1

Powerpoint slides copied from or based upon:

Connally,

Hughes-Hallett,

Gleason, Et Al.

Copyright 2007 John Wiley & Sons, Inc.

Functions Modeling Change

A Preparation for Calculus

Third Edition

SECTION2.6  

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:

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Let's use our calculator:Y= → \Y1= −16x2+64x+3

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Let's use our calculator:Y= → \Y1= −16x2+64x+3

Window ValueXmin -1Xmax 5Xscl 1Ymin -10Ymax 80Yscl 8

Window →

Graph5Page N/A

Although the path of the ball is straight up and down, the graph of its height as a function of time is concave down.

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The ball goes up fast at first and then more slowly because of gravity.

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The baseball height function is an example of a quadratic function, whose general form is y = ax2 + bx + c. 

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Finding the Zeros of a Quadratic Function

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Finding the Zeros of a Quadratic Function

Back to our baseball example, precisely when does the ball hit the ground?

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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?

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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.

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2

2

0 16 64 3

4

2

t t

b b act

a

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2

2

0 16 64 3

4

216

64

3

t t

b b act

aa

b

c

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2

2

2

0 16 64 3

4

2

64 64 4( 16)(3)

2( 16)

t t

b b act

a

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264 64 4( 16)(3)

2( 16)

64 4096 64(3)

32

64 4288

32

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64 4288

3264 65.48282217

32129.48282217

324.046338193

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Let's use our calculator:Y= → \Y1= −16x2+64x+3

Window ValueXmin -1Xmax 5Xscl 1Ymin -10Ymax 80Yscl 8

Window →

Graph18Page N/A

Now let's use the TI to find the zeros of this quadratic function:

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2nd Trace 2: zero

Left Bound ?

Right Bound?

Guess?

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zeroX=4.0463382Y=-1E-11

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Example #1:

Find the zeros of f(x) = x2 − x − 6.

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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

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Example #1:

Find the zeros of f(x) = x2 − x − 6.

Let's use our calculator:

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Let's use our calculator:Y= → \Y1= x2-x-6

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Let's use our calculator:Y= → \Y1= x2-x-6

Window ValueXmin -10Xmax 10Xscl 1Ymin -10Ymax 10Yscl 1

Zoom 6

gives:

Graph26Page N/A

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Now let's use the TI to find the zeros of this quadratic function:

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2nd Trace 2: zero

Left Bound ?

Right Bound?

Guess?

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zerox=-2 y=0

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2nd Trace 2: zero

Left Bound ?

Right Bound?

Guess?

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zerox=3 y=0

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Example #3

Figure 2.29 shows a graph of:

What happens if we try to use algebra to find its zeros?

21( ) 2

2h x x

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Let's try to solve:

21( ) 2

2h x x

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2

2

2 2

1( ) 2

21

0 221

2 42

4

h x x

x

x x

x

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Conclusion?

4 x

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Conclusion?

There are no real solutions, so h has no real zeros. Look at the graph

again...

4 x

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x

y

What conclusion can we draw about zeros and the graph below?

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x

y

h has no real zeros. This corresponds to the fact that the graph of h does not

cross the x-axis.

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Let's use our calculator:Y= → \Y1= (-1/2)x2-2

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Let's use our calculator:Y= → \Y1= (-1/2)x2-2

Window ValueXmin -4Xmax 4Xscl 1Ymin -10Ymax 2Yscl 1

Window

Graph41Page N/A

x

y

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2nd Trace 2: zero

Left Bound ?

Right Bound?

Guess?

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2nd Trace 2: zero

Left Bound ?

Right Bound?

Guess?

ERR:NO SIGN CHNG1:Quit

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Concavity and Quadratic Functions

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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.

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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 ?

47Page 90 Example #4

Let f(x) = x2

Between x = -4 & x = -2:

48

2 2f(-2)-f(-4) ( 2) ( 4) 126

2 ( 4) 2 4 2

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Let f(x) = x2

Between x = -2 & x = 0:

49

2 2f(0)-f(-2) (0) ( 2) 42

0 ( 2) 2 2

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Let f(x) = x2

Between x = 0 & x = 2:

50

2 2f(2)-f(0) (2) (0) 42

2 0 2 2

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Let f(x) = x2

Between x = 2 & x = 4:

51

2 2f(4)-f(2) (4) (2) 126

4 2 2 2

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Let's recap:

52

2 2

2 2

2 2

2 2

f(-2)-f(-4) ( 2) ( 4) 126

2 ( 4) 2 4 2

f(0)-f(-2) (0) ( 2) 42

0 ( 2) 2 2

f(2)-f(0) (2) (0) 42

2 0 2 2

f(4)-f(2) (4) (2) 126

4 2 2 2

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What do these rates tell you about the concavity of the graph of f ?

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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.

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Let's use our calculator:Y= → \Y1= x2

2nd Mode = Quit

( Vars → Enter Enter (-2) - Vars → Enter Enter (-4)) / (-2 - -4) Enter

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( Vars → Enter Enter (-2) - Vars → Enter Enter (-4)) / (-2 - -4) Enter

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( Vars → Enter Enter (0) - Vars → Enter Enter (-2)) / (0 - -2) Enter

-2

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( Vars → Enter Enter (2) - Vars → Enter Enter (0)) / (2- 0) Enter

2

( Vars → Enter Enter (4) - Vars → Enter Enter (2)) / (4 - 2) Enter

6

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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:

(a)  Find and interpret the domain and range of the function and the intercepts of the graph.

(b)  Identify the concavity.

60

2( ) 4.9 8 10h f t t t

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Let's use our calculator:Y= → \Y1= −4.9x2+8x+10

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Let's use our calculator:Y= → \Y1= −4.9x2+8x+10

Window ValueXmin -2Xmax 5Xscl 1Ymin -10Ymax 15Yscl 1

Window

Graph62Page N/A

63

2( ) 4.9 8 10h f t t t

x2 1 0 1 2 3 4 5

10

5

5

10

15Now let's use the TI to find the zeros of this quadratic function.

2nd MODE

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2nd Trace 2: zero

Left Bound ?

Right Bound?

Guess?

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2( ) 4.9 8 10h f t t t

x2 1 0 1 2 3 4 5

10

5

5

10

15Now let's use the TI to find the zeros of this quadratic function.

ZeroX= -.8290322Y= 0

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2nd Trace 2: zero

Left Bound ?

Right Bound?

Guess?

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2( ) 4.9 8 10h f t t t

x2 1 0 1 2 3 4 5

10

5

5

10

15Now let's use the TI to find the zeros of this quadratic function.

ZeroX= 2.4616853Y= 0

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2( ) 4.9 8 10h f t t t

x2 1 0 1 2 3 4 5

10

5

5

10

15So, our zeros (solutions) are:

X= -.8290322Y= 0

X= 2.4616853Y= 0

Which make sense?Page N/A

69

2( ) 4.9 8 10h f t t t

x2 1 0 1 2 3 4 5

10

5

5

10

15Which make sense? Since t ≥ 0:

X= -.8290322Y= 0

X= 2.4616853Y= 0

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70

2( ) 4.9 8 10h f t t t

x2 1 0 1 2 3 4 5

10

5

5

10

15X= 2.4616853Y= 0

Domain?

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2( ) 4.9 8 10h f t t t

x2 1 0 1 2 3 4 5

10

5

5

10

15X= 2.4616853Y= 0

Domain?

The interval of time the diver is in the air, namely 0 ≤ t ≤ 2.462.

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2( ) 4.9 8 10h f t t t

x2 1 0 1 2 3 4 5

10

5

5

10

15X= 2.4616853Y= 0

Range?

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2( ) 4.9 8 10h f t t t

x2 1 0 1 2 3 4 5

10

5

5

10

15X= 2.4616853Y= 0

Range?

Given that the domain is 0 ≤ t ≤ 2.462,what can f(t) be?

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74

2( ) 4.9 8 10h f t t t X= 2.4616853Y= 0

Range?

What you see in yellow.

0 1 2 30

5

10

15

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2( ) 4.9 8 10h f t t t X= 2.4616853Y= 0

Range?

What you see in yellow.

What is the maximum value of f(t)?

0 1 2 30

5

10

15

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2nd Trace 4: maximum

Left Bound ?

Right Bound?

Guess?

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2( ) 4.9 8 10h f t t t X= 2.4616853Y= 0

Range?

What is the maximum value of f(t)?

MaximumX= .81632636Y= 13.265306

0 1 2 30

5

10

15

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2( ) 4.9 8 10h f t t t X= 2.4616853Y= 0

Therefore, the range is:

0 ≤ f(t) ≤ 13.265306

0 1 2 30

5

10

15

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2( ) 4.9 8 10h f t t t What are the intercepts of the graph?

0 1 2 30

5

10

15

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80

2( ) 4.9 8 10h f t t t What are the intercepts of the graph?

How can we calculate?

0 1 2 30

5

10

15

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81

2( ) 4.9 8 10h f t t t What are the intercepts of the graph?

How can we calculate?

We already did

0 1 2 30

5

10

15

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82

2( ) 4.9 8 10h f t t t What are the intercepts of the graph?

How can we calculate?

We already did

t= 2.4616853f(t)= 0 horiz int.

0 1 2 30

5

10

15

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83

2( ) 4.9 8 10h f t t t What are the intercepts of the graph?

How can we calculate?

What about?

0 1 2 30

5

10

15

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84

2( ) 4.9 8 10h f t t t What are the intercepts of the graph?

How can we calculate?

Substitute 0 for t in the above equation...

0 1 2 30

5

10

15

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2( ) 4.9 8 10h f t t t What are the intercepts of the graph?

t= 0, f(t) = 10vert int.

0 1 2 30

5

10

15

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Finally, let's identify the concavity.

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87

2( ) 4.9 8 10h f t t t What can we say about concavity?

0 1 2 30

5

10

15

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88

2( ) 4.9 8 10h f t t t What can we say about concavity?

Concave down.

Let's confirm via a table...

0 1 2 30

5

10

15

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89

t (sec) h (meters) Rate of change Δh/Δt

0 10     

0.5 12.775     

1.0 13.100     

1.5 10.975     

2.0 6.400  Page 91

90

t (sec) h (meters) Rate of change Δh/Δt

0 10      5.55

0.5 12.775      0.65

1.0 13.100      −4.25

1.5 10.975      −9.15

2.0 6.400  Page 91

91

t (sec) h (meters) Rate of change Δh/Δt

0 10      5.55

0.5 12.775      0.65

1.0 13.100      −4.25

1.5 10.975      −9.15

2.0 6.400 decreasing Δh/Δt Page 91

End of Section 2.6

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