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Linear Function

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A Linear Function Is a function of the form

where mand bare real numbers andm is the slope and bis the

y - intercept. The x – intercept is

The domain and range of a linear function are all real numbers.

Graph

The linear function can be graphed using the slope and the y-ntercept

Example If m = 3 b = 2

The linear function can be graphed using the x and the y-intercepts

The average rate of change of a Linear Function is the constant

For example, For f(x)= 5x - 2 , the average rate of change is m =5

- 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

- 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

- 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

- To find the zero of f(x), we set f(x) = 0 and solve.
- x - 8 = 0
- x = 16
- y-intercept
- Will graph in class

- If a function is linear the slope or rate of change is constant
- That is

is always the same

The rate of change is not constant. Not a Linear Function

- Note the rate of change is constant. It is always
- m =.5 thus the function is linear

If g(x) =-2x+30=0

-2x+30=0

-2x = -30, x = 15

y = g(x)

60=-2(-15)+b

b = 30

y =g(x) =-2x +30

If g(x) =-2x+30=20

X = 5

(-15,60)

If g(x) -2x+30=60

-2x+30 60

-2x 30, x 15

(5,20)

(15,0)

If g(x) =-2x+30=60

-2x+30=60

-2x = 30, x = -15

0<2x+30<60

-30 < -2x < 30

15 > x > -15

- 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

- Equilibrium: Supply = Demand
- S(p) = -2000 + 3000p = D(p) =10,000 -1000p
- -2000 + 3000p =10,000-1000p
- 4000p =12000p = 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

- 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

- Book value after 4years = –12000(4)+1210,000 =120000-8000=72000
- After 4 years the machine will be worth $72,000