MTH 209 Week 4
Final Exam logistics
Here is what I've found out about the final exam in
MyMathLab (running from the end of class this
week (week 4 at 10pm) to 11:59pm five days after
the last day of class.
.
Slide 2 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Final Exam logistics There will be 50 questions.
You have only one attempt to complete the exam.
Once you start the exam, it must be completed in that sitting. (Don't start until you have
time to complete it that day or evening.)
You may skip and get back to a question BUT return to it before you hit submit.
You must be in the same session to return to a question.
There is no time limit to the exam (except for 11:59pm five nights after the last class).
You will not have the following help that exists in homework:
Online sections of the textbook
Animated help
Step-by-step instructions
Video explanations
Links to similar exercises
You will be logged out of the exam automatically after 3 hours of inactivity.
Your session will end.
IMPORTANT! You will also be logged out of the exam if you use your back button on
your browser. You session will end.
Slide 3 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Section 6.6
Solving
Equations by
Factoring I
(Quadratics)
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• The Zero-Product Property
• Solving Quadratic Equations
• Applications
Zero-Product Property
To solve equations we often use the zero-product
property, which states that if the product of two
numbers is 0, then at least one of the numbers must
be 0.
Example
Solve each equation.
a. b.
Solution
a. b.
3 ( 4) 0x x (4 1)(3 4) 0x x
3 ( 4) 0x x
04( )3x x
3 0 or 4 0 x x
0 o r 4 x x
(4 1)(3 4) 0x x
3( 4 01)(4 )xx
0 or 31 044 xx
4 1 or 3 4x x
1 4 or
4 3x x
Try Q’s pg 402 13,15,19,23
Solving Quadratic Equations
Any quadratic polynomial in the variable x can be
written as ax2 + bx + c with a ≠ 0.
Any quadratic equation in the variable x can be
written as ax2 + bx + c = 0, with a ≠ 0. This form of
quadratic equation is called the standard form of a
quadratic equation.
Slide 9
Example
Solve each quadratic equation. Check your answers.
a. 36 + x2 = –12x b. x2 − 49 = 0
Solution
a. 36 + x2 = –12x
x2 + 12x + 36 = 0
(x + 6)(x + 6) = 0
x = −6
The only solution is −6.
To check this value,
substitute −6 for x in
the given equation.
36 + x2 = –12x
36 + (−6)2 = –12(−6)
72 = 72
Example (cont)
b. x2 − 49 = 0
(x + 7)(x − 7) = 0
x =
−7
x + 7 = 0 x − 7 =
0 x = 7
To check these values,
substitute 7 and −7 for
x in the given equation. 72 − 49 = 0
0 = 0
(−7)2 − 49 =
0 0 = 0
The solutions are −7 and
7.
Try Q’s pg 403 27,35,43,49
Example
Solve 2x2 − 7x = −5
Solution
The solutions are
(2x – 5)(x − 1) = 0
2x – 5 = 0 or x – 1 = 0
or x = 1
2x2 − 7x = −5
2x2 − 7x + 5 = 0
5
2x
51 and .
2
Try Q’s pg 403 51
Example
If a model rocket is launched at 48 feet per second,
then its height, h, after t seconds is h = 48t – 16t2.
After how long does the rocket strike the ground?
Solution
The rocket strikes the ground when the height is 0.
48t – 16t2 = 0
16t(3 – t) = 0
t = 0
16t = 0 3 – t = 0
t = 3
The rocket strikes the
ground after 3
seconds.
Try Q’s pg 403 67a
Example
A frame surrounding a picture is 2 inches wide. The
picture inside the frame is 7 inches longer than it is
wide. If the overall area of the picture and frame is
198 square inches, find the dimensions of the picture
inside the frame.
Solution
Let x be the width of the picture and x + 7 be its
length.
2
2
2
2
x + 7
x
x + 11
x + 4
Example (cont)
(x + 4)(x + 11) =198
x2 + 15x + 44 = 198
x2 + 15x − 154 = 0
(x − 7)(x + 22) = 0
x − 7 = 0 or x + 22 = 0
x = 7 or x = −22
The only valid solution for x is 7 inches. Because the
length is 7 inches more than the width, the
dimensions are 7 inches and 14 inches.
Try Q’s pg 404 73
Section 8.4
Other
Functions and
Their Properties
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Expressing Domain and Range in Interval Notation
• Absolute Value Function
• Polynomial Functions
• Rational Functions (Optional)
• Operations on Functions
Expressing Domain and Range in
Interval Notation
The set of all valid inputs for a function is called the
domain, and the set of all outputs from a function is
called the range.
Rather than writing “the set of all real numbers” for
the domain of f, we can use interval notation to
express the domain as (−∞, ∞).
Example
Write the domain for each function in interval
notation.
a. f(x) = 3x b.
Solution
a. The expression 3x is defined for all real numbers
x. Thus the domain of f is
b. The expression is defined except when x – 4 = 0
or
x = 4. Thus the domain of f includes all real
numbers except 4 and can be written
1
4f x
x
, .
,4 4, .
Try Q’s pg 563 13,21,25
Absolute Value Function
We can define the absolute value function
by f(x) = |x|.
To graph y = |x|, we begin by making a table of
values.
x |x|
–2 2
–1 1
0 0
1 1
2 2
Example
Sketch the graph of f(x) = |x – 3|. Write its domain
and range in interval notation.
Solution
Start by making a table of values.
x y
0 3
2 1
3 0
4 1
6 3
X
Y
-3 -2 -1 1 2 3 4 5 6 7 8
-5
-4
-3
-2
-1
1
2
3
4
5
0
The domain of f is , .
The range of f is [0, ).
Try Q’s pg 563 39
Polynomial Functions
The following expressions are examples of
polynomials of one variable.
As a result, we say that the following are symbolic
representations of polynomial functions of one
variable.
2 3, , and 51 515 3x xx x
32( ) 3, , and ( 5 1 ( )) 1 5 5f g x x xx xx hx
Example
Determine whether f(x) represents a polynomial
function. If possible, identify the type of polynomial
function and its degree.
a.
b.
c.
3( ) 6 2 7f x x x
3.5( ) 4f x x
4( )
5f x
x
cubic polynomial, of degree 3
not a polynomial function
because the exponent on the
variable is negative
not a polynomial
Try Q’s pg 564 43,45,47,51
Example
A graph of is shown. Evaluate f(1)
graphically and check your result symbolically.
Solution
3( ) 5f x x x
To calculate f(–1) graphically
find –1 on the x-axis and
move down until the graph of
f is reached. Then move
horizontally to the y-axis.
f(1) = –4 3( 1) 5( ( )1 1)f
5 1
4
Try Q’s pg 564 59,73
Example
Evaluate f(x) at the given value of x.
Solution
3 2( ) 4 3 7, 2f x x x x
3 2( ) 4 3 7, 2f x x x x
3 2( 2) 4( 2) 3( 2) 7f
( 2) 4( 8) 3(4) 7f
( 2) 32 12 7f
( 2) 27f
Try Q’s pg 564 55-64
Example
Use and to evaluate each of
the following.
Solution
2( ) 3 1f x x 2( ) 6g x x
a. ( )(1) b. ( )( 2) c. 0f
f g f gg
2a. ( ) 3 11 (1)
4
f
2( ) 61 ( )
5
1g
( )(1) (1) (1)
4 5
9
f g f g
Example (cont)
Use and to evaluate each of
the following.
Solution
2( ) 3 1f x x 2( ) 6g x x
a. ( )(1) b. ( )( 2) c. 0f
f g f gg
2b. ( ) 3( ) 1
3(4
2
13
2
) 1
f2( ) 6 (
6 4
2
2 )2g
( )( 2) ( 2) ( 2)
13 2
11
f g f g
Example (cont)
Use and to evaluate each of
the following.
Solution
2( ) 3 1f x x 2( ) 6g x x
a. ( )(1) b. ( )( 2) c. 0f
f g f gg
c. 0f
g
00
0
ff
g g
2
2
3( ) 1 =
6 )
0
(0
1 =
6
Try Q’s pg 564 64-70
Section 11.1
Quadratic
Functions and
Their Graphs
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Graphs of Quadratic Functions
• Min-Max Applications
• Basic Transformations of Graphs
• More About Graphing Quadratic Functions
(Optional)
The graph of any quadratic function is a
parabola.
The vertex is the lowest point on the graph of
a parabola that opens upward and the highest
point on the graph of a parabola that opens
downward.
The graph is symmetric with respect to the y-
axis. In this case the y-axis is the axis of
symmetry for the graph.
Example
Use the graph of the quadratic function to identify the
vertex, axis of symmetry, and whether the parabola
opens upward or downward.
a. b.
Vertex (0, 2)
Axis of symmetry: x = –2
Open: up
Vertex (0, 4)
Axis of symmetry: x = 0
Open: down
Try Q’s pg 725 29,31
Example
Find the vertex for the graph of
Support your answer graphically.
Solution
a = 2 and b = 8
Substitute into the equation to find the y-value.
2( ) 2 8 3.f x x x
82
2(2)x
2
bx
a
2( ) 2( 2) 8( 2) 3
8 16 3
11
f x
The vertex is (2, 11), which
is supported by the graph.
Try Q’s pg 724 15
Example
Identify the vertex, and the axis of symmetry on the
graph, then graph.
Solution
Begin by making a
table of values.
Plot the points and
sketch a smooth curve.
The vertex is (0, –2)
axis of symmetry x = 0
2( ) 2f x x
x f(x) = x2 –
2
3 7
2 2
1 1
0 2
1 1
2 2
3 7
Example
Identify the vertex, and the axis of symmetry on the
graph, then graph.
Solution
Begin by making a
table of values.
Plot the points and
sketch a smooth
curve.
The vertex is (2, 0)
axis of symmetry x = 2
2( ) ( 2)g x x
x g(x) = (x –
2)2
0 4
1 1
2 0
3 1
4 4
Example
Identify the vertex, and the axis of symmetry on the
graph, then graph.
Solution
Begin by making a
table of values.
Plot the points and
sketch a smooth
curve.
The vertex is (2, 0)
axis of symmetry x = 2
2( ) 2 3h x x x
x h(x) = x2 – 2x –
3
2 5
1 0
0 3
1 4
2 3
3 0
4 5
Example
Find the maximum y-value of the graph of
Solution
The graph is a parabola that opens downward
because a < 0. The highest point on the graph
is the vertex.
a = 1 and b = 2
2( ) 2 3.f x x x
( 2)1
2 2( 1)
bx
a
2( ) ( 1) 2( 1) 3
4
f x
Try Q’s pg 726 81
Example
A baseball is hit into the air and its height h in feet
after t seconds can be calculated by
a. What is the height of the baseball when it is hit?
b. Determine the maximum height of the baseball.
Solution
a. The baseball is hit when t = 0.
b. The graph opens downward
because a < 0. The maximum height occurs at
the vertex. a = –16 and b = 64.
2( ) 16 64 2.h t t t
2( ) 16 64 2h t t t 2(0) 16(0) 64(0) 2
2
h
64 642
2 2( 16) 32
bx
a
Example (cont)
2( ) 16 64 2 h t t t
2(2) 16(2) 64(2) 2
66
h
The maximum height is 66 feet.
2( ) 16 64 2 h t t t
Try Q’s pg 726 87
Basic Transformations of Graphs
The graph of y = ax2, a > 0.
As a increases, the resulting parabola becomes
narrower.
When a > 0, the graph of y = ax2 never lies below the
x-axis.
Section 11.2
Parabolas and
Modeling
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Vertical and Horizontal Translations
• Vertex Form
• Modeling with Quadratic Functions (Optional)
Vertical and Horizontal Translations
The graph of y = x2 is a parabola opening upward
with vertex (0, 0).
All three graphs have the same shape.
y = x2
y = x2 + 1 shifted upward 1 unit
y = x2 – 2 shifted downward 2 units
Such shifts are called translations because they do
not change the shape of the graph only its position
Vertical and Horizontal Translations
The graph of y = x2 is a parabola opening upward
with vertex (0, 0).
y = x2
y = (x – 1)2
Horizontal shift to the right 1 unit
Vertical and Horizontal Translations
The graph of y = x2 is a parabola opening upward
with vertex (0, 0).
y = x2
y = (x + 2)2
Horizontal shift to the left 2 units
Example
Sketch the graph of the equation and identify the
vertex.
Solution
The graph is similar to y = x2 except it has been
translated 3 units down.
The vertex is (0, 3).
2 3y x
Try Q’s pg 739 15,19,27
Example
Sketch the graph of the equation and identify the
vertex.
Solution
The graph is similar to y = x2 except it has been
translated left 4 units.
The vertex is (4, 0).
2( 4)y x
Example
Sketch the graph of the equation and identify the
vertex.
Solution
The graph is similar to y = x2 except it has been
translated down 2 units and
right 1 unit.
The vertex is (1, 2).
2( 1) 2y x
Try Q’s pg 739 39,37
Example
Compare the graph of y = f(x) to the graph of y = x2.
Then sketch a graph of y = f(x) and y = x2 in the
same
xy-plane.
Solution
The graph is translated to the right
2 units and upward 3 units.
The vertex for f(x) is (2, 3) and the
vertex of y = x2 is (0, 0).
The graph opens upward
and is wider.
21( ) ( 2) 3
4f x x
Try Q’s pg 739 41
Example
Write the vertex form of the parabola with a = 3 and
vertex (2, 1). Then express the equation in the form
y = ax2 + bx + c.
Solution
The vertex form of the parabola is
where the vertex is (h, k).
a = 3, h = 2 and k = 1
To write the equation in y = ax2 + bx + c, do the
following:
2( ) ,y a x h k
2)3( 12y x
2)3( 12y x
2( 4 43 1)y x x
23 12 12 1y x x 23 12 13y x x
Try Q’s pg 740 65,68
Example
Write each equation in vertex form. Identify the
vertex.
a. b.
Solution
a. Because , add and subtract 16
on the right.
2 8 13y x x 22 8 7y x x
2 28
162 2
b
2 8 13y x x
2 16 1 68 13y x x
2
4 3y x
The vertex is (4, 3).
Example (cont)
b. This equation is slightly different because the
leading coefficient is 2 rather than 1. Start by
factoring 2 from the first two terms on the right
side.
22 8 7y x x
2 24
42 2
b
2
2
2 8 7
2( 4 ) 7
y x x
x x
2 42 4 74y x x
2 42 4 7 8y x x
2
2 2 1y x The vertex is ( 2, 1).
Try Q’s pg 739 53
Section 11.3
Quadratic
Equations
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Basics of Quadratic Equations
• The Square Root Property
• Completing the Square
• Solving an Equation for a Variable
• Applications of Quadratic Equations
Basics of Quadratic Equations
Any quadratic function f can be represented by
f(x) = ax2 + bx + c with a 0.
Examples:
2 2 21( ) 2 1, ( ) 2 , and ( ) 2 1
3f x x g x x x h x x x
Basics
The different types of solutions to a quadratic
equation.
Example
Solve each quadratic equation. Support your results
numerically and graphically.
a. b. c.
Solution
a. Symbolic: Numerical: Graphical:
23 2 0x 2 9 6x x 2 2 8 0x x
The equation has no real
solutions because x2 ≥ 0 for
all real numbers x.
x y
1 5
0 2
1 5
2
2
2
3 2 0
3 2
2
3
x
x
x
Example (cont)
b. 2 9 6x x
2
2
9 6
6 9 0
( 3)( 3) 0
3 0 or 3 0
3 or 3
x x
x x
x x
x x
x x
The equation has
one real solution.
x y
5 4
4 1
3 0
2 1
1 4
Example (cont)
c. 2 2 8 0x x
2 2 8 0
( 2)( 4) 0
2 0 or 4 0
2 or 4
x x
x x
x x
x x
The equation has two real
solutions.
x y
4 0
2 8
1 9
0 8
2 0
Try Q’s pg 752 29,37,39
The Square Root Property
The square root property is used to solve quadratic
equations that have no x-terms.
Example
Solve each equation.
a. b. c.
Solution
a.
2 10x 225 16 0x 2( 3) 16x
2 10x
2 10x
10x
b. 225 16 0x
225 16x
2 16
25x
16
25x
4
5x
c. 2( 3) 16x
( 3) 16x
3 4x
3 4x
1 or 7x
Try Q’s pg 752 51,53,57
Real World Connection
If an object is dropped from a height of h feet, its
distance d above the ground after t seconds is given
by 2( ) 16d t h t
Example
Find the term that should be added to to form
a perfect square trinomial.
Solution
Coefficient of x-term is –8, so we let b = –8. To
complete the square we divide by 2 and then square
the result.
2 8x x
2 28
162 2
b
2 2168 ( 4)x x x
Try Q’s pg 752 67
Example
Solve the equation
Solution
Write the equation in x2 + bx = d form.
2 8 13 0x x
2 28
162 2
b
2 8 13x x
2 168 3 61 1x x
2( 4) 3x
4 3x
4 3x
5.73 or 2.27x
Try Q’s pg 752 73
Example
Solve the equation
Solution
Write the equation in x2 + bx = d form.
20 2 8 7x x
2 24
42 2
b
22 8 7x x
2 442
47
x x
2 1( 2)
2x
12
2x
2 74
2x x
22
2x
12
2x
22
2x
1.29 or 2.71x
Try Q’s pg 752 81
Example
Solve the equation for the specified variable.
Solution
216 for w x x216w x
16
wx
4
wx
2
16
wx
Try Q’s pg 753 105,107
Section 11.4
Quadratic
Formula
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Solving Quadratic Equations
• The Discriminant
• Quadratic Equations Having Complex Solutions
, 0 and 1,xf x a a a
The solutions to ax2 + bx + c = 0 with a
≠ 0 are given by
QUADRATIC FORMULA
2 4.
2
b b acx
a
Example
Solve the equation 4x2 + 3x – 8 = 0. Support your
results graphically.
Solution
Symbolic Solution
Let a = 4, b = 3 and c = − 8.
2 4
2
b b acx
a
23 3 4 4 8
2 4x
3 137
8x
3 137
8x
or
3 137
8x
1.1x 1.8x or
Example (cont)
4x2 + 3x – 8 = 0
Graphical Solution
Try Q’s pg 764 9
Example
Solve the equation 3x2 − 6x + 3 = 0. Support your
result graphically.
Solution
Let a = 3, b = −6 and c = 3.
2 4
2
b b acx
a
26 6 4 3 3
2 3x
6 0
6x
1x
Try Q’s pg 764 11
Example
Solve the equation 2x2 + 4x + 5 = 0. Support your
result graphically.
Solution
Let a = 2, b = 4 and c = 5.
2 4
2
b b acx
a
24 4 4 2 5
2 2x
4 24
4x
There are no real solutions
for this equation because
is not a real number. 24
Try Q’s pg 764 13
, 0 and 1,xf x a a a
To determine the number of solutions to
the quadratic equation ax2 + bx + c = 0,
evaluate the discriminant b2 – 4ac.
1. If b2 – 4ac > 0, there are two real solutions.
2. If b2 – 4ac = 0, there is one real solution.
3. If b2 – 4ac < 0, there are no real solutions;
there are two complex solutions.
THE DISCRIMINANT AND QUADRATIC
EQUATIONS
Example
Use the discriminant to determine the number of
solutions to −2x2 + 5x = 3. Then solve the equation
using the quadratic formula.
Solution
−2x2 + 5x − 3 = 0
Let a = −2, b = 5 and c = −3.
Thus, there are two
solutions.
b2 – 4ac
= (5)2 – 4(−2)(−3) = 1
2 4
2
b b acx
a
5 1
2 2x
4
4x
1x
or 6
4x
1.5x
Try Q’s pg 764 39a, b
, 0 and 1,xf x a a a
If k > 0, the solution to x2 + k = 0 are given by
THE EQUATION x2 + k = 0
.x i k
Example
Solve x2 + 17 = 0.
Solution
The solutions are
17.i
17 or 17.x i i
Try Q’s pg 764 57
Example
Solve 3x2 – 7x + 5 = 0. Write your answer in standard
form: a + bi.
Solution
Let a = 3, b = −7 and c = 5.
2 4
2
b b acx
a
7 11
6x
and
27 7 4 3 5
2 3x
7 11
6
ix
7 11
6 6x i
7 11
6 6x i
Try Q’s pg 765 73
Example
Solve Write your answer in standard
form:
a + bi.
Solution
Begin by adding 2x to each side of the equation and
then multiply by 5 to clear fractions.
Let a = −2, b = 10 and c = −15.
223 2 .
5
xx
22 10 15 0x x
Example (cont)
Let a = −2, b = 10 and c = −15.
223 2 .
5
xx
2 4
2
b b acx
a
10 20
4x
210 10 4 2 15
2 2x
10 2 5
4
ix
5 5
2 2x i
Try Q’s pg 765 79
Example
Solve by completing the square.
Solution
After applying the distributive property, the equation
becomes
Since b = −4 ,add to each side of the
equation.
4 5x x
2 4 5.x x
2
42
4 2 4 4 5 4x x
2
2 1x
2 1x
2x i
2x i
The solutions are
2 + i and 2 − i.
Try Q’s pg 765 91
Section 11.5
Quadratic
Inequalities
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Basic Concepts
• Graphical and Numerical Solutions
• Symbolic Solutions
Example
Determine whether the inequality is quadratic.
a. 6x + 2x2 – x3 ≥ 0 b. 8 + 7x2 + 2 < 6x2 + x
Solution
a. The inequality 6x + 2x2 – x3 ≥ 0 is not quadratic
because it has an x3-term.
b. Write the inequality as follows.
8 + 7x2 + 2 < 6x2 + x
10 + 7x2 < 6x2 + x
x2 – x + 10 < 0
The inequality is quadratic because it can be written
in the form ax2 + bx + c > 0 with a = 1, b = −1, and c
= 10.
Try Q’s pg 775 7,11
Example
Make a table of values for x2 − 2x – 15 and then
sketch the graph. Use the table and graph to solve
x2 – 2x – 15 ≤ 0. Write your
answer in interval notation.
Solution
Interval notation [−3, 5]
x y = x2 – 2x – 15
−3 0
−2 −7
−1 −12
0 −15
1 −16
2 −15
3 −12
4 −7
5 0
Example (cont)
x2 − 2x – 15
Try Q’s pg 775 19,27
Example
Solve x2 > 4. Write your answer in interval notation.
Solution
The graph of y = x2 – 4 is shown with intercepts −2
and 2. Thus the solution set is given by
x < −2 or x > 2, which can be
written in interval notation as
, 2 2, .
Try Q’s pg 776 49
Example
Solve each of the inequalities graphically.
a. x2 + 3 > 0 b. x2 + 3 < 0 c. (x − 3)2 ≤ 0
Solution
a. Because the graph is always
above the x-axis, x2 + 3 is
always greater than 0. The
solution set includes all real
numbers, or (−∞,∞).
b. Because the graph never
goes below the x-axis, x2 + 3
is never less than 0. Thus
there are no real solutions.
Example (cont)
Solve each of the inequalities graphically.
a. x2 + 3 > 0 b. x2 + 3 < 0 c. (x − 3)2 ≤ 0
Solution
a. Because the graph never
goes below the x-axis, (x −
3)2 ≤ 0 is never less than 0.
When x = 3, y = 0, so 3 is
the only solution to the
inequality (x − 3)2 ≤ 0.
Try Q’s pg 776 51,55,57
Example
Solve each inequality symbolically. Write your
answer in interval notation.
a. x2 – 4x – 12 ≥ 0 b. 10 – x2 > 9x
Solution
a.
x2 – 4x – 12 = 0
(x + 2)(x – 6) = 0
x = −2
x + 2 = 0 x – 6 = 0
x = 6
The solutions lie
outside the values. In
interval notation the
solution set is
( , 2] [6, ).
Example (cont)
Solve each inequality symbolically. Write your
answer in interval notation.
a. x2 – 4x – 12 ≥ 0 b. 10 – x2 > 9x
Solution
b.
– x2 − 9x + 10 < 0
– x2 − 9x + 10 = 0
(x + 10)(x – 1) = 0
x + 10 = 0 x – 1 = 0
x = −10 x = 1
The solutions lie
between these two
values. In interval
notation the solution set
is (−10, 1).
x2 + 9x − 10 = 0
Try Q’s pg 776 31,37
Example
A rectangular building needs to be 9 feet longer than
it is wide. The area of the building must be at least
532 square feet. What widths x are possible for this
building?
Solution
x2 + 9x = 532
x(x + 9) ≥ 532
x2 + 9x – 532 = 0
29 9 4 1 532
2 1x
29 9 4 1 532
2 1
9 2209
2
9 47
2
28, 19
The width is positive, so
the building must be 19
feet or more, x ≥ 19
feet.
Try Q’s pg 776 67
Due for this week…
Homework 4 (on MyMathLab – via the Materials
Link) The fifth night after class at 11:59pm.
Read Chapter 12.1, 12.3 and 14.1-14.3
Do the MyMathLab Self-Check for week 4.
Learning team hardest problem assignment.
Complete the Week 4 study plan after submitting
week 4 homework.
Participate in the Chat Discussions in the OLS
Slide 104 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
End of week 4
You again have the answers to those problems not
assigned
Practice is SOOO important in this course.
Work as much as you can with MyMathLab, the
materials in the text, and on my Webpage.
Do everything you can scrape time up for, first the
hardest topics then the easiest.
You are building a skill like typing, skiing, playing a
game, solving puzzles.
Next Week: Composite Functions, Inverse
Functions, Logarithmic functions,
Arithmetic/Geometric sequences/series.