# CHAPTER 3 - QUADRATIC FUNCTIONS by mujB2kr

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• pg 1
```									                                                      CHAPTER 3 – QUADRATIC FUNCTIONS

DAY 1 and ½ of Day 2

Function
 a set of ordered pairs in which for every x, there is only one y

y  2 x 2  3x  5                                                                             0  2 x2  5x  3
General Form                                                           Standard Form (a.k.a completed square form)
y = f(x) = ax2 + bx + c                                                y = f(x) = a(x  h)2 + k
a0                                                                 10 a  0
Y

9
*    The graph is a parabola. (label vertex, x–int, y–int, axis of symmetry, x = __)
Y                                           8
10                                                              7
9                                                              6
highest or lowest point                 y=0               x=0          x value of vertex
8                                                              5
7                                                              4
6                                                              3
5                                                              2
4                                                              1
X
3
2              -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1
1                                                 -2
X
-3
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1                                                     -4
-2                                                     -5
-3                                                     -6
-4                                                     -7
-5                                                     -8
-6
minimum function                                     maximum function
-9
-7                                                   -10
-8
Domain: set of first members of the ordered pairs of a relation
-9                   Created with an unregistered version of Advanced Grapher - http:/ / www.serpik.com/ agrapher/
Range: -10
set of second members of the ordered pairs of a relation
* Note alphabetic nature (Domain, Range), (x, y), (Horizontal, Vertical), (Across, Up)
with an unregistered version of Advanced Grapher - http:/ / www.serpik.com/ agrapher/
Analyze y = a(x  h) + k    2

Examples:

1.   y = x2                                                         6.  y = x2
2.   y = x2 + 3                                                     7.  y = (x  3)2
3.   y = x2  5                                                     8.  y = (x + 5)2
4.   y = 2x2                                                        9.  y = 2(x  4)2  7
1
5. y = 0.3x2                                                        10. y =  (x + 5)2 + 4
2

*    vertex: (h, k)
*    axis of symmetry: x = h
*    "a" value: affects width, the larger the a is, the narrower the graph
determines direction of opening, + a opens up,  a opens down

Page 1                       Math 20 IB Ch 3 Quadratic Functions
Examples:          (Do about 4 of the 10)(Focus is on using 2 nd Calc)

Find:

a.    vertex                                   2nd calc max or min
b.    equation of axis of symmetry             x = value from vertex
c.    direction of opening
d.    width compared to y = x2                 Same, Wider or Narrower
e.    domain & range                           always x  R , y  or y  value from vertex
f.    max. or min. and value                   vertex
g.    y–int                                    2nd calc value x=0
h.    x–int(s)  graphing calculator for most questions       2nd calc zero

* find x–intercepts without calculator.
** find x–intercepts with calculator.

Direction    Width         Domain   Max or
Axis of               Compare
Vertex                  Of                       And      Min     y–int   x–int
Sym.
to y  x
2
opening                    Range    Value
*   1. y = x2 + 5

2. y = 3x2

3. y = 2x2  4

*   4. y = (x  2)2

1
*   5. y =        (x + 3)2
2

6. y = 2(x         5 )2  3

7. 2y  8 = 6(x + 3)2

8. y =        3 (x  1)2 +     3

** 9. y = 5x2  7x  11
(calculator)

**10. y =  2     x2 + 17x + 
(calculator)

Page 2                   Math 20 IB Ch 3 Quadratic Functions

Width
Direct   Compare
Axis of                       Domain     Max or Min
Vertex                   Of        to                                     y–int      x–int
Sym.                        And Range     Value
open
yx   2

2
*1. y = x + 5                  (0,5)         x 0        Up     S          xR            Min           (0,5)      none
y 5          (0,5)
2. y = 3x2                     (0,0)         x 0        Up      N         xR            Min           (0,0)      (0,0)
y 0          (0,0)
3. y = 2x2  4               (0,–4)         x 0     Down       N         xR            Max          (0,–4)      none
y  –4        (0,–4)
*4. y = (x  2)2               (2,0)         x 2        Up      S         xR            Min           (0,4)      (2,0)
y 0          (2,0)
1                 (–3,0)        x  –3    Down       W         xR            Max          (0,–4.5)   (–3,0)
*5. y =      (x + 3)2
2                                                               y 0         (–3,0)

6. y = 2(x  5 )2  3         5, 3       x 5         Up      N         xR
y  –3
Min           (0,7)     (1.01,0)

    5, 3                  (3.46,0)

7. 2y  8 = 6(x + 3)2        (–3,4)        x  –3    Down       N        xR             Max          (0,–23)    (–4.15,0)
y 4          (–3,4)                   (–1.85,0)

8.y = 3 (x                   1, 3         x 1        Up      N        xR             Min          (0,3.46)    None
1)2+ 3                                                                    y 3

1, 3 
** 9.y=5x27x11 (0.7,–13.45)               x  0.7      Up      N         xR            Min          (0,–11)    (–0.94,0)
(calculator)                                                                y
–13.45   (0.7,–13.45)          (2.34,0)
**10.                    (6.01,54.23)       x  6.01 Down        N         xR         Max      (0,3.14) (–0.18,0)
2
y = 2 x +17x +                                                          y  54.23 (6.01,54.23)           (12.2,0)
 (calculator)

Page 3                   Math 20 IB Ch 3 Quadratic Functions
Assignment
pg. 109 # 1, 4, 19, (33, 34)  includes x & y–int
pg. 119 # 3, 5, 17, 21, 25, (39, 40)  includes x & y–int

Vertex          A.S.       Range        y–int          x–int       max or     Width
min &    compared
value     to y = x2
y = (x  11 )2

1
y=    (x  2)2 + 2
2
y = 3(x + 1)2 + 4

1
y=    (x  6)2  8
2
y = (x + 2)2 + 2

y = 4x2  12x + 9
calculator only
y = 0.6x2 + 5x + 13
calculator only
Questions
2
Y = x2  7x  13
5
y = 7x2 + 11x + 23

1. Calculate the x and y intercepts for y = 3(x + 4)2  12 (algebraically & graphically)

x–int (–2,0) (–6,0)     y–int (0,36)

2. Calculate the x and y intercepts for y = 4(x  1)2 + 1 (algebraically & graphically)

x–int (0.5,0) (1.5,0)    y–int (0,–3)

2
(        )
3. Find exact y–intercept for y = - 2x x + 2 5 + 7 11 .                            (
y–int 0, 7 11   )

Page 4                 Math 20 IB Ch 3 Quadratic Functions

Vertex         A.S.        Range         y–int         x–int      max or       Width
min &      compared
value       to y = x2
y = (x  11 )2           ( 11 , 0)      x = 11        y0         (0, 11)      ( 11 , 0)    min (0)         S
1                    (2, 2)         x=2        y  2     (0, 2 + 2)      none       min (2)       W
y = (x  2)2 + 2
2
y = 3(x + 1)2 + 4        (1, 4)       x = 1        y4          (0, 1)      (0.15, 0),   max (4)        N
(2.15, 0)

1                     (6, 8)        x=6         y  8       (0, 10)       (10, 0),  min (8)        W
y=    (x  6)2  8                                                               (2, 0)
2
y = (x + 2)2 + 2        (2, 2)      x = 2       y  2     (0, 4 + 2)   (0.51, 0), max (2)         S
(4.51, 0)
y = 4x2  12x + 9         (1.5, 0)      x = 1.5       y0          (0, 9)       (1.5, 0)  min (0)          N
calculator only
y = 0.6x2 + 5x + 13   (4.17, 23.42)    x = 4.17    y  23.42     (0, 13)      (2.08, 0)     max         W
calculator only                                                                (10.41, 0)   (23.42)
Questions
2                (8.75, 43.63)                                          (1.69, 0)
Y = x2  7x  13
5                                                                        (19.19, 0)
y = 7x2 + 11x + 23    (0.79, 27.32)                                            (2.76, 0)
(1.19, 0)

Page 5                   Math 20 IB Ch 3 Quadratic Functions
DAY ½ of Day 2 and Day 3

Determining the Equation of a Quadratic Function Given Some Information               y = a(x  h)2 + k

Examples:

1. a = 5 and vertex is (4, 1)                                                ANS y  5( x  4) 2  1

y  a ( x  h) 2  k
y  5( x   4) 2  1
y  5( x  4) 2  1

2. congruent to y = 3x2, opens down, and vertex is (4, 9)                    ANS y  3( x  4) 2  9

congruent means has the same “a” value (same shape)
y  a ( x  h) 2  k
y  3( x  4)2  9            opens down thus  a

2
ANS y       x  2  5
2
3. Vertex is (2, 5) and passes thru (1, 11)                    must find a
3
y  a ( x  h) 2  k
y  a ( x  2)  5
find a
11  a (1  2) 2  5
11  a (3) 2  5
6  9a
2
a
3
2
thus y       ( x  2) 2  5
3

4. congruent to y = 6x2, range is y  3, and axis of symmetry is x = 5      ANS y  6( x  5)2  3

y  a ( x  h) 2  k
y  6( x  5)2  3            opens down since y  3,  a

Page 6                Math 20 IB Ch 3 Quadratic Functions
5. axis of symmetry is x = 2, passes thru (3, 20) and (4, 22)                                  ANS y = 2(x + 2)2  30

y  a ( x  h) 2  k
y  a ( x  2) 2  k

20  a (3  2) 2  k                    22  a (4  2) 2  k
20  a (5) 2  k                        22  a (2) 2  k
20  25a  k                            22  4a  k

20  25a  k
eqn 2  1          22  4a  k
42  21a
a2                 substitute a into either
equation to get k
k  30

y  2( x  2) 2  30

6. quadratic function passes thru (1, 12), (1, 6), and (2, 9)
* use y = ax2 + bx + c

12 = a  b + c    18 = 2a + 2c                            21 = 3c
6 = a + bY c
+      33 = 6a + 3c                           c = 7, a = 2, b = 3
10
9 = 4a + 2b + c
9
8
7                                                                                       ANS y = 2x2  3x + 7
6
7.   Given5the     following graph, find the equation.
4
3
y  a ( x  3) 2  1
2
1
X       1  a(2) 2  1
0 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
-1         1 2 3 4 5 6 7 8 9 10                        1  4a  1
-2                                                     2  4a
-3
1
-4                                                     a
-5                                                         2
-6
-7
ANS y = 0.5(x  3)2 + 1
-8
-9
-10

reated with an unregistered version of Advanced Grapher - http:/ / www.serpik.com/ agrapher/

Page 7                         Math 20 IB Ch 3 Quadratic Functions
Assignment
pg. 109 # 48, 50, 57, 62
pg. 119 # 47, 48, 53, 54, 60, 67, 75

plus:

1. quadratic function passes thru (1, 9), (2, 7), and (2, 9). Find the equation in general form.
ANS y = 2x2 + 4x + 7

Y
10
9
1.   Write an equation for each parabola:
8
3
a. vertex7is (0, ) passing thru (3, 3)
6      2
5                                                                                          1 2 3
4                                                                              ANS y =       x +
3                                                                                          6     2
2
b.       1
X

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
-1        1 2 3 4 5 6 7 8 9 10
-2
-3
-4
-5
-6
-7
-8
-9
-10                                                                                       1
Y                                                 ANS y =  (x  3)2  2
10
Created with an unregistered version of Advanced Grapher - http:/ / www.serpik.com/ agrapher/                         3
9
c.                                8
7
6
5
4
3
2
1
X

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1
-2
-3
-4
-5                                                         ANS y = (x + 4)2 + 1
-6
d. passing thru (1, 3), (1, 7), and (2, 8)
-7
-8                                                         ANS y = 3x2  2x + 8
-9
-10
e. vertex is (2, 1), y–int is 13
ANS
Created with an unregistered version of Advanced Grapher - http:/ / www.serpik.com/ agrapher/   y = 3(x + 2)2  1

Page 8                            Math 20 IB Ch 3 Quadratic Functions
f. congruent to y = 0.4x2, maximum at (2, 3)
ANS y = 0.4(x  2)2  3

g. axis of symmetry is x = 6, range is y  2, congruent to y = 4x2
ANS y = 4(x + 6)2 + 2

h. minimum at (3, 2), passes thru (5, 6)
1
ANS y =      (x + 3)2  2
8

2. Two points on a parabola are (5, 2) and (9, 2). What is the equation of the axis of symmetry?
(Can not be done with equation. Must use logic and symmetry.)              ANS x = 2

3. Find the value of "a" and "q" for y = a(x + 3)2 + q passing thru (1, 2) and (3, 34).
ANS a = 1, q = 2

4. The vertex is (3, 12), x–int is 5, what is the other x–int?
ANS 1

y  a ( x  3) 2  12
0  a (5  3) 2  12                    X int of 5 means (5,0) for x and y
Solve for a
12  a (2) 2
12  4a
a  3

y  3( x  3) 2  12                   Rewrite equation with a value.
Insert 0 for y to solve for x intercepts.
0  3( x  3) 2  12
12  3( x  3) 2
4  ( x  3) 2
4  x3                               Square root of 4 is positive 2 and negative 2.
2  x3           2  x  3
x5               x 1                  X intercepts are 5 (known) and 1.

Page 9                    Math 20 IB Ch 3 Quadratic Functions
DAY 4 and ½ of Day 5

Graphing without a calculator

The graph of y = x2 is on a vertex (0, 0) and the points (1, 1), ( 2, 4) and (3, 9).
The graph of y = ax2 is on a vertex (0, 0) and the points (1, 1a), and( 2, 4a).
The graph of y = a(x – p)2 + q has the same shape as the graph of y = ax2 but we go 1 unit left or right of
the vertex and ‘a’ units up; 2 units left or right and ‘4a’ units up.
Sketch y = 2x2       and y = 2(x  3)2  11 and y = – 2(x + 4)2 + 5

Changing Forms

Standard  General                            y = a(x  h)2 + k  y = ax2 + bx + c

Examples:        FOIL the bracket and simplify.

1. y = 2(x  3)2  11                2. y  ( x  1) 2  5                    3. y  3( x  2) 2  5

y  2( x 2  6 x  9)  11          y  ( x 2  2 x  1)  5                   y  3( x 2  4 x  4)  5
y  2 x 2  12 x  18  11          y   x2  2x 1  5                        y  3 x 2  12 x  12  5
y  2 x 2  12 x  9                y   x2  2x  4                           y  3 x 2  12 x  7

General  Standard                            y = ax2 + bx + c  y = a(x  h)2 + k

What is so special about?                     y  x 2  10 x  25       and y  x 2  6 x  9
Answer, they are both perfect squares         y  ( x  5) 2                  y  ( x  3) 2

Steps: (Optional)

1. Bracket x2 and x term                     If no x term, steps are done.
2
2. Factor out the coefficient of x term from bracketed terms
3. Take the coefficient on the x term, divided it by two, then square it. Add this number inside
the brackets to make a perfect square trinomial.
4. Multiply this number with the coefficient in front of the brackets. Take its opposite and add it
to the end of the equation. (Because what you add inside the bracket must be taken away
outside)
5. Factor the perfect square trinomial.

Examples:
Determine:
a.   vertex                                        b. equation of axis of symmetry
c.   range                                         d. max. or min. and value
e.   y–int                                         f. x–int ins some (using calculator)
g.   sketch

1. y = 3x2                                    2. y = x2  5
No x term so equation is done.                No x term so equation is done.

Page 10                       Math 20 IB Ch 3 Quadratic Functions
3. y = x2 + 6x  2

y  ( x 2  6 x)  2                   Bracket the x2 and x terms.
y  ( x 2  6 x  9)  2  9           Divide the coefficient of the x term by 2 and square it.
Complete the square.
y  ( x  3) 2  11

4. y = 2x2  12x + 7

y  (2 x 2  12 x)  7                 Bracket the x2 and x terms.
Factor out the 2 in front of the x2 term.
y  2( x 2  6 x)  7
Divide the coefficient of the x term by 2 and square it.
y  2( x 2  6 x  9)  7  18         Subtract 18 because you added 9 inside a bracket that is
y  2( x  3) 2  11                          multiplied by 2.
Complete the square.

5. y = 3x2 + 12x  15

y  (3x 2  12 x)  15                Bracket the x2 and x terms.
Factor out the –3 in front of the x2 term.
y  3( x 2  4 x)  15
Divide the coefficient of the x term by 2 and square it.
y  3( x 2  4 x  4)  15  12       Add 12 because you added 4 inside a bracket that is
y  3( x  2)2  3                           multiplied by –3.
Complete the square.

6. y = x2 + 10x  3
Bracket the x2 and x terms.
y  ( x 2  10 x)  3
Factor out the –1 in front of the x2 term.
y  ( x 2  10 x)  3                 Divide the coefficient of the x term by 2 and square it.
y  ( x 2  10 x  25)  3  25       Add 25 because you added 25 inside a bracket that is
multiplied by –1.
y  ( x  5) 2  22                   Complete the square.

7. y + 5x2 + 8x = 0

y  5 x 2  8 x                       Manipulate the equation to y = format
y  (5 x  8 x)
2
Bracket the x2 and x terms.
8
y  5( x 2  x)                       Factor out the –5 in front of the x2 term.
5
8   16 16
y  5( x 2  x  )                   Divide the coefficient of the x term by 2 and square it.
5   25   5                Add 16/5 because you added 16/25 inside a bracket that is
    4  16
2
multiplied by –5.
y  5  x                          Complete the square.
    5   5

Page 11                Math 20 IB Ch 3 Quadratic Functions
2 2 3
8. y =     x  x2
3      5
2      3                                 Bracket the x2 and x terms.
y   x2  x   2
3      5 
Factor out the 2/3 in front of the x2 term.
y   x 2  0.9 x   2
2
3                                          Divide the coefficient of the x term by 2 and square
y   x 2  0.9 x  0.2025   2  0.135
2                                          it.
3                                          Subtract 0.135 because you added 0.2025 inside a
2                                                 bracket that is multiplied by 2/3.
y   x  0.45   2.135
2

3                                          Complete the square.

9. y = 0.1x2 + x                                                y  0.1 x  5  2.5
2

10. y = 30x2 + 1200x + 1600                                     y  30  x  20   13600
2

11. y = 3x2  30x
y  3 ( x 2  10 x)
y  3 ( x 2  10 x  25)  75
y  3 ( x  5) 2  75

12. y = 2x2  20x + p
y  2( x 2  10 x)  p
y  2( x 2  10 x  25)  p  50
y  2( x  5) 2  p  50

13. y =    5 x 2  4 10 x  2 5 (IB)
          
y  5 x2  4 2x  2 5

y    5  x  4 2 x  8  2
2
5 8 5

5x  2 2 6 5
2
y

Page 12               Math 20 IB Ch 3 Quadratic Functions
Assignment
pg. 131 # 17, 26, 28, 39, 47, 48, 52, 54, 64

plus:

2     4                                        2     7     39
1. y =  x 2  x  1                          ANS y =  ( x  )2 
7     5                                        7     5     25

39
Find the range as well.                   y
25

2. y = x2 + 8x
ANS y = (x + 4)2  16

3. y = 7x2  28x + m
ANS y = 7(x  2)2 + m  28

4. y =     7 x 2  2 21x  5 7 (IB)
ANS y =     7( x  3) 2  2 7

14 2 4
5. y =       x  x  1 (IB)
3    3
14      1     19
ANS y =       ( x  )2 
3      7     21

6. A quadratic function passes thru (1, 0), (–2, –6), and (3, 14).
Find the equation in general form.

ANS: y = 1x2 + 3x – 4

Page 13                  Math 20 IB Ch 3 Quadratic Functions
1/2 of Day 5 and Day 6                              Do on Calculator to 2 decimal places.

Problem Solving

A. Function is Given

Examples:

1. The height (h(d) in metres) of a missile for a horizontal distance (d) is given by the function
h(d) = 0.05d2 + 4d.

Find:
Calc.
a. height at d = 15

h(d) = 0.05d2 + 4d          Insert 15 for d
h(15) = 0.05(15)2 + 415
ANS 48.75 m

b. maximum height and horizontal distance that gives this height

h(d) = 0.05d2 + 4d            Complete the square
h(d)=0.05(d  80d + 1600) + 80
2

h(d)= 0.05(d  40)2 + 80
ANS h = 80 m, d = 40 m

c. horizontal distance when the missile hits the ground

h(d) = 0.05d2 + 4d             height = 0 if it hits the ground
0 = 0.05d + 4d
2

0 = 0.05(d2  80d + 1600) + 80
0 = 0.05(d  40)2 + 80
1600 = (d  40)2
40 = d  40,
d = 0, 80 m
ANS 80 m                        0 means when the rocket is launched

Page 14                 Math 20 IB Ch 3 Quadratic Functions
2. The path of a basketball shot is modelled by h(t) = 1.2t2 + 3.3t + 2.2 where
h(t) = height of ball in m, and t = time in seconds.
Calc.
Find:

a. max. height reached                                              ANS 4.47 m
K value of completed square
h(t) = 1.2t2 + 3.3t + 2.2
h(t )  (1.2t 2  3.3t )  2.2
h(t )  1.2(t 2  2.75t )  2.2
h(t )  1.2(t 2  2.75t  1.890625)  2.2  2.26875
h(t )  1.2(t  1.375) 2  4.46875

b. time when max is reached                                         ANS 1.37 s
h value of completed square

c. distance ball is from floor when player releases it              ANS 2.2 m

h(t) = 1.2t2 + 3.3t + 2.2              player releases at time = 0
h(0) = 1.2(0)2 + 3.3(0) + 2.2
h(0) = 2.2

d. time when the ball hits the floor.                               ANS 3.30 s

type h(t) = 1.2t2 + 3.3t + 2.2 into y =
2nd calc 2 (zero) and determine the right x intercept

Assignment
pg. 132 # 76 a – c, 79 a – c,                       extra if needed 80 a – b

Read 129–130 eg #4, 5, 6

Be sure to ask the Plus question.

Plus: The height (above the green) of a golf ball hit from an elevated tee is given by the function
h(t) = 2.5t2 + 20t + 42. Find:
calc.
a. max height and time.        b. height of elevated tee.     c. time when the ball hits the green.
(82 m; 4 s)                    (42 m)                           (9.73 s)

Page 15                  Math 20 IB Ch 3 Quadratic Functions
B. Number Functions

Algebraic – max or min is the “k” value and it happens at “h” after completing the square.

Examples:

1. 30 times a number minus the square of the number multiplied by 3 gives some value. Find:
a. function to represent value              ANS V = 30x  3x2
b. value when the number is 2               ANS 48

V = 30x  3x2                           Insert 2 into the equation for x.
V = 30(2)  3(2)2
V= 48

c. maximum value                               ANS 75

V  3x 2  30 x
V  3( x 2  10 x)
max value (a.k.a. K) is 75
V  3( x 2  10 x  25)  75
V  3( x  5)2  75

2. The sum of 2 numbers is 10. The product of the 2 numbers plus twice one number gives a value.
Find max. value and the 2 numbers.

x + y = 10
V = xy + 2x
V = x(10  x) + 2x          Substitute y = 10 – x in for y
V = x2 + 12x
V = (x2  12x + 36) + 36 Complete the square
V = (x  6)2 + 36
ANS x = 6 and y = 4, max. = 36

3. A second number is twice the first number. The third number is 18 less than the second number.
A minimum value can be obtained if you multiply the first and third numbers, then subtract
the second number and then add 73. Find this minimum value and the three numbers

1st = x
2nd = 2x
3rd = 2x – 18
V = x(2x  18)  2x + 73
V = 2x2  18x  2x + 73
V = 2x2  20x + 73                            The minimum value is 23
V = 2(x2  10x + 25) + 73  50
V = 2(x  5)2 + 23                            The numbers are 5, 10 and 8

Page 16                 Math 20 IB Ch 3 Quadratic Functions
4. Given two number x & y, we know 3x + 5y = 50. Ten times the product of the numbers minus 16
times the 1st number gives a maximum. Find numbers and max. (IB)

M = 10xy  16x
 50  3 x 
M = 10x              16x
 5 
M = 100x  6x2  16x
M = 6x2 + 84x
M = 6(x2  14 + 49) + 294
M = 6(x  7)2 + 294
29
x = 7,     , max. = 294
5
Assignment
pg. 132 – 133 # 72 a, b, 75

plus:

1. A second number is twice the 1st number. A 3rd number is 3 more than the 1st number.
The square of the 3rd number minus the product of the 1st two gives a value. Find the max.
value and numbers that give this max.
ANS max. = 18, numbers = 3, 6, 6

2. 7 times one number plus twice a 2nd number gives 60. Twice the 1st times the 2nd plus the
square of the 1st gives some value. Find the max. value and numbers that give this max.
(IB)

7x + 2y = 60
V = 2xy + x2
 60  7 x     2
V = 2x            +x
 2 
V = 6x2 + 60x
V = 6(x2  10x + 25) + 150
V = 6(x  5)2 + 150
25
ANS max. = 150, numbers = 5,
2

Page 17                 Math 20 IB Ch 3 Quadratic Functions
C. Area Problems

Examples:

1.              y                               240 m of fence. Find dimensions so that
area is a maximum.
x
3x + 2y = 240
A = xy
 240  3 x 
A = x            
    2      
3
A =  x2 + 120x
2
3
A =  (x2  80x + 1600) + 2400
2
3
A =  (x  40)2 + 2400
2

ANS x = 40 m, y = 60 m, area = 2400 m2

2.       existing fence                         80 m of fence. Find dimensions so that area is a
maximum.
x
5x + y = 80
A = xy = x(80  5x) = 5x2 + 80x
y
A = 5(x2  16x + 64) + 320
A = 5(x  8)2 + 320

ANS x = 8 m, y = 40 m, area = 320 m2

3.              y                               200 m of fence. Find dimensions so that area is a
maximum.

x                     4x + 5y = 200
 200  4 x     4 2
A = xy = x             =  x + 40x
    5          5
4                            4
A =  (x2  50x + 625) + 500 =  (x  25)2 + 500
5                            5

ANS x = 25 m, y = 20 m, area = 500 m2

4. Determine the maximum area of a triangle if the sum of its base and height is 10 m.
b + h = 10
1      1            1               1                     25     1             25
A = bh = b(10  b) =  b2 + 5b =  (b2  10b + 25) +              =  (b  5)2 +
2      2            2               2                     2      2             2

25 2
ANS b = 5 m, h = 5 m, area =     m
2

Page 18                Math 20 IB Ch 3 Quadratic Functions
Assignment
pg. 133 # 82, 83

plus:

1.                 y                     180 m of fence. Find dimensions so that area is a
maximum.
x
ANS 30 m by 15 m, area = 450 m2

Extra Question

300 m of fence. Find dimensions so that area is a
Fence                     maximum.

x             5x + 2y = 300
 300  5 x     5 2
A = xy = x             =  x + 150x
    2          2
5                             5
y                     A =  (x2  60x + 900) + 2250 =  (x  30)2 + 2250
2                             2

ANS x = 30 m, y = 75 m, area = 2250 m2

Page 19               Math 20 IB Ch 3 Quadratic Functions
D. Income Problems

Examples:

1. You sell 60 books if the price is \$8 or less. For every \$2 increase, sales drop by 6. Find:
a. income if price is \$12
b. max. income
c. price and number of books to yield above

let x = number of increases

X                  # of books           price/book
0                      60                    \$8
1                  54 (60  6)         \$10 (8 + 2(1))
2                 48 (60  2(6))       \$12 (8 + 2(2))
.                       .                     .
.                       .                     .
.                       .                     .
X                    60  6x              8 + 2x

I = # of books  price/book
I = (60  6x)(8 + 2x)
I = 12x2 + 72x + 480
I = 12(x2  6x + 9) + 480 + 108
I = 12(x  3)2 + 588

After 3 increases, # of books = 42, price/book = \$14, income = \$588

2. 400 articles bought if price is \$5.00. Each increase of \$0.05 results in 2 fewer sales. Find
maximum income and price and # of articles that yield this max.

I = (400  2x)(5 + 0.05x)
I = 0.10x2 + 10x + 2000
I = 0.10(x2  100x + 2500) + 2000 + 250
I = 0.10(x  50)2 + 2250

After 50 increases, # of articles = 300, price/article = \$7.50, income = \$2 250

3. A riverboat cruise ride is \$36/person. Cruise averages 300 people a day. Each \$2 increase results
in 10 fewer customers. What increase in price would yield maximum income?

I = (36 + 2x)(300  10x)
I = 20x2 + 240x + 10 800
I = 20(x2  12x + 36) + 10 800 + 720
I = 20(x  6)2 + 11 520

ANS \$12 increase

Assignment
pg. 133 # 84, 85

Page 20                 Math 20 IB Ch 3 Quadratic Functions
DAY 7 and Quiz

Using Symmetry to Find Other Points

Examples:

1. A quad. function has its vertex at (3, 4). If one point of the function is (7, 20), find the other
point.

Think logic. From x = –3 to x = –7, you moved 4 to the left, so you must move 4 to the right
to be symmetrical. Thus, x = 1. Same logic, from y = 4 to y = 20, you moved up 16. Since
the two points are symmetrical about the axis of symmetry, y must also be 20 on the new
point.
ANS (1, 20)

2. Two points of a quad. function are (1, 6) and (5, 6). Its range is y  12. Find its vertex and
equation.

Again, think logic. The two points are (1, 6) and (5, 6). Thus, the middle x must be 3.
1+5 / 2 = 3. Since range is y  12, the highest y value must be 12. So vertex is (3, 12).

Start with vertex and solve for “a” using either of the known points.

y  a ( x  h) 2  k
y  a ( x  3) 2  12
I will use (1, 6)
6  a(1  3) 2  12
6  a(2) 2  12
6  4a
3
a
2
3
ANS (3, 12), y =      (x  3)2 + 12
2

3. The vertex of a quadratic function is (3, 8). One x–int is 4. What is the other one?

A little more logic. From x = 3 to x = –4, you moved 7 to the left, so you must move 7 to the
right to be symmetrical. Thus, x = 10.

ANS 10

Page 21                 Math 20 IB Ch 3 Quadratic Functions
Axis

The general form of the quad. function y = ax2 + bx + c when expressed in
standard form y = a(x  p)2 + q has a vertex (p, q) and an axis of x = p.
2
æ      b÷ö
2
Expressed in standard form y = ax + bx + c becomes y = a ç    çx -     ÷ - etc
è        ÷
2a ø
-b                            æb
-      ö
ie The axis is always x =       and the vertex is always ç , etc ÷
ç 2a    ÷
÷
2a                           è       ø

Calculator use

1.      Find the Maximum height and the time it hits the ground if a projectile has a path of
h = –x2 + 228x + 5 ( nearest hundredth)
h max =413680.53 at x = 362.87              h = 0 at t = 725.75s
- 5 2
2.      What is the range of y =      x + 71 11x + 47.32 (nearest hundredth)
27
range: y  74906.17

Assignment

1. A quad. function has its vertex at (1, 11). One x–int is 6. Find a second x–int.   ANS (8, 0)

2. Two points of a quad. function are (3, 5) and (7, 5). If its range is y  45, find vertex and
equation. ANS (2, 45), y = 2(x  2)2 + 45

Page 22                 Math 20 IB Ch 3 Quadratic Functions
Day 8        Review

Day 9        Exam

Extra Questions to show when to use calculator.

1.        The coordinates of the vertex of y = x2 – 8x + 2 are:

A.   (–4, –14)
B.   (–4, 2)
C.   (4, –14)
D.   (4, 2)

2.        Sketch y = –2x2 – 8x –1 on the grid below. You must show at least 5 points including the vertex.

3.        A cannon ball is fired from a cannon and follows a path given by h = –0.2t2 + 4t + 2.3 where h is
height in metres and t is time in seconds. What is the maximum height? Or, when does the
maximum height occur?

Page 23                Math 20 IB Ch 3 Quadratic Functions
Review Questions

1. Which parabola is the narrowest?                                                     ANS b

9
a. y =   (x  3)2 + 4                          b. y =  75 x2 + 2
2
c. y = 3(x + 1)2                              d. y = 0.1(x + 5)2  10

2. The height of a ball (h in metres) after t seconds is given by h = 3t2 + 12t + 5. Find:

a. height after 2 seconds                                                            ANS 17

b. time when ball hits ground (nearest hundredth)                                    ANS 4.38

c. max. height                                                                       ANS 17

1 2 1                                         1     1     95
3. Complete the square for y =      x  x + 2.                            ANS     (x  )2 +
3    6                                        3     4     48

4. Find the vertex of y = 2mx2  12mx + 4.                                       ANS (3, 4  18m)

5. The range of y = x2  20x + k is:                                            ANS y  k + 100

6. Which of the following is a quadratic function?                                      ANS d

15
a. y =                                         b. y = x3 + 2x + 4
x2

c. y = x  10                                  d. y = x2 + x

7. The y–int of y = 2 11 (x  10)2 + 71 is:                               ANS 200 11 +71 or 734.32

3                                                                            9
8. If y = x2 +     x + m is a perfect square, then find the value of m.                 ANS
4                                                                            64

9. Find the equation of a function that passes through (1, 20), (1, 4), and (4, 5).
Use y = ax2 + bx + c                                                 ANS y = 3x2 + 12x + 5

Page 24                  Math 20 IB Ch 3 Quadratic Functions
9 2 25
10. Write y = 2x2  9x + 7 in completed square form.                     ANS y = 2(x      ) 
4     8

11. State the equation of the axis of symmetry for y = x2  16x + 15.          ANS x = 8 or x  8 = 0

1
12. Find the negative x–int of y =      (x  1)2  4. (nearest tenth)                    ANS (16.9, 0)
80

13. Find the equation of a function with vertex (2, 8) and y–int of 12.      ANS y = 5(x + 2)2 + 8

14. Find the equation of a function congruent to y = 5x2, with range of y  2, and equation of axis of
symmetry x + 3 = 0.                                                         ANS y = 5(x + 3)2  2

15. Find the vertex of y =   0.03 x2 – 7x + 3. (0.01)
ANS (20.21, –67.73)

Text Review p 144 # 9, 18, 19, 26, 36, 56, 63

Page 25                  Math 20 IB Ch 3 Quadratic Functions

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