Day 1 Notes
Note: There aren’t very many notes from Day 1, since we spent time discussing class
expectations, syllabus, and doing a group review of algebra.
Definition 1 A function f is a rule that assigns to each value of x a unique(!) value of y.
We write y = f (x) to indicate y is a function of x.
x - independent variable (cause) - horizontal axis
y - dependent variable (effect) - vertical axis
domain - allowed values of x
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Definition 2 A linear function is a function of the form f (x) = mx + b (m, b constant).
Linear functions are used to model many common business situations.
Interpreting the constants:
∆y
m = slope = ∆x
represents “rate of change” :
This tells how much y changes if x is increased by 1.
b = y-intercept
often represents“starting point” or “baseline” (when x = 0)
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Businesses sometimes use linear depreciation to describe assets which lose value over
time.
Example: A company car has an initial value of $22,000 and will be depreciated linearly
over 5 years with a scrap value of $7,000.
1. What is the rate of depreciation of the car?
We set x = number of years passed and y = value of car.
The depreciation rate is how much the value of the car drops each year, so
this question is asking for
rate of depreciation =
δy
change in value of car
=
.
change in years
δx
This is the same as the slope. We have two points: (0,22000) and (5, 7000), so we can
plug these into the slope formula. This gives:
δy
22000 − 7000
15000
=
=
= −3000
δx
0−5
−5
So the car loses $3000 per year.
2. Find a linear function giving the book value V (in dollars), as a function of time t (in
years).
Note: The term “book value” is the same as “value” in this instance.
For a linear function, we want to write V (t) = mt + b.
We found slope in the previous problem. (The fact that we changed the letters of our
variables doesn’t change anything.)
Also, we know that when 0 years have passed, the car is worth $22000, so the yintercept is (0, 22000).
Therefore, the equation is V (t) = −3000t + 22000.
3. What is the book value of the car after 2 years?
Since t = 2 years, we plug this into V , yielding
V (3) = −3000(2) + 22000.
= 16000
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Other linear functions used in business
Let x be the total number of units of a product manufactured or sold.
• Total cost function:
C(x) = Total cost of manufacturing x units of the product
The cost function must take into account
– fixed costs (which the company pays no matter what)
– variable costs (which change as more units are manufactured).
Fixed costs often (but not always!) correspond to the y-intercept in the cost function.
Variable costs often (but not always!) correspond to the slope of a linear cost function.
• Revenue function:
R(x) = Total revenue from sale of x units of product
This formula often (but not always!) leads to R(x) = (unit price) timesx.
• Profit function:
P (x) = R(x) − C(x) (Total profit = revenue minus cost)
NOTE: Be careful! When you subtract, you should put parentheses around the entire
C(x) expression. Otherwise, there may be algebra erros.
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Example: TeddyBearCo manufactures teddy bears.
The company has fixed costs of $18,000.
It costs the company $14 to produce one bear.
Each bear is sold for $42.
• Find the cost, revenue, and profit functions for TeddyBearCo.
The total cost is C(x) = 14x + 18000.
The revenue function is R(x) = 42x.
The profit function is
P (x) = R(x) − C(x)
= 42x − (14x + 18000)
= 28x + 18000
NOTE: Again, note the parentheses around the C(x) expression. Without them, the
answer would be wrong!
• How many bears must the company sell to break even?
We complete this on Day 2. You have reached the end of the Day 1 notes.
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