Download 1.1 Functions Cartesian Coordinate System

1.1 Functions
• This section deals with the topic of
functions, one of the most important topics
in all of mathematics. Let’s discuss the
idea of the Cartesian coordinate system
first.
Cartesian Coordinate System
• The Cartesian coordinate system was named after
Rene Descartes. It consists of two real number
lines which meet at a point called the origin. The
two number lines which meet at a right angle
divide the plane into four areas called quadrants.
• The quadrants are numbered using Roman
numerals as shown. Each point in the plane
corresponds to one and only one ordered pair of
numbers (x , y). Two ordered pairs are shown.
I
II
III
(3,1)
x
(-1,-1)
IV
y
1
Graphing an equation
• To graph an equation in x and y, we need to find ordered
pairs that solve the equation and plot the ordered pairs on
a grid.
For example, let’s plot the graph of the equation
y = x2 + 2
Making a table of ordered pairs
• Let’s make a table of ordered pairs that
satisfy the equation y = x2 + 2
x
-3
-2
-1
0
1
2
y
( −3) 2 + 2
(−2)2 + 2 = 6
(−1) 2 + 2 = 3
(0) 2 + 2 = 2
(1) 2 + 2 = 3
(2) 2 + 2 = 6
Plotting the points
• Next, plot the points and connect them with a smooth
curve. You may need to plot additional points to see the
pattern formed.
2
Function
• The previous graph is the graph of a function. The idea of
a function is this: a relationship between two sets D and R
• such that for each element of the first set, D, there
corresponds one and only one element of the second set, R.
For example, the cost of a pizza (C) is related to the size of
the pizza. A 10 inch diameter pizza costs 9.00 while a 16
inch diameter pizza costs 12.00.
Function definition
• You can visualize a function by the following diagram which shows
a correspondence between two sets, D, the domain of the function
and R, the range of the function. The domain gives the diameter of pizzas
range
and the range gives the cost of the pizza.
domain
10
12
9.00
10.00
12.00
16
Functions specified by equations
• Consider the previous equation that was
graphed
-2
Input x = -2
Process: square (–2)
then subtract 2
(-2,2) is an ordered
pair of the function.
2
Output: result is 2
3
Function Notation
• The following notation is used to describe functions The
variable y will now be called
• 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
can now be expressed as
Function evaluation
• Consider our function
• What does
mean?
Replace x with the
value –3 and evaluate the expression
• The result is 11 . This means that the point (-3,11) is on
the graph of the function.
Some Examples
• 1.
f (a) = 3(a) − 2
f (6 + h) = 3(6 + h) − 2 = 18 + 3h − 2
= 16 + 3h
4
Domain of a Function
• Consider
f ( x) = 3 x − 2
f (0) = ?
f (0) = 3(0) − 2 = −2
• which is not a real number. Question: for
what values of x is the function defined?
Domain of a function
• Answer:
• is defined only when the radicand (3x-2)
• is greater than or equal to zero. This implies that
3x-2
0
or
2
≥
x≥ 3
Domain of a function
• Therefore, the domain of our function is the set of real
numbers that are greater than or equal to 2
3
• Examples. Find the domain of the following functions.
f ( x) =
• Answer:
1
x−4
2
{ x x ≥ 8} , [8, ∞)
5
More examples
• Find the domain of
f ( x) =
1
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
Mathematical modeling
• The price-demand function for a company is given by
•
p ( x) = 1000 − 5 x,
•
0 ≤ x ≤ 100
where P(x) represents the price of the item and x
represents the number of items. Determine the revenue
function and find the revenue generated if 50 items are
sold.
Solution
• Revenue = price x quantity so
• R(x)= p(x)*x = (1000 − 5 x)i x
• When 50 items are sold, x = 50 so we will evaluate the
revenue function at x = 50
R (50) = (1000 − 5(50))i50 = 37,500
•
The domain of the function has already been specified.
We are told that
0 ≤ x ≤ 100
6
1.2
Elementary Functions; Graphs and
Transformations
• In this presentation, you will be given an equation of a
function and asked to draw its graph. You should be able
to state how the graph is related to a “standard” function.
It is not important that you plot a great many points for
each graph. It IS important that you recognize the general
shape of the graph. You can verify your answers using a
graphing calculator, but only after you have attempted to
construct the graph by hand.
Problem 1
• Construct the graph of
Solution
12
10
8
Series1
6
4
2
0
-4
-2
0
2
4
1
Problem 2
• Now, sketch the related graph given by the
equation below and explain, in words, how it is
related to the first function you graphed.
Solution: Problem 2
• The graph has the same shape as the original
function. The difference is that the original graph
has been translated two units to the right on the
x-axis. Conclusion: The graph of the
function f(x-2) is the graph of f(x) shifted
horizontally two units to the right on the x-axis.
• Notice that replacing x by x-2 shifts the graph
horizontally to the right and not the left.
Problem 3
• Now, graph the following “standard”
function: Complete the table:
-3
-2
-1
0
1
2
3
2
Solution to problem 3
30
20
10
Ser ies1
0
-4
-2
0
2
4
- 10
- 20
- 30
Problem 4
• Now, graph the following related function:
Solution to problem 4
30
20
10
Ser ies1
0
-4
-2
0
2
4
-10
- 20
- 30
3
Problem 4 solution
• The graph of
• is obtained from the graph of
• by translating the graph of the original
function up one unit vertically on the
positive y-axis.
Problem 5
• Graph:
• What is the domain of this function?
Solution to problem 5
• The domain is all non-negative real
numbers. Here is the graph:
6
5
4
3
Series1
2
1
0
0
5
10
15
20
25
30
4
Problem 6
• Graph:
• Explain, in words, how it compares to
problem 5.
Problem 6 solution
(Notice that the graph lies entirely within the fourth
quadrant)
0
-1
0
5
10
15
20
25
30
-2
-3
Series1
-4
-5
-6
Graph of –f(x)
• The graph of the function –f(x) is a reflection of
the graph of f(x) across the x-axis. That is, if the
graphs of f(x) and –f(x) are folded along the xaxis, the two graphs would coincide.
5
Cube root function
• Sketch the graph of the cube root function. Complete the
table of ordered pairs:
x
y
-27
-8
-1
0
1
8
27
Variation of cube root function
• Sketch the following variation of the cube root function:
6
Same graph as graph of cube root function. Shifted
horizontally to the left one unit.
Graph of f(x+c) compared to graph of f(x):
• The graph of f(x+c) has the same shape as the
graph of f(x) with the exception that the graph of
f(x+c) is translated horizontally to the left c units
when c >0 and is translated horizontally to the
right c units when c < 0.
Absolute Value function
• Now, graph the absolute value function. Be sure
to choose x values that are both positive and
negative as well as zero.
a ( x) = x
7
Graph of absolute value
function
Notice the symmetry of the graph.
Variation of absolute value function
a ( x + 1) − 2 = x + 1 − 2
Shift absolute value graph to the left one unit and down two
units on the vertical axis.
8
Linear functions and Straight Lines
Linear Functions
• The equation f(x) = mx+b m and b are real
numbers is the equation of a linear
function. The domain is the set of all real
numbers. The graph of a linear function is
a straight line. Some examples of graphs
will follow in the next few slides.
f ( x) =
1
x+8
4
1
f(x)= -2x+3
More examples
f ( x ) = −2
Graphing
• Graph
f ( x) =
3
x+2
4
• using a table of
values for x and y
x y
-4
0
4
8
2
Solution:
Graphing using intercepts
• Graph 5x+6y = 30 using the x and y
intercepts:
• 1. Set x = 0 and solve for y
•
5(0) + 6y = 30
•
y=5.
• 2. Now, let y =0 and solve for x:
•
5x + 6(0) = 30, x = 6
• 3. Plot the two ordered pairs (0, 5) and (6,0) and connect the points
with a straight line.
solution
3
Special cases
• 1. The graph of x=k is the graph of a
vertical line k units from the y-axis.
• 2. The graph of y=k is the graph of the
horizontal line k units from the x-axis.
Some examples:
• 1. Graph x=-7
• 2. Graph y = 3
solutions
X=7
Y=3
4
Slope of a line
• Slope of a line:
• = rise
run
m=
D
( x1 , y1 )
y2 − y1
x2 − x1
Rise
D ( x2 , y2 )
run
Slope-intercept form
• The equation y = mx + b
• is called the slope-intercept form of an
equation of a line .
• The letter m represents the slope and b
represents the y intercept.
Find the slope and intercept from
an equation of a line
• 1. Find the slope and y
intercept of the line
whose equation is
• 5x – 2y = 10
5x − 2 y = 10
Solution: Solve the equation for y in
terms of x. Identify the coefficient
of x as the slope and the yintercept is the constant term.
Therefore: the slope is 5/2 and the y
intercept is -5
−2 y = −5x +10
y=
−5x 10 5
+ = x −5
−2 −2 2
-
5
Point-slope form
• The point- slope form of
the equation of a line is
y − y1 =
as follows:
m( x − x1 )
• It is derived from the
definition of the slope of a
line:
y −y
2
1
x2 − x1
=m
Examples
• Find the equation of the line through the
points (-5, 7) and (4, 16) :
• Solution:
( −5, 7 )
(4,16)
16 − 7
9
= =1
4 − (−5) 9
y − 16 = 1( x − 4) → y = x − 4 + 16 = x + 12
m=
Applications
• Office equipment was purchased for $20,000 and will
have a scrap value of $2,000 after 10 years. If its value
is depreciated linearly , find the linear equation that
relates value (V) in dollars to time (t) in years:
Solution: when t = 0 , V = 20,000 and when t = 10, V = 2,000. Thus, we
have two ordered pairs (0, 20,000) and (10, 2000). We find the
slope of the line using the slope formula. The y-intercept is already
known (when t = 0, V = 20,000, so the y-intercept is 20,000) . The
slope is : (2000-20,000)/(10 – 0) = -1,800.
Therefore, our equation is V(t)=
- 1,800t + 20,000
6
Quadratic functions
• If a, b, c are real numbers with a not equal
to zero, then the function
f ( x) = ax 2 + bx + c
• is a quadratic function and its graph is a
parabola.
Vertex form of the quadratic
function
• It is convenient to convert the general form
of a quadratic equation
f ( x) = ax 2 + bx + c
• to what is known as the vertex form.
f ( x ) = a ( x − h) 2 + k
Completing the square to find the
vertex of a quadratic function
• The example below illustrates the procedure:
Consider
f ( x ) = −3 x 2 + 6 x − 1
Complete the square to find the
vertex:
f ( x) = −3( x 2 − 2 x +
_____
) −1
f ( x) = −3( x − 2 x + 1) − 1 − 3
2
f ( x) = −3( x − 1) 2 − 4
1
Completing the square, continued
• The vertex is (1 , 2)
• The quadratic function opens down since
the coefficient of the x squared term is
negative (-3) .
Intercepts of a quadratic function
• Find the x-intercepts of.
f ( x) = −3x 2 + 6 x − 1
• Set f(x) = 0
0 = −3 x 2 + 6 x − 1
• Use the quadratic formula:
X=
−b ± b 2 − 4ac
2a
Intercepts of a quadratic function
2
• X= −6 ± 6 − 4(−3)(−1) = −6 ± 24 ≈ 0.184,1.816
2(−3)
−6
f ( x) = −3x 2 + 6 x − 1
•
f ( x) = −3(0) 2 + 6(0) − 1
Find the y-intercept : Let x = 0 and solve for y:
We have (0, -1)
2
Generalization
• Summary:
f ( x ) = a ( x − h) 2 + k
• where a is not equal to zero.
Graph of f is a parabola:
if a > 0, the graph opens
upward
if a < 0 , the graph opens
downward.
Generalization, continued
•
•
•
•
Vertex is (h , k)
Line or axis of symmetry: x = h
f(h) = k is the minimum if a > 0, otherwise, f(h) = k is
the maximum
Domain : set of all real numbers
•
Range:
{ y y ≤ k}
{ y y ≥ k}
if a < 0. If a > 0, the range is
Application of Quadratic Functions
•
A Macon Georgia peach orchard farmer now
has 20 trees per acre. Each tree produces, on
the average, 300 peaches. For each
additional tree that the farmer plants, the
number of peaches per tree is reduced by 10.
How many more trees should the farmer plant
to achieve the maximum yield of peaches?
What is the maximum yield?
•
3
Solution
• Solution: Yield= number of peaches per
tree x number of trees
• Yield = 300 x 20 = 6000 ( currently)
• Plant one more tree: Yield = ( 300 – 1(10))
* ( 20 + 1) = 290 x 21 = 6090 peaches.
• Plant two more trees:
• Yield = ( 300 – 2(10)* ( 20 + 2) = 280 x 22
= 6160
Solution, continued
• Let x represent the number of additional
trees. Then Yield =( 300 – 10x) (20 + x)=
• Y(x)= −10 x 2 + 100 x + 6000
• To find the maximum yield, note that the Y(x) function is
a quadratic function opening downward. Hence, the
vertex of the function will be the maximum value of the
yield.
•
Solution, continued
• Complete the square to find the vertex of the parabola:
•
2
• Y(x) =
−10( x − 10 x + 25) + 6000 + 250
• (we have to add 250 on the outside since we multiplied
• –10 by 25 = -250. The equation is unchanged, then.
4
Solution,continued
• Y(x)= −10( x − 5) 2 + 6250
• Thus, the vertex of the quadratic function is ( 5 , 6250) .
So the farmer should plant 5 additional trees and obtain
a yield of 6250 peaches. We know this yield is the
maximum of the quadratic function since the the value of
a is -10. The function opens downward so the vertex
must be the maximum.
•
5