Chapter 2 Graphing an Equation
Functions and Graphs To sketch the graph an equation in x and y,y we need to find
ordered pairs that solve the equation and plot the ordered
pairs on a grid. This process is called point-by-point
plotting.
Section 1
For example, let’s plot the graph of the equation
Functions
y = x −2 2
Graphing an Equation: Graphing an Equation:
Making a Table of Ordered Pairs Plotting the points
Make a table of ordered x y Next,
Next plot the points and connect them with a smooth
pairs that satisfy the curve. You may need to plot additional points to see the
–3 2
(–3) +2 =
equation pattern formed.
11
y = x2 − 2 –2 (–2)2+2 = 6
–1 (–1)2+2 = 6
0 (0)2+2 = 2
1 (1)2+2 = 3
2 (2)2+2 = 6
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Functions Function Definition
You can visualize a function by the following diagram which
The previous graph is the graph of a function. The idea of shows a correspondence between two sets: D, D the domain of
a function is this: a correspondence between two sets D the function, gives the diameter of pizzas, and R, the range of
and R such that to each element of the first set, D, there
the function gives the cost of the pizza.
corresponds one and only one element of the second set,
R.
The first set is called the domain, and the set of 10
9.00
corresponding elements in the second set is called the 12 10.00
range
range. 16
For example, the cost of a pizza (C) is related to the size of 12.00
the pizza. A 10 inch diameter pizza costs $9.00, while a 16
inch diameter pizza costs $12.00. domain D
range R
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Functions Specified by Equations Functions Specified by Equations
If in an equation in two variables,
variables we get exactly one Consider the equation that was graphed on a previous slide
output (value for the dependent variable) for each input
(value for the independent variable), then the equation
specifies a function. The graph of such a function is just
y = x2 − 2 –2
Input:
x = –2
the graph of the specifying equation. Process:
If we get more than one output for a given input
input, the
(–2,2) is an (−2 ) 2
−2 square (–2),
then subtract 2
ordered pair of
equation does not specify a function.
the function.
Output:
2 result is 2
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Vertical Line Test for a Function
Vertical Line Test for a Function
(continued)
If y
you have the graph
g p of an equation,
q , there is an easyy This g
graph
p is not the graph
g p of a
way to determine if it is the graph of an function. It is function because you can draw a
called the vertical line test which states that: vertical line which crosses it
twice.
An equation specifies a function if each vertical line in
the coordinate system passes through at most one
point on the graph of the equation.
If any vertical line passes through two or more points Thi is
This i the
h graphh off a
on the graph of an equation, then the equation does function because any vertical
not specify a function. line crosses only once.
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Function Notation Function Evaluation
Thee following
o ow g notation
otat o iss used to describe
desc be functions.
u ct o s. Thee Consider our function f (x) = x2 −2
variable y will now be called f (x).
What does f (–3) mean?
This is read as “ f of x” and simply means the y coordinate
of the function corresponding to a given x value.
Our previous equation
y = x2 − 2
can now be expressed as
f ( x) = x 2 − 2
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Function Evaluation Some Examples
Consider our function f (x) = x2 −2 1.
1
( ) = 3x
f (x) 3 −2
What does f (–3) mean?
Replace x with the value –3 and evaluate the expression f (2) = 3(2) − 2 = 4 = 2
f (−3) = (−3) 2 + 2 f (a) = 3(a) − 2
The result is 11 . This means that the point (–3,11) is on
the graph of the function. f (6 + h) = 3(6 + h) − 2 = 18 + 3h − 2
= 16 + 3h
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Domain of a Function Domain of a Function
Consider f ( x) = 3 x − 2 Answer: f ( x) = 3 x − 2
f (0) = ? is defined only when the radicand (3x – 2) is equal to
or greater than zero. This implies that
f (0) = 3(0) − 2 = −2
2
which is not a real number. x≥
Question: for what values of x is the function defined? 3
15 16
Domain of a Function Domain of a Function
(continued) (continued)
Therefore
Therefore, the domain of our function is the set of real Therefore,
Therefore the domain of our function is the set of real
numbers that are greater than or equal to 2/3. numbers that are greater than or equal to 2/3.
Example: Find the domain of the function Example: Find the domain of the function
1 1
f ( x) = x−4 f ( x) = x−4
2 2
Answer:
{ x x ≥ 8} , [8, ∞)
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Domain of a Function: Domain of a Function:
Another Example Another Example
Find the domain of 1 Find the domain of 1
f ( x) = f ( x) =
3x − 5 3x − 5
In this case, the function is defined for all values of x
except where the denominator of the fraction is zero. This
means all real numbers x except 5/3.
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Mathematical Modeling Solution
The p
price-demand function for a company
p y is ggiven byy Revenue = Price · Quantity, so
R(x)= p(x) · x = (1000 – 5x) · x
p( x) = 1000 − 5 x, 0 ≤ x ≤ 100 When 50 items are sold, x = 50, so we will evaluate the
revenue function at x = 50:
where x represents the number of items and P(x) represents the
price of the item. Determine the revenue function and find the R(50) = (1000 − 5(50))i50 = 37,500
revenue generated if 50 items are sold.
The domain of the function has already been specified.
specified We
are told that
0 ≤ x ≤ 100
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Break-Even and Profit-Loss
Break-Even and Profit-Loss
Analysis
Analysis
(continued)
Any y manufacturingg company
p y has costs C and revenues R. Price demand functions,
Price-demand functions usually determined by financial
The company will have a loss if R < C, will break even departments, play an important role in profit-loss analysis.
if R = C, and will have a profit if R > C. p = m – nx
Costs include fixed costs such as plant overhead, etc. and variable (x is the number of items than can be sold at $p per item.)
costs, which are dependent on the number of items produced.
C = a + bx The revenue function is
(x is the number of items produced) R = (number of items sold) · (price per item)
= xp = x(m – nx)
The profit function is
P = R – C = x(m – nx) – (a + bx)
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Example of Profit-Loss Analysis Answer to Revenue Problem
A company manufactures notebook computers.
computers Its Since Revenue = Price · Quantity,
marketing research department has determined that the
data is modeled by the price-demand function R( x) = x • p( x) = x • (2000 − 60 x) = 2000 x − 60 x 2
p(x) = 2,000 – 60x, when 1 < x < 25, (x is in thousands).
The domain of this function is the same as the domain
What is the company’s revenue function and what is its of the price-demand function, which is 1 ≤ x ≤ 25 (in
domain? thousands.)
25 26
Profit Problem Answer to Profit Problem
The financial department for the company in the preceding Since Profit = Revenue – Cost,, and our revenue function
problem has established the following cost function for from the preceding problem was R(x) = 2000x – 60x2,
producing and selling x thousand notebook computers:
P(x) = R(x) – C(x) = 2000x – 60x2 – (4000 + 500x)
C(x) = 4,000 + 500x (x is in thousand dollars). = –60x2 + 1500x – 4000.
Write a profit function for producing and selling x thousand The domain of this function is the same as the domain of
notebook computers, and indicate the domain of this function. the original price-demand function, 1< x < 25 (in
thousands.) 5000
Thousand dollars
25
Thousand cameras
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