Linear Function
A Linear Function
Is a function of the form
f ( x )  mx  b
where m and b are real numbers and m is the slope and b is the
y - intercept. The x – intercept is
b
m
The domain and range of a linear function are all real numbers.
Graph
Graph of a Linear Function
The linear function can be graphed using the slope
and the y-ntercept
f (x)  3x  2 m = 3 b = 2
Example If
The linear function can be graphed using the x and
the y-intercepts
Average Rate of Change
The average rate of change of a Linear Function is
the constant
m 
y
x
For example, For f(x)= 5x - 2 , the average rate of
change is m =5
Page 121 #15
• f(x) = -3x+4
• The slope is m = -3, the y-intercept b = 4
• The average rate of change is the
constant m = -3
• Since m =-3 is negative the graph is
slanted downwards. Thus the function is
decreasing
Page 121 #19
•
•
•
•
•
•
f(x) = 3
f(x)=0x + 3
m=0 b=3
The average of change is 0
Since the average rate of change, m = 0
The function is constant neither increasing
or decreasing
Page 121 #21
• To find the zero of f(x), we set f(x) = 0 and
solve.
• 2x - 8 = 0
• x = 8/2 = 4
• y-intercept
• Will graph in class
Page 121 #25
• To find the zero of
f(x), we set f(x) = 0
and solve.
1
• 2 x-8=0
• x = 16
• y-intercept
• Will graph in class
Linear or Non Linear Function
• If a function is linear the slope or rate of
change is constant
• That is  y
is always the same
y
Page 121 #28
X
Y=f(x)
-2
¼
-1
½
0
1
1
2
2
4
y
x
f (  1)  f (  2 )

y
x
y
x
 2  (  1)



1/ 2 1/ 4
f ( 0 )  f (  1)
0  (  1)
f (1)  f ( 0 )
1 0
1 2


1/ 2
 1/ 2
1
11/ 2
 .1 / 2
1

2 1
1
1
The rate of change is not
constant. Not a Linear
Function
Page 121 #32
y
x
-2
-1
0
1
2
y = f(x)
-4
-3.5
-3
-2.5
-2
x
y
x
x
x
 1  (2)
f ( 0 )  f (  1)

y
y
f (  1)  f (  2 )

0  (  1)
f (1)  f ( 0 )


1 0
f ( 2 )  f (1)
2 1



 3 .5  (  4 )
 1  2)
 3  (  3 .5 )
0 1
 .5
 2 .5  (  3)
1

 2  (  2 .5 )
 .5
1)
•Note the rate of change is constant. It is always
m =.5 thus the function is linear
 .5
 .5
m 
20  60
5  (  15 )

 40
 2
20
60=-2(-15)+b
Page 121 #38
(-15,60)
y = g(x)
If g(x) =-2x+30=0
-2x+30=0
-2x = -30, x = 15
b = 30

y =g(x) =-2x +30
(5,20)
If g(x) =-2x+30=20
X=5
(15,0)
If g(x) -2x+30=60
-2x+30  60
-2x 30, x  15
If g(x) =-2x+30=60

-2x+30=60
-2x = 30, x = -15
0<2x+30<60
-30 < -2x < 30
15 > x > -15
Page 122 # 44
• Cost Function: C(x) = 0.38x + 5 in dollars
• Find Cost for x = 50 minutes
• C(50) = .38(50) + 5 = 19+5= $24
• Given Bill, find cost
• C(x) = 0.38x + 5 = 29.32
• 0.38x = 24.32 x = 24.32 /.38
x = 62
• Estimated Cost of Monthly Bill, find Maximum minutes
• 0.38x + 5 = 60
0.38x = 55 x = 55 /.38 x = 144.8
• Can use as many as 144 minutes
Page 122 # 48
Supply(S) and Demand(D)
• Equilibrium: Supply = Demand
• S(p) = -2000 + 3000p = D(p) =10,000 -1000p
• -2000 + 3000p =10,000-1000p
• 4000p =12000
p = 3000
•
• Quantity sold if Demand is less than Supply
• If 10,000 -1000p < -2000 +3000p 4000p > 12000 p > 3000
• The price will decrease if the quantity of demand is less than the
quantity of supply
Page 122 # 54
Straight Line Depreciation
•
•
•
•
Straight Line Depreciation = Book Value / approximate life
Let V(x) be value of machine after x years
Cost of machine = Book Value = V(0)
V(x) = 120,000 – ($120,000 / 10)x = -12,000x +120,000
Page 122 # 54
Straight Line Depreciation(cont)
• Book value after 4years = –12000(4)+1210,000 =1200008000=72000
• After 4 years the machine will be worth $72,000