Powerpoint slides copied from or based upon:
Functions Modeling Change
A Preparation for Calculus
Third Edition
Connally,
Hughes-Hallett,
Gleason, Et Al.
Copyright  2007 John Wiley & Sons, Inc.
1
SECTION
2.6
QUADRATIC FUNCTIONS
2
A baseball is “popped” straight up by a
batter. The height of the ball above the
ground is given by the function:
y = f(t) = −16t2 + 64t + 3,
where t is time in seconds after the ball
leaves the bat and y is in feet.
Let's use our calculator:
Page 88
3
Let's use our calculator:
Y= → \Y1= −16x2+64x+3
Page N/A
4
Let's use our calculator:
Y= → \Y1= −16x2+64x+3
Window →
Page N/A
Window
Xmin
Xmax
Xscl
Ymin
Ymax
Yscl
Value
-1
5
1
-10
80
8
Graph
5
Although the
path of the
ball is
straight up
and down, y
the graph of
f
its height as e
a function of e
t
time is
concave
down.
80
70
60
50
40
30
20
10
0
-1
Page 88
-10
0
1
2
t seconds
3
4
5
6
The ball
goes up fast
at first and
then more
slowly
y
because of
f
gravity.
e
80
70
60
50
40
30
e
t
20
10
0
-1
Page 88
-10
0
1
2
t seconds
3
4
5
7
The baseball
height function is
an example of a
quadratic
function, whose
general form is
y = ax2 + bx + c.
80
70
60
y
50
f
e
e
t
40
30
20
10
0
-1 -10 0
Page 89
1
2
3
4
5
t seconds
8
Finding the Zeros of a Quadratic Function
Page 89
9
Finding the Zeros of a Quadratic Function
Back to our baseball example, precisely
when does the ball hit the ground?
Page 89
10
Finding the Zeros of a Quadratic Function
Back to our baseball example, precisely
when does the ball hit the ground?
Or:
For what value of t does f(t) = 0?
Page 89
11
Finding the Zeros of a Quadratic Function
Back to our baseball example, precisely
when does the ball hit the ground?
Or:
For what value of t does f(t) = 0?
Input values of t which make the output
f(t) = 0 are called zeros of f.
Page 89
12
0  16t  64t  3
2
b  b  4ac
t
2a
2
Page N/A
13
0  16t  64t  3
2
b  b  4ac
t
2a
a  16
2
b  64
c3
Page N/A
14
0  16t  64t  3
2
b  b  4ac
t
2a
2
64  64  4(16)(3)

2(16)
2
Page N/A
15
64  64  4( 16)(3)

2( 16)
2
64  4096  64(3)

32
Page N/A
64  4288

32
16
64  4288

32
64  65.48282217

32
129.48282217

32
 4.046338193
Page N/A
17
Let's use our calculator:
Y= → \Y1= −16x2+64x+3
Window →
Page N/A
Window
Xmin
Xmax
Xscl
Ymin
Ymax
Yscl
Value
-1
5
1
-10
80
8
Graph
18
Now let's
use the TI
to find the
zeros of
this
quadratic
function:
80
70
60
50
y
40
f
e
e
t
30
20
10
0
-1
Page N/A
-10
0
1
2
t seconds
3
4
5
19
2nd
Trace
2: zero
Left Bound ?
Right Bound?
Guess?
Page N/A
20
80
70
60
y
f
e
e
t
zero
X=4.0463382
-1
Y=-1E-11
Page N/A
50
40
30
20
10
0
-10
0
1
2
t seconds
3
4
5
21
Example #1:
Find the zeros of f(x) = x2 − x − 6.
Page 89
22
Example #1:
Find the zeros of f(x) = x2 − x − 6.
Set f(x) = 0 and solve by factoring:
x2 − x − 6 = 0
(x-3)(x+2) = 0
x = 3 & x = -2
Page 89
23
Example #1:
Find the zeros of f(x) = x2 − x − 6.
Let's use our calculator:
Page 89 Example #1
24
Let's use our calculator:
Y= → \Y1= x2-x-6
Page N/A
25
Let's use our calculator:
Y= → \Y1= x2-x-6
Zoom 6
gives:
Page N/A
Window
Xmin
Xmax
Xscl
Ymin
Ymax
Yscl
Value
-10
10
1
-10
10
1
Graph
26
-5
Page N/A
-4
-3
-2
8
7
6
5
4
3
2
1
0
-1 -1 0
-2
-3
-4
-5
-6
-7
-8
1
2
3
4
5
27
Now let's use
the TI to find the
zeros of this
quadratic
function:
Page N/A
-5
-4
-3
-2
8
7
6
5
4
3
2
1
0
-1 -1 0
-2
-3
-4
-5
-6
-7
-8
1
2
3
4
5
28
2nd
Trace
2: zero
Left Bound ?
Right Bound?
Guess?
Page N/A
29
-5
-4
zero
x=-2 y=0
Page N/A
-3
-2
8
7
6
5
4
3
2
1
0
-1 -1 0
-2
-3
-4
-5
-6
-7
-8
1
2
3
4
5
30
2nd
Trace
2: zero
Left Bound ?
Right Bound?
Guess?
Page N/A
31
-5
-4
zero
x=3 y=0
Page N/A
-3
-2
8
7
6
5
4
3
2
1
0
-1 -1 0
-2
-3
-4
-5
-6
-7
-8
1
2
3
4
5
32
Example #3
Figure 2.29 shows a graph of:
1 2
h( x )   x  2
2
What happens if we try to use
algebra to find its zeros?
Page 89 Example #3
33
Let's try to solve:
1 2
h( x )   x  2
2
Page 89
34
Page 90
1 2
h( x )   x  2
2
1 2
0   x 2
2
1 2
2
2   x  4  x
2
  4  x
35
Conclusion?
 4  x
Page 90
36
Conclusion?
 4  x
There are no real solutions, so h has
no real zeros. Look at the graph
again...
Page 90
37
What conclusion can we draw about
zeros and the graph below?
-4
Page 89
-3
-2
2
y
1
0
-1 -1 0
1
-2
-3
-4
-5
-6
-7
-8
-9
-10
x
2
3
4
38
h has no real zeros. This corresponds to
the fact that the graph of h does not
cross the x-axis.
-4
Page 89
-3
-2
2
y
1
0
-1 -1 0
1
-2
-3
-4
-5
-6
-7
-8
-9
-10
x
2
3
4
39
Let's use our calculator:
Y= → \Y1= (-1/2)x2-2
Page N/A
40
Let's use our calculator:
Y= → \Y1= (-1/2)x2-2
Window
Page N/A
Window
Xmin
Xmax
Xscl
Ymin
Ymax
Yscl
Value
-4
4
1
-10
2
1
Graph
41
2
y
1
0
-4
-3
-2
-1
-1 0
1
2
3
x
4
-2
-3
-4
-5
-6
-7
-8
-9
Page N/A
-10
42
2nd
Trace
2: zero
Left Bound ?
Right Bound?
Guess?
Page N/A
43
2nd
Trace
2: zero
Left Bound ?
Right Bound?
Guess?
ERR:NO SIGN CHNG
1:Quit
Page N/A
44
Concavity and Quadratic Functions
Page 90
45
Concavity and Quadratic Functions
Unlike a linear function, whose graph is a
straight line, a quadratic function has a graph
which is either concave up or concave down.
Page 90
46
Example #4
Let f(x) = x2. Find the average rate of
change of f over the intervals of length 2
between x = −4 and x = 4.
What do these rates tell you about the
concavity of the graph of f ?
Page 90 Example #4
47
Let f(x) = x2
Between x = -4 & x = -2:
f(-2)-f(-4) (2)  (4)
12


 6
2  (4)
2  4
2
2
Page 90
2
48
Let f(x) = x2
Between x = -2 & x = 0:
f(0)-f(-2) (0)  (2)
4


 2
0  (2)
2
2
2
Page 90
2
49
Let f(x) = x2
Between x = 0 & x = 2:
f(2)-f(0) (2)  (0)
4

 2
20
2
2
2
Page 90
2
50
Let f(x) = x2
Between x = 2 & x = 4:
f(4)-f(2) (4)  (2)
12


6
42
2
2
2
Page 90
2
51
Let's recap:
f(-2)-f(-4) ( 2)  ( 4)
12


 6
2  (4)
2  4
2
2
2
f(0)-f(-2) (0)  ( 2)
4


 2
0  (2)
2
2
2
f(2)-f(0) (2)  (0)
4

 2
20
2
2
2
2
f(4)-f(2) (4)  (2)
12


6
42
2
2
2
Page 90
2
2
52
What do these rates tell you about the
concavity of the graph of f ?
Page 90
53
What do these rates tell you about the
concavity of the graph of f ?
Since these rates are increasing, we expect
the graph of f to be bending upward. Figure
2.30 confirms that the graph is concave up.
Page 90
54
Page 90
55
Let's use our calculator:
Y= → \Y1= x2
2nd Mode = Quit
( Vars → Enter Enter (-2) - Vars →
Enter Enter (-4)) / (-2 - -4) Enter
Page N/A
56
Let's use our calculator:
Y= → \Y1= x2
2nd Mode = Quit
( Vars → Enter Enter (-2) - Vars →
Enter Enter (-4)) / (-2 - -4) Enter
-6
Page N/A
57
( Vars → Enter Enter (-2) - Vars →
Enter Enter (-4)) / (-2 - -4) Enter
-6
( Vars → Enter Enter (0) - Vars → Enter
Enter (-2)) / (0 - -2) Enter
-2
Page N/A
58
( Vars → Enter Enter (2) - Vars → Enter
Enter (0)) / (2- 0) Enter
2
( Vars → Enter Enter (4) - Vars → Enter
Enter (2)) / (4 - 2) Enter
6
Page N/A
59
Example #5
A high diver jumps off a 10-meter springboard.
For h in meters and t in seconds after the diver
leaves the board, her height above the water is
in Figure 2.31 and given by:
h  f (t )  4.9t  8t  10
2
(a) Find and interpret the domain and range of
the function and the intercepts of the graph.
(b) Identify the concavity.
Page 91 Example #5
60
Let's use our calculator:
Y= → \Y1= −4.9x2+8x+10
Page N/A
61
Let's use our calculator:
Y= → \Y1= −4.9x2+8x+10
Window
Page N/A
Window
Xmin
Xmax
Xscl
Ymin
Ymax
Yscl
Value
-2
5
1
-10
15
1
Graph
62
h  f (t )  4.9t  8t  10
2
Now let's use
the TI to find
the zeros of
this quadratic
function.
2nd MODE
15
10
5
2
0
1
1
2
3
4
5
x
5
Page N/A
10
63
2nd
Trace
2: zero
Left Bound ?
Right Bound?
Guess?
Page N/A
64
h  f (t )  4.9t  8t  10
2
Now let's use
the TI to find
the zeros of
this quadratic
function.
Zero
X= -.8290322
Y= 0
Page N/A
15
10
5
2
0
1
1
2
3
4
5
x
5
10
65
2nd
Trace
2: zero
Left Bound ?
Right Bound?
Guess?
Page N/A
66
h  f (t )  4.9t  8t  10
2
Now let's use
the TI to find
the zeros of
this quadratic
function.
Zero
X= 2.4616853
Y= 0
Page N/A
15
10
5
2
0
1
1
2
3
4
5
x
5
10
67
h  f (t )  4.9t  8t  10
2
So, our zeros
(solutions)
are:
15
10
X= -.8290322
Y= 0
X= 2.4616853
Y= 0
Which make
sense?
Page N/A
5
2
0
1
1
2
3
4
5
x
5
10
68
h  f (t )  4.9t  8t  10
2
Which make
sense? Since
t ≥ 0:
15
10
X= -.8290322
Y= 0
X= 2.4616853
Y= 0
5
2
0
1
1
2
3
4
5
x
5
Page 91
10
69
h  f (t )  4.9t  8t  10
2
X= 2.4616853
Y= 0
15
10
Domain?
5
2
0
1
1
2
3
4
5
x
5
Page 91
10
70
h  f (t )  4.9t  8t  10
2
X= 2.4616853
Y= 0
15
10
Domain?
The interval
of time the
diver is in the
air, namely
0 ≤ t ≤ 2.462.
Page 91
5
2
0
1
1
2
3
4
5
x
5
10
71
h  f (t )  4.9t  8t  10
2
X= 2.4616853
Y= 0
15
10
Range?
5
2
0
1
1
2
3
4
5
x
5
Page 91
10
72
h  f (t )  4.9t  8t  10
2
X= 2.4616853
Y= 0
15
10
Range?
Given that the
domain is
0 ≤ t ≤ 2.462,
what can f(t)
be?
Page 91
5
2
0
1
1
2
3
4
5
x
5
10
73
h  f (t )  4.9t  8t  10
2
X= 2.4616853
Y= 0
15
Range?
10
What you see
in yellow.
5
0
Page 91
0
1
2
3
74
h  f (t )  4.9t  8t  10
2
X= 2.4616853
Y= 0
15
Range?
10
What you see
in yellow.
What is the
maximum
value of f(t)?
5
0
Page 91
0
1
2
3
75
2nd
Trace
4: maximum
Left Bound ?
Right Bound?
Guess?
Page N/A
76
h  f (t )  4.9t  8t  10
2
X= 2.4616853
Y= 0
15
Range?
10
What is the
maximum
value of f(t)?
Maximum
X= .81632636
Y= 13.265306
Page 91
5
0
0
1
2
3
77
h  f (t )  4.9t  8t  10
2
X= 2.4616853
Y= 0
Therefore, the
range is:
0 ≤ f(t) ≤
13.265306
15
10
5
0
Page 91
0
1
2
3
78
h  f (t )  4.9t  8t  10
2
What are the
intercepts of
the graph?
15
10
5
0
Page 91
0
1
2
3
79
h  f (t )  4.9t  8t  10
2
What are the
intercepts of
the graph?
How can we
calculate?
15
10
5
0
Page 91
0
1
2
3
80
h  f (t )  4.9t  8t  10
2
What are the
intercepts of
the graph?
How can we
calculate?
We already
did
15
10
5
0
Page 91
0
1
2
3
81
h  f (t )  4.9t  8t  10
2
What are the
intercepts of
the graph?
How can we
calculate?
We already did
15
10
5
t= 2.4616853
f(t)= 0 horiz int.
0
Page 91
0
1
2
3
82
h  f (t )  4.9t  8t  10
2
What are the
intercepts of
the graph?
How can we
calculate?
What about?
15
10
5
0
Page 91
0
1
2
3
83
h  f (t )  4.9t  8t  10
2
What are the
intercepts of
the graph?
How can we
calculate?
Substitute 0
for t in the
above
equation...
Page 91
15
10
5
0
0
1
2
3
84
h  f (t )  4.9t  8t  10
2
What are the
intercepts of
the graph?
t= 0, f(t) = 10
vert int.
15
10
5
0
Page 91
0
1
2
3
85
Finally, let's identify the concavity.
Page 91
86
h  f (t )  4.9t  8t  10
2
What can we
say about
concavity?
15
10
5
0
Page 91
0
1
2
3
87
h  f (t )  4.9t  8t  10
2
What can we
say about
concavity?
15
10
Concave
down.
5
Let's confirm
via a table...
0
Page 91
0
1
2
3
88
Page 91
t (sec)
h (meters)
0
10
0.5
12.775
1.0
13.100
1.5
10.975
2.0
6.400
Rate of change
Δh/Δt
89
t (sec)
h (meters)
0
10
Rate of change
Δh/Δt
5.55
0.5
12.775
0.65
1.0
13.100
−4.25
1.5
10.975
−9.15
2.0
Page 91
6.400
90
t (sec)
h (meters)
0
10
Rate of change
Δh/Δt
5.55
0.5
12.775
0.65
1.0
13.100
−4.25
1.5
2.0
Page 91
10.975
6.400
−9.15
decreasing Δh/Δt
91
End of Section 2.6
92