides 2y = x – 7 2 7x 2 y2 − = Both sides are divided by 2 2 7 x 2 1 y −= 4. )3,2(, 3 4 slope −−
  • 15. 15 Since the slope is 3 4 − , m = 3 4 − and the given point is (-2, 3), x1 = -2 and y1 = 3. Hence, y – y1 = m(x – x1) [ ])2(x 3 4 3y −−−=− )2x( 3 4 3y +−=− 3 8 x 3 4 3y −−=− (Distributive property is applied) 3 3 8 x 3 4 y +−−= (3 is added to both sides) 3 9 3 8 x 3 4 y +−−= (Renaming 3) 3 1 x 3 4 y +−= Try this out Determine f(x) = mx + b, given the slope and a point of the linear function. Set A 1. slope 3, (2, 5) 2. slope 4, (3, -4) 3. slope 2, (-5, 1) 4. slope -2, (-4, -2) 5. slope 5, (-7, 0) Set B 1. slope 5, (-2, 2) 2. slope -3, (3, 1) 3. slope 1, (6, -1) 4. slope -9, (-5, 0) 5. slope )3,2(, 3 1 − Set C 1. slope )1,4(, 3 2 − 2. slope       − 2 1 ,4, 2 1
  • 16. 16 3. slope       − 2 1 , 3 2 , 3 2 4. slope ( )3.0,1,2.0 5. slope ( )2.0,1.0,3.0− Lesson 8 Determining f(x) = mx + b, Given Any Two Points Let y = f(x). To determine f(x) = mx + b, given any two points of the linear function, substitute the coordinates of the first point to x1 and y1, and the coordinates of the second point to x2 and y2 in the form of the equation ( )1 12 12 1 xx xx yy yy − − − =− and then, solve for y in terms of x. Examples: Determine f(x) = mx + b, given two points of the linear function. 1. (2, 1), (3, 4) Since the first point is (2, 1), x1 = 2 and y1 = 1 and the second point is (3, 4), x2 = 3 and y2 = 4. Hence, ( )1 12 12 1 xx xx yy yy − − − =− ( )2x 23 14 1y − − − =− ( )2x 1 3 1y −=− y – 1 = 3(x – 2) y – 1 = 3x – 6 y = 3x – 6 + 1 y = 2x – 5 2. (-2, 5), (0, -3) Since the first point is (-2, 5), x1 = -2 and y1 = 5 and the second point is (0, -3), x2 = 0 and y2 = -3. Hence,
  • 17. 17 ( )1 12 12 1 xx xx yy yy − − − =− ( )[ ]2x )2(0 53 5y −− −− −− =− ( )2x 2 8 5y + − =− y – 5 = -4(x + 2) y – 5 = -4x – 8 y = -4x – 8 + 5 y = -4x – 3 3. (-2, -7), (-3, -1) Since the first point is (-2, -7), x1 = -2 and y1 = -7 and the second point is (- 3, -1), x2 = -3 and y2 = -1. Hence, ( )1 12 12 1 xx xx yy yy − − − =− ( )[ ]2x )2(3 )7(1 )7(y −− −−− −−− =−− ( )2x 1 6 7y + − =− y – 7 = -6(x + 2) y – 7 = -6x – 12 y = -6x – 12 + 7 y = -6x – 5 Try this out Determine f(x) = mx + b, given two points of the linear function. Set A 1. (1, 3), (-2, 6) 2. (3, 0), (0, -2) 3. (-5, 0), (-4, 7) 4. (0, 4), (4, 0) 5. (0, 1), (1, -2) Set B 1. (1, 4), (5, 6) 2. (8, -2), (6, -4) 3. (9, 0), (5, -2) 4. (5, -6), (6, -5) 5. (9, 3), (5, -6)
  • 18. 18 Set C 1. (-3, -2), (2, 1) 2. (-1, 3), (1, 0) 3. (-2, -3), (3, -1) 4. (-1, 3), (3, -3) 5. (-1, -1), (-3, -3) Let’s summarize 1. In the linear function f(x) = mx + b, m is the slope and b is the y-intercept. 2. The trend of a linear function is said to be increasing if the slope is positive. The trend of a linear function is said to be decreasing if the slope is negative. 3. To determine the y-intercept of a linear function f(x) = y, substitute x = 0 and solve for the value of y. This value is the y-intercept. To determine the x-intercept of the linear function, substitute y = 0 and solve for the value of y. This value is the x-intercept. 4. To determine a point on the graph of a given linear function y = f(x): a. Get any element from the domain of the linear function. Recall that the domain of any linear function is the set of real numbers. This element is the abscissa of the point. b. Substitute this abscissa to x in the linear function. c. Solve for the corresponding value of y. This value is the ordinate of the point. d. Repeat steps a to c using other element from the domain of the linear function to determine other points on the graph of the linear function. 5. To determine f(x) = mx + b, given the slope and y-intercept of the linear function, substitute the given slope to m and the given y-intercept to b. 6. To determine f(x) = y = mx + b, given the x- and y-intercept of the linear function, substitute the given x-intercept to a and the given y-intercept to b in the form of the equation 1 b y a x =+ and then, solve for y in terms of x. 7. To determine f(x) = y = mx + b, given the slope and one point of the linear function, substitute the given slope to m and the coordinates of the given
  • 19. 19 point to x1 and y1 in the form of the equation y – y1 = m(x – x1)and then, solve for y in terms of x. 8. To determine f(x) = y = mx + b, given any two points of the linear function, substitute the coordinates of the first point to x1 and y1, and the coordinates of the second point to x2 and y2 in the form of the equation ( )1 12 12 1 xx xx yy yy − − − =− and then, solve for y in terms of x. What have you learned 1. What is the slope of the f(x) = -5x + 4? 2. True or false? The trend of a linear function with positive slope is increasing. 3. Solve for the y-intercept of 6x – 3y = 4. 4. Which of the following step will solve for the y-intercept of the linear function y = 3x + 9? a. Substitute zero to x and solve for y. b. Substitute zero to y and solve for x. c. Take the square root of the constant term. d. Take the negative reciprocal of the coefficient of x. 5. What is the x-intercept of 12 = 9y + 18x? 6. Name three points on the graph of y – 2x + 4 = 0. 7. Determine the linear function with slope -3 and y-intercept -4. 8. The x- and y-intercepts of a linear function are both -2. Find the linear function. 9. The graph of a linear function passes through the point (-3, 7), and its slope is -4. Express the linear function in the form f(x) = mx + b. 10.What linear function has (-2, -5) and (-3, 1) as points on its graph?
  • 20. 20 Answer Key How much do you know 1. 2 2. False 3. -3 4. b 5. -4 6. (-1, -5), (0, -2), (1, 1) 7. y = 5x – 2 8. y = -x + 3 9. y = -2x – 8 10. 5 14 x 5 3 y += Try this out Lesson 1 Set A 1. The slope is 4 and the y-intercept is 5. 2. The slope is -7 and the y-intercept is 1. 3. The slope is 4 and the y-intercept is 0. 4. The slope is -4 and the y-intercept is 3. 5. The slope is 7 2 − and the y-intercept is 2. Set B 1. The slope is 8 and the y-intercept is -9. 2. The slope is -4 and the y-intercept is -6. 3. The slope is 1 and the y-intercept is 0. 4. The slope is 8 3 − and the y-intercept is 3. 5. The slope is 3 and the y-intercept is 17 5 − . Set C 1. The slope is 1 and the y-intercept is 2 1 . 2. The slope is 4 9 and the y-intercept is 2 3 − . 3. The slope is -0.3 and the y-intercept is 0.1. 4. The slope is -5.2 and the y-intercept is -4.4. 5. The slope is 8 9 and the y-intercept is 2 3 .
  • 21. 21 Lesson 2 Set A 1. increasing 2. decreasing 3. decreasing 4. decreasing 5. increasing Set B 1. increasing 2. decreasing 3. decreasing 4. decreasing 5. increasing Set C 1. decreasing 2. increasing 3. decreasing 4. increasing 5. increasing Lesson 3 Set A 1. The x-intercept is -2 and the y-intercept is 16. 2. The x-intercept is 2 1 and the y-intercept is -2. 3. The x-intercept is 4 3 and the y-intercept is 3. 4. The x-intercept is 2 5 and the y-intercept is -5. 5. The x-intercept is -2 and the y-intercept is 2. Set B 1. The x-intercept is 2 1 and the y-intercept is 7. 2. The x-intercept is 2 3 − and the y-intercept is -3. 3. The x-intercept is 2 3 − and the y-intercept is 3. 4. The x-intercept is 4 3 and the y-intercept is -3.
  • 22. 22 5. The x-intercept is 7/6 and the y-intercept is - 11 7 . Set C 1. The x-intercept is 5 and the y-intercept is 5 3 − . 2. The x-intercept is 8 3 − and the y-intercept is 2 3 . 3. The x-intercept is 7 1 and the y-intercept is 1 3 − . 4. The x-intercept is 3 1 and the y-intercept is 0.1. 5. The x-intercept is 13 11 − and the y-intercept is -4.4. Lesson 4 Set A (Answers may vary) 1. (-1, 5), (0, 6), (1, 7) 2. (-1, 9), (0, 7), (1, 5) 3. (-1, -2), (0, 0), (1, 2) 4. (-1, 7), (0, 4), (1, 1) 5. (-1, 2), (0, 4), (1, 6) Set B (Answers may vary) 1. (-1, -9), (0, -4), (1, 2) 2. (-1, -2), (0, -6), (1, -8) 3. (-1, 1), (0, 3), (1, 1) 4.       − 11 1 ,1 ,       11 7 ,0 ,       − 11 13 ,1 5. ( )1, 1− − , 1 0, 3  −    , 1 1, 3       Set C (Answers may vary) 1. (-1, 7), (0, 0), (1, -7) 2. (-4, 3), (0, 2), (4, 1) 3. ( )1, 3− − , 2 0, 3  −    , 5 1, 3       4.       −− 4 15 ,1 ,       − 2 3 ,0 ,       4 3 ,1 5. (-1, 0.4), (0, 0.1), (1, -0.2) Lesson 5 Set A 1. f(x) = 3x + 2
  • 23. 23 2. f(x) = -2x + 4 3. f(x) = 5x – 7 4. f(x) = -8x – 1 5. f(x) = 6x Set B 1. f(x) = x – 1 2. f(x) = -9x 3. ( ) 2x 3 4 xf −= 4. ( ) 1 1 f x x 3 3 = − − 5. ( ) 3 4 x7xf −= Set C 1. ( ) 4 1 x 3 2 xf −= 2. ( ) 2 1 f x x 5 3 = − − 3. f(x) = 0.2x + 1 4. f(x) = x – 0.1 5. ( )f x 2x 2= + Lesson 6 Set A 1. ( ) 2x 3 2 xf +−= 2. ( ) 4x2xf += 3. ( ) 3x3xf −= 4. ( ) 1x 8 1 xf −−= 5. ( ) 8x2xf −−= Set B 1. ( ) 6x 5 6 xf +−= 2. ( ) 12x3xf += 3. ( ) 16x 3 4 xf −−= 4. ( ) 18x 5 6 xf −−= 5. ( ) 7 1 x 7 3 xf +=
  • 24. 24 Set C 1. ( ) 5 1 x 65 1 xf +−= 2. ( ) 3x27xf += 3. ( ) 3x 8 9 xf +−= 4. ( ) 2 1 x 8 7 xf += 5. ( ) 12 13 x 8 39 xf += Lesson 7 Set A 1. ( ) 1x3xf −= 2. ( ) 16x4xf −= 3. ( ) 11x2xf += 4. ( ) 10x2xf −−= 5. ( ) 35x5xf += Set B 1. ( ) 12x5xf += 2. ( ) 10x3xf +−= 3. ( ) 7xxf −= 4. ( ) 45x9xf −−= 5. ( ) 3 11 x 3 1 xf −= Set C 1. ( ) 3 11 x 3 2 xf −= 2. ( ) 2 5 x 2 1 xf +−= 3. ( ) 18 17 x 3 2 xf +−= 4. ( ) 1.0x2.0xf += 5. ( ) 23.0x3.0xf +−= Lesson 8 1. y = -x + 4 2. y = 2 3 x - 2
  • 25. 25 3. y = 7x + 35 4. y = -x + 4 5. y = -3x + 1 Set B 1. y = x + 7 2 2 2. y = x - 10 3. y = x – 9 2 2 4. y = x - 11 5. y = 9x - 69 4 4 Set C 1. y = 3x – 1 5 5 2. y = - 2 3 x + 2 3 3. y = 2x – 11 5 5 4. y = - 2 3 x + 2 3 5. y = x What have you learned 1. -5 2. true 3. 3 4 − 4. a 5. 2 3 6. (0, -4), (1, -2), (2, 0) 7. y = -3x - 4 8. y = -x - 2 9. y = -4x - 5 10.y = -6x - 17