Chabot Mathematics
§8.3 Quadratic
Fcn Applications
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Review § 8.3
MTH 55
 Any QUESTIONS About
• §8.3 → Quadratic Eqn Graphs
 Any QUESTIONS About HomeWork
• §8.3 → HW-39
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Parabolas with Vertical Shifts
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Parabolas with Vertical Shifts
 The Graph of
F(x) = x2 + k has the
same SHAPE as
f(x) = x2, with the
shape shifted
VERTICALLY:
k > 0 produces
shift up k units
• k units UP when k > 0
– i.e.; k is POSITIVE
• |k| units DOWN
when k < 0
k < 0 produces
shift down k units
– i.e., k is NEGATIVE
• The Vertex is (0, k)
Chabot College Mathematics
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Parabolas with Horizontal Shifts
Chabot College Mathematics
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Parabolas with Horizontal Shifts
 The Graph of
F(x) = (x − h)2 has the
same SHAPE as
f(x) = x2, with the shape
shifted HORIZONTALLY:
h > 0 produces
shift right h units
• h units RIGHT when
h>0
– POSTIVE h
• |h| units LEFT
when h < 0
– NEGATIVE h
h < 0 produces
shift left h units
• The Vertex is (h,0)
Chabot College Mathematics
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Caveat: Shifting Parabolas
 Horizontal Shifts are EASY to move in
the WRONG Direction
 To determine the size & direction of a
Horizontal shift, find the value of x that
makes x−h = 0. Some Examples
• F(x) = (x − 5)2 shifts RIGHT 5-units as
x = +5 causes x − 5 to be zero
• F(x) = (x + 7)2 shifts LEFT 7-units as
x = −7 causes x + 7 to be zero
Chabot College Mathematics
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Parabolas with ↨ & ↔ Shifts
Chabot College Mathematics
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  F(x) = (x+3)2 − 2
 Cast F(x) = (x+3)2 − 2 into
the form F(x) = (x−h)2 + k
 Since h < 0, there is a shift to the left,
and since k < 0, there is a shift down.
Chabot College Mathematics
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Graphs of f(x) = a(x – h)2 + k
 The Graph of f(x) = a(x – h)2 + k has the
same shape as the graph of y = a(x – h)2
• If k is positive, the graph of y = a(x – h)2 is
shifted k units up.
• If k is negative, the graph of y = a(x – h)2 is
shifted |k| units down
• The vertex is (h, k), and the
axis of symmetry is x = h.
• For a > 0, k is the MINimum function value
• For a < 0, k is the MAXimum function value
Chabot College Mathematics
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
1
2
f
(
x
)


(
x

3)
 1.
Example  Graph
2
 SOLUTION: Make T-Table, ID Vertex
y
and Maximum
LoS
x
y
3
2
(x, y)
maximum (−1)1
0
–1
–2
–3
–4
–5
-11/2
–3
–3/2
–1
–3/2
–3
(0, -11/2)
(–1, –3)
(–2, –3/2)
(–3, –1)
(–4, –3/2)
(–5, –3)
-5 -4 -3 -2 -1
-1
-2
-3
-4
-5
-6
1
2 3 4 5
Concave
DOWN
-7
-8
vertex
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
x
Maximum & Minimum Probs
 We have seen that for any quadratic
function f, the value of f(x) at the vertex
is either a maximum or a minimum.
 Thus problems in which a quantity must
be maximized or minimized can be
solved by finding the coordinates of
the vertex, assuming the problem can
be modeled with a quadratic function.
Chabot College Mathematics
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Maximum

A farmer has 200 ft of fence with
which to form a rectangular pen on his
farm. If an existing fence forms one
side of the rectangle, what dimensions
will MAXIMIZE the size of the area?
1. Familiarize - make a drawing and
label it, letting
•
w = Rectangle Width
•
l = Rectangle Length
Chabot College Mathematics
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w
l
Existing fence
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
w
Example  Maximum
 Recall that for Rectangles
• Area = lw
• Perimeter = 2w + 2l
 Since the existing fence forms one
length of the rectangle, the fence will
comprise three sides. Thus
2w  l  200
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Maximum
2. Translate. Now have two equations:
One guarantees that all 200 ft of fence
will be used; the other expresses area
in terms of length and width.
2w  l  200
w
l
A  lw
w
Existing fence
Chabot College Mathematics
15
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Maximum
3. CarryOut - Now need to express A as
a function of l or w but not both. To do
so, we solve for l in the first equation
to obtain l = 200 – 2w. Substituting for
l in the second equation, yields a
quadratic function.
A 

Chabot College Mathematics
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200  2ww
200w  2w
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Maximum
 Factoring and completing the square to
2
Obtain
A  2w  50  5000
 Then by the
Max-at-Vertex Criteria:
wmax  50
 Find lmax from the Perimeter Constraint
lmax  200  2wmax  100
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Maximum
4. Check - The check is left for us to do
Later.
5. State - The dimensions for the largest
rectangular area for the pen that can
be enclosed is 50 ft by 100 ft.
100’
50’
50’
Existing fence
Chabot College Mathematics
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Amax



200  2wmax wmax
200  2  5050
2
5000 ft 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Sniping Siblings
 A widower with 10 children marries a
widow who also has children. After their
marriage, they have their own children.
If the total number of children is 24, and
we assume that the children of the
same parents do not fight.
 Find the maximum possible number of
fights among the children.
• In this example, a fight between Sean and
Misty, no matter how many times they
fight, is considered as ONE fight.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Sniping Siblings
 SOLUTION: Suppose the widow had x
number of children before marriage. Then
the couple has 24 − 10 − x = 14 − x
additional children after their marriage.
 Since the children of the same parents do
not fight, there are no fights among the 10
children the widower brought into the
marriage, or among the x children the
widow brought into the marriage, or
among the 14 − x children of the parents
Chabot College Mathematics
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Sniping Siblings

Then The possible number of fights
among the children of
1. the widower (10 children) and the widow
(x children) is 10x.
2. the widower (10 children) and the couple
(14 – x children) is 10(14 – x), and
3. the widow (x children) and the couple
(14 – x children) is x(14 – x)
Chabot College Mathematics
21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Sniping Siblings
 TRANSLATE: The possible number y of
all fights is given by
y  10x  10 14  x   x 14  x 
 10x  140  10x  14x  x 2
 140  14x  x 2
 In the Quadratic y = f (x) = –x2 +14x + 140
Function:
• a = –1, b = 14, and c = 140.
Chabot College Mathematics
22
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Sniping Siblings
 The vertex (h, k) is given by
b
14
h

7
2a
2 1
k  f 7    7   14 7   140  189
2
 Since, a = −1 < 0, the function f opens
DOWN and has MAXIMUM value k.
Hence, the maximum possible number
of fights among the children is 189.
Chabot College Mathematics
23
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Rocket Ballistics
 A Model Rocket is launched straight up
with an initial velocity of 60 feet per sec.
 The equation h ≈ −16t2 + 60t describes
the height, h, of the rocket, t seconds
after launch, FIND:
• the maximum height that the rocket
reaches.
• the amount of time that the rocket is in the
air; i.e., find the total flite-time
Chabot College Mathematics
24
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Rocket Ballistics
 SOLUTION - Since the graph of
h ≈ −16t2 + 60t is a parabola that opens
down, the maximum height occurs at
its vertex:
b
60
t

 1.875
2a
2(16)
 Use the quadratic equation to find the
height at the vertex time of 1.875 sec
h = −16(1.875)2 + 60(1.875) = 56.25
Chabot College Mathematics
25
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Rocket Ballistics
 SOLUTION: for h ≈ −16t2 + 60t the
Vertex is at (1.875, 56.25)
Time at Which
Max Height
Occurs
Max Height
 Interpreting Vertex Information find:
• The maximum height is 56.25 feet,
• The max height occurs 1.875 seconds
after rocket launch
Chabot College Mathematics
26
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Rocket Ballistics
 The rocket Flite-Time is from launch
until it returns to the ground.
 At launch and upon returning to the
ground, the rocket’s height is 0, so we
need to find t when h = 0
0 = −16t2 + 60t
0 = − 4t(4t − 15)
−4t = 0
or
4t − 15 = 0
t=0
or
t = 3.75
Chabot College Mathematics
27
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Rocket Ballistics
 The height is 0 when t = 0 and
when t = 3.75 seconds, so the rocket
is in the air for
3.75 seconds.
Max Hgt ≈ 56ft
 Check by
Graphing
h = −16t2 + 60t
Crash-Down
@ ≈ 3.8 sec
Chabot College Mathematics
28
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Graph InEquality
 Graph the quadratic function
f(x) = x2 + 2x + 2 and solve each
quadratic inequality.
a. x2 + 2x + 2 > 0
b. x2 + 2x + 2 < 0
 SOLUTION – Analyze Eqn Parameters
• Step 1: a = 1, b = 2, and c = 2
• Step 2: a = 1 > 0, the parabola opens UP
• Step 3: Find (h, k) → Next Slide
Chabot College Mathematics
29
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Graph InEquality
b
2
b
2
h





1
 Find h by Formula h  



1
2a
2 1 2 1
2a
2
 Find k = f(h) k  f 1

1

2
1
 1




k  f 1  1  2 21
2
 With the Vertex of (−1,1) The Max for
f(x) = 1, which occurs at x = −1
 Next find x-intercepts → f(x) = 0
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Graph InEquality
 Setting f(x) = 0:
x
2 
x 2  2x  2  0
2   4 12  2 

2 1
2
2
4
x-intercepts are  1  i .
 The solutions are not real, the graph
does not intersect the x-axis
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Graph InEquality
 Find y-intercept by setting x = 0 in f(x)
f 0   0   2 0   2  2
2
y-intercept is 2 .
 From the (−1,1) Vertex recognize the
Line of Symmetry (LoS) at x = −1
• Moving 1-unit to the Left & Right of the LoS
produces points (−2, 2), and (0, 2) which
are symmetric with respect to the axis of
symmetry
Chabot College Mathematics
32
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Graph InEquality
 To Draw Graph use
• Opens UP
• Vertex (−1,1)
• NO x-intercepts
• y-intercept = 2
• Pts Symmetric
about the LoS
– (−2, 2)
– (0, 2)
Chabot College Mathematics
33
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Graph InEquality
 Use Graph to
Assess InEqualities

The entire graph
lies above the
x-axis. Thus,
y is always > 0.
a) x2 +2x + 2 > 0 is
always true, the
solution is (−∞,∞)
b) x2 +2x + 2 < 0 is
never true, the
solution is Ø
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Fitting Quadratic Functions
 Whenever a certain quadratic
function fits a situation, that
function can be determined if
THREE inputs and their outputs
are known.
 Each of the given ordered pairs is
called a DATA POINT
Chabot College Mathematics
35
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Quadratic Fitting
 Use the data points (0, 10.4), (3, 16.8),
and (6, 12.6) to find a quadratic function
that fits the data.
 SOLUTION – Need to Find a function of
the form f(x) = ax2 + bx + c given that
f(0) = 10.4, f(3) = 16.8, and f(6) = 12.6
 Thus Need
a(0)2 + b(0) + c = 10.4
a, b, & c
a(3)2 + b(3) + c = 16.8
Such That:
a(6)2 + b(6) + c = 12.6
Chabot College Mathematics
36
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Quadratic Fitting
 This amounts to Solving a SYSTEM of
Three Eqns for Unknowns a, b, & c
c = 10.4
9a + 3b + c = 16.8
36a + 6b + c = 12.6
(1)
(2)
(3)
 Substituting c = 10.4 into eqns (2) & (3)
and solving the resulting system yields
53
39
a    0.59, b 
 3.9.
90
10
Chabot College Mathematics
37
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Example  Quadratic Fitting
 Thus the data set (0, 10.4), (3, 16.8),
and (6, 12.6) produces a Quadratic Fit:
53 2
f ( x)   x  3.9 x  10.4.
90
Chabot College Mathematics
38
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
WhiteBoard Work
 Problems
From §8.3
Exercise Set
• 58, 68
 A Quadratic
(and Linear)
Fit for Fish
Chabot College Mathematics
39
Relationship between centrum radius and precaudal length
for eastern North Pacific salmon sharks (Lamna ditropis),
showing significant fits given by linear and quadratic
equations (sexes combined, n=182).
PCL = precaudal length, CR = centrum radius
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
All Done for Today
Quadratic
Production
Function
Chabot College Mathematics
40
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
41
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Chabot College Mathematics
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5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
4
5
6
5
5
y
4
4
3
3
2
2
1
1
0
-10
-8
-6
-4
-2
-2
-1
0
2
4
6
-1
0
-3
x
0
1
2
3
4
5
-2
-1
-3
-2
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
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-4
M55_§JBerland_Graphs_0806.xls
-5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt
8
10