Download Contents

Contents
4
Advanced Algebra
4.1 Functional definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.1 CBa1: Evaluate a function with literal and numerical arguments. . .
4.1.2 CBa2: Find domain and range from defining equation or graph. . . .
4.1.3 CBa3: Know and apply vertical line test for functions to the graphs
of relations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.4 CBa4: Interpret the value of f(a) in applications. . . . . . . . . . . . .
4.1.5 CBa5: Qualitative properties and transformations of a function and
its graph. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Algebra of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.1 CBb1: Perform arithmetic operations, difference quotient type. . . .
4.2.2 CBb2: Form and evaluate composite functions. . . . . . . . . . . . . .
4.2.3 CBb3: Identify whether an increase in value of one variable in an
expression will cause increase or decrease in the value of the expression. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Inverse function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3.1 CBh1: Recognize domain and range relationship between f and f −1 .
4.3.2 CBh2: Recognize relationship between f and f −1 (given f as graph
or as rule). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3.3 CBh3: Identify one-to-one functions from graph. . . . . . . . . . . . .
4.3.4 CBh4: Find f −1 given f . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Exp and log functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.1 CBi1: Recognize graphs of exponential and logarithmic functions
and their inverse relations . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.2 CBi2: Manipulate expressions using logarithmic functions: log(ab),
log(ap ). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.3 CBi3: Write logarithmic and exponential functions in equivalent
forms: y = bx ≡ logb (y) = x. . . . . . . . . . . . . . . . . . . . . . . . .
4.4.4 CBi4: Solve literal equations using properties of exponents or logs. .
4.5 Manipulations of Absolute Values . . . . . . . . . . . . . . . . . . . . . . . .
4.5.1 CBc1: Solve |ax + b| < c. . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5.2 CBc2: Transform a < x < b into |cx + d| < e. . . . . . . . . . . . . . .
4.6 Quadratic Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.1 CBe1: Solve simple quadratic inequalities (e.g., ax2 + bx + c > 0). . .
1
5
5
5
7
11
14
17
20
20
23
24
27
27
29
33
37
41
41
45
47
49
51
51
55
57
57
CBe2: Solve inequalities that lead to simple quadratic (e.g., (ax +
b)/(cx + d) < ex + f . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Nonlinear systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.7.1 CBk1: Solve nonlinear systems of equations . . . . . . . . . . . . . . .
Binomial theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.8.1 CBg1: Determine coefficients of particular terms in the expansion
of (a + b)n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.8.2 CBg2: Factor difference of two cubes. . . . . . . . . . . . . . . . . . .
4.8.3 CBg3: Factor an ± bn when n > 3. . . . . . . . . . . . . . . . . . . . . .
Summation notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.9.1 CBj1: Recognize and find sums of arithmetic and geometric series. .
4.9.2 CBj2: Understand and use summation notation. . . . . . . . . . . . .
Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.10.1 CBd1: Perform arithmetic operations. . . . . . . . . . . . . . . . . . .
4.10.2 CBd2: Recognize and use the graphical representation of complex
numbers and their sums. . . . . . . . . . . . . . . . . . . . . . . . . .
4.10.3 CBd3: Manipulate expressions involving conjugates or magnitude. .
Theory of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.11.1 CBf1: Recognize and use relationship between roots and factors. . .
4.11.2 CBf2: Recognize relationship between degree of a polynomial and
maximum possible number of roots and conjugate pairs of complex
roots. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.11.3 CBf3: Find, characterize, or interpret solutions to quadratic equations with negative discriminant. . . . . . . . . . . . . . . . . . . . . .
Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.12.1 CBw1: General story/word/application problem involving sub score
1 objectives. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.12.2 CBw2: General story/word/application problem involving sub score
2 objectives. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Qualitative Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.13.1 CBq1: Graph interpretation. . . . . . . . . . . . . . . . . . . . . . . . .
4.13.2 CBq2: Algebraic Proportionality and Scaling. . . . . . . . . . . . . . .
4.6.2
4.7
4.8
4.9
4.10
4.11
4.12
4.13
5
Analytic Geometry
5.1 Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.1 CGa1: Recognize and use relationship between slope and parallel
or perpendicular lines. . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.2 CGa2: Find the equation of a line through a point parallel or perpendicular to a given line. . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.1 CGb1: Recognize and apply the distance formula (two dimensions).
5.3 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3.1 CGc1: Recognize and apply the definition of a circle as locus of
points equidistant from a fixed point. . . . . . . . . . . . . . . . . . .
61
65
65
72
72
75
77
78
78
83
86
86
91
93
95
95
100
101
103
103
107
112
112
116
119
119
119
122
126
126
133
133
5.3.2
5.4
CGc2: Find center and radius of a circle from its equation or vice
versa. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Conics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4.1 CGd1: Given an equation of the form Ax2 + By 2 + Cx + Dy + E = 0,
be able to identify the graph or name of the conic or vice versa. . .
5.4.2 CGd2: Given an equation of a parabola in the form Ax2 + Bx + Cy +
D = 0 or Ay 2 + By + Cx + D = 0, describe the parabola, giving its
vertex, axis, and the direction it opens, and vice versa. . . . . . . .
5.4.3 CGd3: Recognize the graph of y = (ax + b)/(cx + d) from equation
and vice versa. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4.4 CGd4: Solve verbal problems. . . . . . . . . . . . . . . . . . . . . . .
3
. 136
. 139
. 139
. 144
. 148
. 152
4
Chapter 4
Advanced Algebra
4.1
Functional definition, notation and interpretation
4.1.1 CBa1: Evaluate a function with literal and numerical arguments.
Given a function defined by an expression, evaluate the function at the given value in the
domain.
Example: f (x) = x2 . If x = 2, then f (2) = 4. If x = a + 1, then f (a + 1)
= (a + 1)2 = a2 + 2a + 1.
1. f (z) = 2z 2 + 3. Evaluate f (z) when z = −3 and when z = 2x + 1.
z = −3:
z = 2x + 1:
!
"
f (−3) = 2 × (−3)2 + 3 = 18 + 3 = 21.
f (2x + 1) = 2(2x + 1)2 + 3
= 2(4x2 + 4x + 1) + 3
= 8x2 + 8x + 5.
2. f (z) = 2z 2 + z −1 . Evaluate f (z) when z = 1 and when z = a + 1.
Recall that if x $= 0 then x−1 = 1/x.
z = 1:
f (1) = 2 × 12 + 1−1 = 2 + 1 = 3.
z = a + 1:
f (a + 1) = 2(a + 1)2 + (a + 1)−1
1
= 2(a2 + 2a + 1) +
a+1
1
= 2a2 + 4a + 2 +
a+1
5
4.1. FUNCTIONAL DEFINITION
CHAPTER 4. ADVANCED ALGEBRA
(2a2 + 4a + 2)(a + 1) + 1
=
a+1
3
2
2a + 2a + 4a2 + 4a + 2a + 2 + 1
=
a+1
3
2a + 6a2 + 6a + 3
=
.
a+1
3. f (z) = (z + 1)3 . Evaluate f (z) when z = −3 and when z = x − 1.
z = −3:
z = x − 1:
f (−3) = (−3 + 1)3 = (−2)3 = −8.
f (x − 1) = [(x − 1) + 1]3
= (x − 1 + 1)3 = x3 .
4. f (y) = 2y 2 − 3y. Evaluate f (y) when y = 0 and when y = x + 2.
y = 0:
y = x + 2:
f (0) = 2 × 02 − 3 × 0 = 0.
f (x + 2) = 2(x + 2)2 − 3(x + 2)
= 2(x2 + 4x + 4) − 3x − 6
= 2x2 + 8x + 8 − 3x − 6 = 2x2 + 5x + 2.
Note, that there is another way of solving the this problem. We can first factor out,
simplify and then evaluate.
f (x + 2) = 2(x + 2)2 − 3(x + 2)
= (x + 2)[2(x + 2) − 3]
= (x + 2)(2x − 1) = 2x2 + 5x + 2.
5. h(u) = −u2 + 4u − 7. Evaluate h(u) when u = −1 and when u = 1 − a.
u = −1:
u = 1 − a:
h(−1) = −(−1)2 + 4(−1) − 7 = −1 − 4 − 7 = 12.
f (1 − a) = −(1 − a)2 + 4(1 − a) − 7
= −(1 − 2a + a2 ) + 4 − 4a − 7
= −1 + 2a − a2 − 4a − 3 = −a2 − 2a − 4.
6
CHAPTER 4. ADVANCED ALGEBRA
4.1. FUNCTIONAL DEFINITION
4.1.2 CBa2: Find domain and range from defining equation or graph.
Find the domain and range for a function given by an expression or by a graph.
√
Example: f (x) = 1 − x2 . The domain is the closed interval [−1, 1] and the range is the
closed interval [0, 1].
√
1. f (x) = 1/ 1 − x2 .
√
For this function we need to find for which values of x the expression 1/ 1 − x2
is defined (this is the domain of the function). The expression has a denominator
which cannot be zero (recall we cannot divide by zero). There is, moreover, an even
radical which is defined only for positive values of the radicand.
The denominator is zero when x = 1 and x = −1. This means that these points must
be excluded from the domain of the function.
We need to find where the radicand (1 − x2 ) is positive. We then solve
1 − x2 > 0
(1 − x)(1 + x) > 0
The solution for this inequality is the interval (−1, 1). A general strategy for solving
this kind of inequalities is given at the end of the section.
Combining all of the above, we find the domain: it is the interval (−1, 1).
We look at the range now. We make a table for this function and plot the points:
0
x
f(x) 1
0.2
1.02
0.6
1.25
0.9 0.99
2.29 7.58
We can continue the table finding points closer and closer to 1. The value of the
function will increase without bound (for example, f (0.999) = 22.36). It can get as
much big as we please. So, all numbers bigger or equal than one are in the range.
Therefore, the range is all positive numbers bigger or equal to one.
This is a guess. But actually we let y ≥ 1 and solve for x.
y=√
y2 =
7
1
,
1 − x2
1
,
1 − x2
4.1. FUNCTIONAL DEFINITION
CHAPTER 4. ADVANCED ALGEBRA
1 − x2 =
1
,
y2
1
y2 − 1
=
,
y2
y2
#
y2 − 1
.
x=±
y
x2 = 1 −
The square root of y 2 − 1 is always smaller than y so the fraction has a numerator
smaller by absolute value than the denominator. Therefore, this number is always
between minus one and one. With other words there is a number x such that f (x) =
y.
Answer: the domain is the interval (−1, 1) and the range is [1, ∞).
2. f (x) = x2 + 1.
We see that this function is defined for all real numbers because there is no denominator or radical sign. We see that we have sum of two nonnegative numbers, x2 and
one. The first one is at least zero, and the second is 1. Therefore, their sum is bigger
or equal to one. It follows that the range of the function is [1, ∞).
Let y ≥ 1. We solve for x:
and
y = x2 + 1,
x2 = y − 1,
#
x1 = + y − 1
#
x2 = − y − 1.
With other words if we choose a number x there exists y such that f (x) = y.
Answer: The domain of the function is all real numbers and the range is interval
[1, +∞).
√
3. f (x) = 1 − x.
We know that the square root is defined only for the nonnegative values of the
radicand. Therefore, we need to solve the inequality
1 − x ≥ 0.
The solution of this inequality is
x ≤ 1.
Therefore, the domain is the interval (−∞, +1].
For the range we need to see which values the function might take. If we choose
a positive number, for example 10, there is a value of x, for which f (x) = 10 —
8
CHAPTER 4. ADVANCED ALGEBRA
4.1. FUNCTIONAL DEFINITION
it happens when x = −99. Therefore, the range is all positive numbers. On the
other hand, we cannot obtain negative values for square root function because when
taking square root we consider only the positive, the principal square root.
Let y ≤ 1. We solve for x:
√
y = 1 − x,
y 2 = 1 − x,
x = 1 − y2.
With other words there we can find a number x such that f (x) = y.
Answer: the domain of this function is the interval (−∞, +1] and the range [0, +∞).
4. We can find easily the domain and the range of a function using the graph of it. If we
project the graph onto x-axis, we find the domain of the function. When projecting
onto y-axis, we find the range. For this function they are [−10, 10) and [−2, 2). The
right endpoint for the domain and the upper endpoint for the range are excluded
because the function is not defined for them. Recall that filled point means that the
endpoint is included in the interval and not filled — excluded.
3 y
2
y
1
DOMAIN
Projection on the
x axis
We draw a vertical
line (parallel to y
axis) untill it crosses
the x axis
-10
-8
-6
-4
Projection on the
y axis.
-2 0
2
4 x 6
10 x
8
-1
RANGE
We draw a horizontal
line (parallel to x
axis) untill it crosses
the y axis.
-2
-3
The domain is the interval [-10, 10).
Ther range is the interval [-2, 2.2).
Answer: the domain is [−10, 10) and the range is [−2, 2 15 ) or [−2, 2.2).
5. Answer: domain [−10, 10), range (−2 14 , 2] or (−2.25, 2].
3
DOMAIN [-10, 10)
RANGE [-2.25, +2.0)
2
y
1
-10
-8
-6
-4
-2 0
-1
-2
-3
6. Answer: domain [−10, 2], range [0, 3 12 ].
9
2
4 x 6
8
10
4.1. FUNCTIONAL DEFINITION
CHAPTER 4. ADVANCED ALGEBRA
5
4
DOMAIN [-10, 2].
RANGE [0, 3.5].
3
y
2
1
-10
-8
-6
-4
-2 0
-1
4 x 6
2
8
10
-2
-3
7. This is a step-function. It is defined in an interval, but its functional values are
constants in the open or other kinds of intervals. The domain is [−12, 12]. the range
however is the set {1, 3, 6}.
12
10
8
y6
4
2
-12 -10 -8 -6 -4
0
-2
-4
-6
-8
-10
-12
2
4
6
x
8 10 12
THE DOMAIN IS THE INTERVAL [-12, 12].
THE RANGE IS THE SET OF THE NUMBERS
{1, 3, 6}.
Answer: domain [−12, 12], range {1, 3, 6}.
10
CHAPTER 4. ADVANCED ALGEBRA
4.1. FUNCTIONAL DEFINITION
4.1.3 CBa3: Know and apply vertical line test for functions to the graphs
of relations.
Use the vertical line test to determine if a graph is the graph of a function.
Recall the vertical line test: A curve is a function if for any value of x, there is only one
value of y corresponding to it. Geometrically, if we draw a vertical line, it will cross the
curve only ones. This means the curve is a function. If the curve is crossed more than
once by at least one vertical line, this is not a function. We make a counter example
with the definition: we found a value of x for which there are more than one value of y
corresponding.
a.
b.
12
10
8
y6
4
2
10
TEST LINE
8
6
y
4
-12 -10 -8 -6 -4
2
-10
-8
-6
-4
-2 0
-2
2
4 x 6
8
10
-4
-6
0
-2
-4
-6
-8
-10
-12
2
4
6
x
8 10 12
TEST LINE IS x=0, THE y AXIS. IT
CROSSES THE GRAPH TWICE:
AT y=3 AND y=6.
-8
-10
Solution: a. graph of a function,
b. not a graph of function
For b) there is only one spot where vertical line crosses the graph of the relation twice (at
x = 0 — y-axis). We can choose any point on the x-axis and to draw a vertical line, so we
choose x = 0. This line crosses the graph we check at y = 3 and y = 6. These cross-points
correspond to the x value we have chosen. Therefore, this is not a function.
11
4.1. FUNCTIONAL DEFINITION
CHAPTER 4. ADVANCED ALGEBRA
1. It is a function.
10
TEST LINE
8
6
4
2
-10
-8
-6
-4
-2 0
-2
2
4 x 6
8
10
2
4 x 6
8
10
2
4 x 6
8
10
-4
-6
-8
-10
2. It is a function.
10
8
6
4
2
-10
-8
-6
-4
-2 0
-2
-4
-6
-8
-10
3. It is a function.
10
8
6
y
4
2
-10
-8
-6
-4
-2 0
-2
-4
-6
-8
-10
12
CHAPTER 4. ADVANCED ALGEBRA
4.1. FUNCTIONAL DEFINITION
4. It is not a function. There are infinitely many lines at which vertical line crosses the
graph twice. Example: x = 4.4.
10
8
6
y
4
2
-10
-8
-6
-4
-2 0
-2
2
4 x 6
8
10
-4
-6
-8
-10
5. It is not a function. The line x = 0 crosses the graph many times.
10
8
6
4
2
-10
-8
-6
-4
-2 0
-2
2
4
6
8
10
2
4 x 6
8
10
-4
-6
-8
-10
6. It is a function.
10
8
6
4
2
-10
-8
-6
-4
-2 0
-2
-4
-6
-8
-10
13
4.1. FUNCTIONAL DEFINITION
CHAPTER 4. ADVANCED ALGEBRA
4.1.4 CBa4: Interpret the value of f(a) in applications.
1. The cost, C(b), to produce b algebra textbooks is given by C(b) = 35b + 1100 dollars.
The publisher gets 60 dollars for each text sold. If 115 texts are published and 100
are sold, what is the publisher’s profit or loss?
Let discuss what will happen if the publisher produces and sells 20 books. We see
on the graph that the cost for producing 20 books is C(20) = (35×20)+1100 = $1800.
cost (dollars)
revenue (dollars)
revenue line
cost line
algebra books
produced
C (20) = 1100 + 20x35 = 1800
revenue (20) = 20x60 = 1200
At the same time, if he sells all these books its revenue will be Revenue = 60 × 20 =
$1200. If the publisher produces 60 books the cost is C(60) = 35 × 60 + 1100 = 3200
and so on. We see that the cost is a linear function of number of books printed. The
same, the revenue is a linear function of numbers of books sold. We know that 115
books were published and 100 were sold. Then,
Profit/loss = Revenue(100) − C(115) =
= 60 × 100 − (35 × 115 + 1100) =
6000 − 5125 = 875.
The publisher has profit of $875 if he publishes 115 and sells 100 algebra books.
2. A balloon is being inflated. After t minutes its volume is V (t) = t/(1 + t) cubic
meters. What is the volume of the balloon after 10 minutes?
Lets analyze what happens after t = 2 minutes. The volume is V (2) = 2/(1+2) = 2/3
14
CHAPTER 4. ADVANCED ALGEBRA
4.1. FUNCTIONAL DEFINITION
volume
cubic meters. After t = 6 the volume is 6/7 cubic meters. In ten minutes the volume
is 10/11 meters. If this continues more time and the relation is the same, after 99
minutes the balloon volume is 0.99 cubic meters and so on. The volume gets closer
and closer to 1 cubic meter but never is equal to 1 cubic meter. The graph below
shows the process for the first 12 minutes and the volume in 2, 6 and 10 minutes.
time
Balloon volume - time
Answer:In ten minutes the volume of the balloon is 10/11 cubic meters.
S(t)
(dollars)
3. Cellphone service costs S(t) = 30 + 0.05t dollars per month, where t is the connect
time in minutes. How much will 200 minutes of connect time cost if it is all in one
month?
connection time (min)
S(50) = 30 + 0.05x50 = $32.50
S(200) = 30 +0.05x200 = $40.00
If all the minutes are in one month, so that we need to use the formula once. We
compute then:
S(200) = 30 + 0.05 × 200 = $40.00.
15
4.1. FUNCTIONAL DEFINITION
CHAPTER 4. ADVANCED ALGEBRA
4. Cellphone service costs S(t) = 30 + 0.05t dollars per month, where t is the connect
time in minutes. How much will 200 minutes of connect time cost if it is 99 minutes
are in May and 101 minutes are June?
We have a different situation now. In May, there were 99 minutes. Their cost is
S(99) = $34.95. The other 101 minutes were used in June. We need to use the same
formula again to find the cost of the cell phone service in June: S(101) = $35.05. All
200 minutes connection time cost $70.00.
5. The
√ approximate number of runs a softball team scores in a game is given R(w) =
4 + w, where w is the number of walks in a game. About how many runs are
scored if there are 5 walks?
# of points
This formula is an approximation of the number of runs scored related to the number of walks in the game. If there are 2 walks, the team will score about 2.45 runs.
We use the formula (or the graph) to see that if there are w = 5 walks, the team could
make about 3 runs.
# of runs
If there are 2 runs the team scores approximately 2.45 points. For 5 runs - about 3.
16
CHAPTER 4. ADVANCED ALGEBRA
4.1. FUNCTIONAL DEFINITION
4.1.5 CBa5: Qualitative properties and transformations of a function
and its graph.
Match the functions with their graphs.
8
6
4
2
–3
–2
–1
0
y = f(x)
1
x
2
3
–2
–4
–6
–8
Example. The graph of y = f (x) is given. Which is the graph of y = 2f (x)?
Solution:
8
6
4
2
–3
–2
–1
0
–2
–4
–6
–8
17
y = 2f(x)
1
x
2
3
4.1. FUNCTIONAL DEFINITION
CHAPTER 4. ADVANCED ALGEBRA
Match the function on the left with its graph on the right.
1.
y = −2f (x)
a.
8
6
4
2
–3
–2
0
–1
1
2
x
–2
3
4
–4
–6
–8
2.
y = f (x) + 2
b.
8
6
4
2
–3
–2
–1
0
1
x
2
3
1
x
2
3
1
x
2
3
–2
–4
–6
–8
3. y = f (x − 2)
c.
8
6
4
2
–3
–2
–1
0
–2
–4
–6
–8
4.
y = f (|x|)
d.
8
6
4
2
–3
–2
–1
0
–2
–4
–6
–8
Explanation and key:
Function y = −2f (x):
For each value of x the function y takes value of (−2) times the value of function f (x).
We can choose any point of the number line and compare the values. We use x = 1 and
compute y = −2f (1) = −2 × 1 = −2 Note, f (1) = 1. This is the graph c).
Function y = f (x) + 2:
Now, the function value is shifted up two units for each value of f (x). Then, we can
choose point x = 0: f (0) = 0 and y = f (0) + 2 = 0 + 2 = 2. We look for a graph, which
has value of 2 at x = 0. That is the graph d).
Function y = f (x − 2):
What we compute is that for the graph of y, the value of it at x=2 is the value of f (x) at
x = 0. If we denote y = y(x), then y(2) = f (2 − 2) = f (0) = 0. We look at the graphs and
find the one with value of 0 at x = 2. This is graph a).
18
CHAPTER 4. ADVANCED ALGEBRA
4.1. FUNCTIONAL DEFINITION
Function y = f (|x|):
This is the most difficult example. Recall that the absolute value function for any
positive number a is the number a, and for any negative value b is (−b) — its opposite
number. For our function y it means that for all positive values of x it has to be the same
graph as the graph of f (x), and for any negative values of x it must be the function
y = f (−x). For x = −1 we have y(−1) = f (| − 1|) = f (1) = 1. The only graph with these
properties is graph b).
Answer: 1-c), 2-d), 3-a), 4-b).
19
4.2. ALGEBRA OF FUNCTIONS
4.2
CHAPTER 4. ADVANCED ALGEBRA
Algebra of Functions
4.2.1 CBb1: Perform arithmetic operations, difference quotient type.
Given a function defined by an expression, simplify a difference quotient for the function.
Example: f (x) = x2 . Simplify (f (x) − f (2))/(x − 2). We compute the ratio
f (x) − f (2)
x−2
x2 − 22
.
x−2
We factor the numerator of the above fraction (x2 − 22 ) = (x − 2)(x + 2) and get
=
=
(x − 2)(x + 2)
.
x−2
We divide out the factor (x − 2) in the numerator and denominator and find the answer
= x + 2 for x $= 2.
Answer: x + 2 for x $= 2.
1. f (z) = 2z 2 + 3. Simplify (f (x) − f (2))/(x − 2).
We use the same method to simplify the expression
2x2 + 3 − (2 × 22 + 3)
2(x2 − 22 )
f (x) − f (2)
=
=
x−2
x−2
x−2
We factor out the numerator 2(x2 − 22 ) = 2(x − 2)(x + 2) :
=
2(x − 2)(x + 2)
= 2(x + 2).
x−2
Answer: 2(x + 2) for x $= 2.
2. f (z) = 1/z. Simplify (f (1 + h) − f (1))/h.
We write the ratio as follows:
=
1
(f (1 + h) − f (1))
h
%
$
1
1
−1
=
h 1+h
=
1 − (1 + h)
(1 + h)h
20
CHAPTER 4. ADVANCED ALGEBRA
=
4.2. ALGEBRA OF FUNCTIONS
−h
1
=−
.
(1 + h)h
1+h
Recall that fractions could be added or subtracted only if they have one and the
same denominator. We find the least common denominator for fractions 1/(1 + h)
and 1/1. It is 1(1 + h). So, we do not need to multiply the first fraction but we do
multiply the second one by (1+h)/(1+h). We have the fraction (−h)/(1+h) divided
by h, so we simplify the expression writing both denominators as one
Answer: −1/(1 + h) for h $= 0.
3. f (z) = 2z + 1. Simplify (f (2 + h) − f (2))/h.
We compute the above ratio
=
=
1
[f (2 + h) − f (2)]
h
1
[2(2 + h) + 1 − (2 × 2 + 1)]
h
2(2 + h − 2)
=
= 2.
h
Answer: 2.
4. f (z) = 1/z 2 . Simplify (f (x) − f (−1))/(x + 1)
%
$
1
1
1
=
−
x + 1 x2 (−1)2
%
$
1
1 − x2
=
=
x+1
x2
(1 − x)(1 + x)
1−x
=
.
2
x (1 + x)
x2
We remove equal factors from the numerator and the denominator (1 + x).
Answer: (1 − x)/x2 for x $= −1.
5. f (x) = 3x2 + 4x + 9. Simplify (f (3 + h) − f (3))/h
1
[3(3 + h)2 + 4(3 + h) + 9 − (32 − 4 × 3 + 9)]
h
=
1
{3[(3 + h)2 − 32 )] + 4[(3 + h) − 3] + 9 − 9}
h
1
= [3(3 + h − 3)(3 + h + 3) + 4(3 + h − 3)]
h
22h + 4h2
1
= [3h(6 + h) + 4h] =
h
h
=
21
4.2. ALGEBRA OF FUNCTIONS
CHAPTER 4. ADVANCED ALGEBRA
We remove a factor of h and find:
= 22 + 4h.
Answer: 22 + 4h for h $= 0.
22
CHAPTER 4. ADVANCED ALGEBRA
4.2. ALGEBRA OF FUNCTIONS
4.2.2 CBb2: Form and evaluate composite functions.
Given two functions, evaluate their composites numerically and graphically.
Example: If f (2) = 4 and g(4) = 8 then g(f (2)) = 8. If f (x) = x2 and g(x) = x + 3 then
g(f (x)) = x2 + 3.
1. If f (3) = 12 and g(12) = 19, what is f (g(12) − 16)?
We replace g(12) with its equivalent, 19 and compute
f (g(12) − 16) = f (19 − 16) = f (3) = 12.
2. If f (4) = −1 and g(x) = x2 + 11, what is g(2f (4))?
Even we do not know the expression the function f (x), we know that f (4) = −1.
We insert this value in the expression g(2f (4)) and evaluate it:
g(2f (4)) = g(2 × (−1)) = g(−2)
= (−2)2 + 11 = 15.
3. If f (4) = −1, f (−1) = 4 and g(3) = −1, what is f (g(3))?
We use the value of g(3) in the expression f (g(3)) and evaluate it
f (g(3)) = f (−1) = 4.
4. If f (4) = −1, f (−1) = 4, g(3) = −1 and g(−1) = 41, what is g(f (4))?
Using the same idea we write
g(f (4)) = g(−1) = 41.
5. If f (40) = −1, f (−10) = 4, g(4) = −1 and g(−1) = 41, what is g(f (−10))?
g(f (−10)) = g(4) = −1.
23
4.2. ALGEBRA OF FUNCTIONS
CHAPTER 4. ADVANCED ALGEBRA
4.2.3 CBb3: Identify whether an increase in value of one variable in
an expression will cause increase or decrease in the value of the
expression.
Given an equation involving two variables, how does increasing or decreasing one variable affect the value of the other variable?
Example: x + y = 2. Answer: As x increases y decreases.
1. x − y = 7.
We solve the above for x.
x = y + 7.
We investigate what happens if we increase y. Let y = 2 then x = 2 + 7 = 9. Let now
we change x, by using increased value of y = 3. Then x = 10. Since 3 > 2 and 10 > 9
x increases as y increases.
Answer: as y increases x increases.
2. xy = 7.
We express one variable in terms of the other.
y=
7
.
x
We can choose two values of x and check that when one of them is bigger than the
other, then the corresponding values of y will increase too. Let x = 1 then y = 71 = 7.
If x = 2 then, y = 3.5 and x = 7, y = 1. Therefore as x increase, y decrease.
We can check negative values as well. Let x = −7, then y = −1. We choose second
value: x = −1, then y = −7. In this case, the function decrease too. This function
is not defined for x = 0 because if we multiply zero by any number, it is still zero.
We can make a conclussion that the variable y decrease when x increase through
negative values and it does so for positive values. (But!)When x is negative, y is
negative too, and when x is positive, y is positive too. For a function, to be increasing
one for any values x1 < x2 we need to have y1 < y2 . Choosing one negative number
(say, -7) and one positive (say 2) we have this inequality true. Choosing two positive
numbers, it is opposite. This means that the expression xy = 7 decreases in intervals
(−∞, 0) and (0, ∞) but if we consider all real numbers, it neither.
Answer: As x increases through positive values y decrease.
3. x − y 3 = 7.
We express x in terms of y.
x = y 3 + 7.
We choose some values of y and compute the values of x.
y = 0 then x = 7;
y = 1 then x = 8;
24
CHAPTER 4. ADVANCED ALGEBRA
4.2. ALGEBRA OF FUNCTIONS
y = 2 then x = 15.
y = −1 then x = 6.
y = −2 then x = −1.
We see that as y increases, x increases.
Answer: As y increases x increases.
4. y = x/(x + 1).
We choose some positive values of x and compute corresponding values of y.
x = 1 then y = 21 ;
x = 2 then y = 23 ;
x = 3 then y = 34 .
Aside of above computations, we can use systematic way of solving this problem.
We can express the algebraic expression in a following way:
y=
=
x+1−1
x
=
x+1
x+1
1
1
x+1
−
=1−
.
x+1 x+1
x+1
In this expression, we have the constant 1 reduced by the fraction 1/(x + 1). When
we increase x, the denominator becomes bigger, and therfore, the fraction becomes
smaller (see also Chapter 1.3.1 ARc2). We subtract smaller quantity if x increases,
therefore the value of the expression increases.
We see that when x increases y increases too. For this problem, we can use the same
reasoning as in problem #1: for all real numbers, the expression is neither.
Answer: As x increases through positive values, y increases.
5. x2 + y 2 = 1.
We express x in terms of y:
x=
#
1 − y2.
We try to find x, therefore this is an equation for x. This means, we need to find
all values of#x for all possible y. The equation for x has real solution only if the
expression 1 − y 2 > 0. That is when −1 ≤ y ≤ +1. The square root funciton
returns only positive values. We know that there are always two numbers with one
and the same square. Therefore, we have two expressions for x, not one:
x=
#
#
1 − y 2 or x = − 1 − y 2 ,
25
y ∈ [−1, 1].
4.2. ALGEBRA OF FUNCTIONS
CHAPTER 4. ADVANCED ALGEBRA
y
y
We see for any value of y there are two corresponding values:
one bigger than zero and one smaller. We cannot make any
conclusions.
y
Combining the two graphs, we see that this is an
equation of a circle.
Answer: It cannot be drown any conclussion for the expression x2 + y 2 = 1.
26
CHAPTER 4. ADVANCED ALGEBRA
4.3
4.3. INVERSE FUNCTION
Inverse function
4.3.1 CBh1: Recognize domain and range relationship between f and
f −1 .
Recognize the relationship between the domain and range of a function and its inverse.
Example: The domain of f is [0, 5] and the range of f is [9, 22]. The domain of f −1 is
[9, 22] and the range of f −1 is [0, 5]. f (4) = 7 so f −1 (7) = 4.
Recall that f −1 means the inverse of function f, not the reciprocal of f. Next, the function has an inverse if it is a one-to-one correspondence (it means that any two different
elements in the domain of the function correspond different elements in the range).
1. The domain of f is [0, 20], the range of f is [9, 22], and f (12) = 15.
If f (12) = 15 it means that for the element of the domain 12 the corresponding element of the range is 15, i.e. the our function machine transforms 12 into 15. If we
take the inverse function machine, it will send 15 into 12.
Recall that the range of the original function is the domain of its inverse and the
range of the inverse is the domain of the original function.
f −1 (15) = 12,
the domain of f −1 is [9, 22], and the range of f −1 is [0, 20].
2. The domain of f is [5, 20], the domain of f −1 is [9, 22], and f −1 (12) = 7.
We are not given any ranges but we do know the domains of the function and its
inverse. We also know that the domain of the inverse is the same as the range of the
function and the range of the inverse is the domain of the function.
Now, we know what the inverse of 12 is. The inverse function machine transforms
12 into 7, so the function machine must transform 7 into 12.
f (7) = 12,
the range of f is [9, 22], and the range of f −1 is [0, 20].
3. The range of f is [3, 20], the range of f −1 is [0, 22], and f (9) = 5.
27
4.3. INVERSE FUNCTION
CHAPTER 4. ADVANCED ALGEBRA
We know all the ranges but not the domains. We make use of the fact that the range
of the function is the domain of the inverse, and the range of the inverse is the same
as the domain of the function.
The function machine sends 9 into 5 so that the inverse function machine is sending
5 into 9.
f −1 (5) = 9,
the domain of f is [0, 22], and the domain of f −1 is [3, 20].
4. The domain of f is [−20, 20], the range of f is [9, 22], and f (12) = 13.
Using the same reasoning as above, we write:
f −1 (13) = 12,
the domain of f −1 is [9, 22], and the range of f −1 is [−20, 20].
5. The domain of f −1 is [0, 20], the range of f −1 is [9, 22], and f −1 (5) = 12.
f (5) = 12,
the domain of f is [9, 22], and the range of f is [0, 20].
28
CHAPTER 4. ADVANCED ALGEBRA
4.3. INVERSE FUNCTION
4.3.2 CBh2: Recognize relationship between f and f −1 (given f as graph
or as rule).
Recognize the inverse of a function given by a rule or by a graph.
Example: Suppose f (x) = 3x + 1. Then f −1 (x) = (x − 1)/3 because f ((x − 1)/3) = x and
f −1 (3x + 1) = x.
1. Suppose f (x) = 5x − 4. Which of the following is f −1 (x)?
(a) 4 + (x/5); (b) (x+4)/5; (c) 5x + 4;
(d)
4x + 5;
(e) (x-4)/5.
We use the strategy for finding inverses:
Step 1. We want to see how x depends on y. For this reason, we make new y (former
x) and new x (former y).Thus, we interchange x and y:
y = 5x − 4 → x = 5y − 4.
Step 2. Solve for y:
5y = x + 4,
1
y = (x + 4).
5
Step 3. Replace y with f −1 :
f −1 =
x+4
.
5
Answer: The inverse of f (x) = 5x − 4 is f −1 = (x + 4)/5 or b).
2. Suppose f (x) = (5x + 2)/(7x + 3). Which of the following is f −1 (x)?
(a) (7x+3)/(5x + 2); (b) (x-2)/(-7x + 5); (c) (3x-2)/(5-7x);
(d) (2x+5)/(3x+7); (e) (5-2x)/(7-3x).
We use the same strategy:
Step 1:
y=
5x + 2
7x + 3
→
Step 2:
x=
x=
5y + 2
.
7y + 3
5y + 2
,
7y + 3
We multiply both sides by (7y + 3):
x(7y + 3) = 5y + 2,
29
4.3. INVERSE FUNCTION
CHAPTER 4. ADVANCED ALGEBRA
We subtract from both sides 5y and open the parenthesis on the left side:
7xy + 3x − 5y = 2,
We subrtact 3x from both sides and factor out y on the left side:
y(7x − 5) = 2 − 3x,
We divide both sides by (7x − 5):
y=
2 − 3x
,
7x − 5
or, equivalently,
3x − 2
.
5 − 7x
The last result is the fraction, both the numerator and the denominator multiplied
by (-1).
Step 3:
3x − 2
.
f −1 =
5 − 7x
y=
Answer: the inverse of f (x) = (5x + 2)/(7x + 3) is f −1 = (3x − 2)/(5 − 7x).
30
CHAPTER 4. ADVANCED ALGEBRA
4.3. INVERSE FUNCTION
Match the graphs in the first column with their inverses in the second column.
1.
a.
10
10
8
8
6
y
4
6
y
4
2
–10 –8
–6
–4
2
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
2.
2
4 x 6
8
10
2
4 x 6
8
10
2
4 x 6
8
10
2
4 x 6
8
10
b.
10
10
8
8
6
y
4
6
y
4
2
–10 –8
–6
–4
2
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–4
–2 0
–2
–4
–6
–6
–8
–8
–10
–10
3.
c.
10
10
8
8
6
y
4
6
y
4
2
–10 –8
–6
–4
2
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
4.
d.
10
10
8
8
6
y
4
6
y
4
2
–10 –8
–6
–4
–2 0
–2
2
2
4 x 6
8
10
–10 –8
–4
–6
–4
–2 0
–2
–4
–6
–6
–8
–8
–10
–10
(answer on the next page)
31
4.3. INVERSE FUNCTION
CHAPTER 4. ADVANCED ALGEBRA
Recall that the graph of the inverse is symmetric to the graph of the function with respect to the line y = x. The way to find the graph of the inverse is to find a graph which
is symmetric to the graph of the function with respect to the line y = x.
The other way of checking your answer is to use the fact that all the functions have
different domains and ranges. We can use these to match the functions and their inverses.
For example, function (2) has domain [-4, 5] and range [0, 3]. The only function with
domain [0, 3] and range [-4, 5] is (d). In the last right colunn we graph both functions. We
see that these are inverses of each other!
1.
c.
1c
10
10
8
8
6
y
4
6
y
4
2
–10 –8
–6
–4
–2 0
–2
10
8
6
y
4
2
2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
2.
2
4 x 6
8
10
-10
-8
-6
-4
-2 0
-2
–4
10
10
8
8
6
y
4
6
y
4
–2 0
–2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
3.
–6
–4
10
8
8
6
y
4
6
y
4
2
2
–2 0
–2
–2 0
–2
4 x 6
8
10
–10 –8
–6
–4
–4
–4
–6
–6
–8
–8
–10
–10
4.
–6
–4
10
8
8
6
y
4
6
y
4
2
2
–2 0
–2
–2 0
–2
4 x 6
8
10
4 x 6
8
10
4 x 6
8
10
2
2
4 x 6
8
10
-10
-8
-6
-4
-2 0
-2
2
-4
-6
-8
-10
10
8
6
y
4
2
2
4 x 6
8
10
-10
-8
-6
-4
-2 0
-2
2
-4
-6
-8
-10
4a
10
2
10
8
6
y
4
a.
–10 –8
8
10
3b
10
2
4 x 6
-8
b.
–10 –8
10
-6
2
2
8
2d
2
–6
4 x 6
-10
d.
–10 –8
2
-4
–10 –8
–6
–4
–4
–4
–6
–6
–8
–8
–10
–10
32
10
8
6
y
4
2
2
4 x 6
8
10
-10
-8
-6
-4
-2 0
-2
-4
-6
-8
-10
2
CHAPTER 4. ADVANCED ALGEBRA
4.3. INVERSE FUNCTION
4.3.3 CBh3: Identify one-to-one functions from graph.
Identify the graphs of one-to-one functions.
Example: Which of the following graphs are those of one-to-one functions? Circle your
answers.
Definition: a function is one-to-one if for any two different inputs we have different outputs.
If there is at least one output to which there are at least two different inputs pointing, the
function is not one-to-one. In practice, we draw horizontal line and cross the graph. If
there is any such line crossing the graph more than once, the function is not one-to-one.
Algebraically we interchange x and y and solve for y. If we get only one expression for y
this means the function is one-to-one.
a.
b.
10
10
8
8
TEST LINE
6
y
4
6
4
2
2
-10
-8
-6
-4
-2 0
-2
2
4 x 6
8
-10
10
-8
-6
-4
-2 0
-2
-4
-4
-6
-6
-8
-8
-10
-10
Solutions: a. function is not one-to-one.
2
b. function is one-to-one.
33
4 x 6
8
10
4.3. INVERSE FUNCTION
CHAPTER 4. ADVANCED ALGEBRA
1.
2.
–10 –8
–6
–4
10
10
8
8
6
6
4
4
2
2
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
3.
2
4 x 6
8
10
2
4 x 6
8
10
2
4 x 6
8
10
4.
–10 –8
–6
–4
10
10
8
8
6
y
4
4
2
2
–2 0
–2
6
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
5.
6.
–10 –8
–6
–4
10
10
8
8
6
6
4
4
2
2
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
7.
8.
10
8
4
6
4
2
2
–10 –8
–6
–4
–2 0
–2
2
4 x 6
8
10
–2
–4
–1
0
–2
–6
–8
–4
–10
34
1
x
2
CHAPTER 4. ADVANCED ALGEBRA
4.3. INVERSE FUNCTION
The criterion for check one-to-one function is the following: if we can draw a horizontal line and it crosses the graph more than once, then the function is not one-to-one. Such
functions in the set are 3, 4, 6, 7.
3.
4.
10
10
8
8
6
y
4
4
6
2
–10 –8
–6
–4
–2 0
–2
2
2
4 x 6
8
–10 –8
10
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
6.
2
4 x 6
8
10
2
4 x 6
8
10
7.
10
10
8
8
6
6
4
4
2
–10 –8
–6
–4
–2 0
–2
2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
The other ones 1, 2, 5, 8 are one-to-one:
35
4.3. INVERSE FUNCTION
CHAPTER 4. ADVANCED ALGEBRA
1.
2.
–10 –8
–6
–4
10
10
8
8
6
6
4
4
2
2
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
5.
2
4 x 6
8
10
8.
10
8
4
6
4
2
2
–10 –8
–6
–4
–2 0
–2
2
4 x 6
8
10
–2
–4
–1
0
–2
–6
–8
–4
–10
36
1
x
2
CHAPTER 4. ADVANCED ALGEBRA
4.3. INVERSE FUNCTION
4.3.4 CBh4: Find f −1 given f .
Given f (x), find f −1 (x).
Example: f ( x) = 3x + 7. Answer: f −1 (x) = (x − 7)/3. Recall the steps given in CBh2:
Step 1: Interchange x and y.
Step 2: Solve for y.
Step 3: Replace y with f −1 .
1. f (x) = 5x − 9 :
Step 1:
y = 5x − 9
→
x = 5y − 9,
Step 2:
y=
x+9
,
5
Step 3:
f −1 =
x+9
.
5
We check computing f (f −1 (x)):
f (f −1 (x)) = f ((x + 9)/5)
$
%
x+9
=5
− 9 = x + 9 − 9 = x.
5
We proved that f (f −1 (x)) = x so the above functions are indeed inverses of each
other.
Answer: f −1 (x) = (x + 9)/5.
2. f (x) = (5x − 9)/(2x − 1) :
Step 1:
y=
5x − 9
2x − 1
→
x=
5y − 9
,
2y − 1
Step 2:
x=
5y − 9
,
2y − 1
x(2y − 1) = 5y − 9,
37
4.3. INVERSE FUNCTION
CHAPTER 4. ADVANCED ALGEBRA
2xy − x − 5y = −9,
y(2x − 5) = x − 9,
y=
x−9
.
2x − 5
Step 3:
f −1 =
We check computing f (f −1 (x)):
x−9
.
2x − 5
f (f −1 (x)) = f ((x − 9)/(2x − 5))
=
5[(x − 9)/(2x − 5)] − 9
2[(x − 9)/(2x − 5)] − 1
=
5(x − 9) − 9(2x − 5)
2(x − 9) − (2x − 5)
−13x
5x − 45 − 18x + 45
=
= x.
2x − 18 − 2x + 5
−13
The concussion is the same as in #1: the functions are inverses of each other.
Answer: f −1 (x) = (x − 9)/(2x − 5).
=
3. f (x) = 5x3 − 9 :
Step 1:
y = 5x3 − 9
→
x = 5y 3 − 9,
Step 2:
x = 5y 3 − 9,
1
y 3 = (x + 9),
5
&
3 x + 9
,
y=
5
#
Step 3: f −1 = 3 (x + 9)/5.
We check the same way:
#
f (f −1 (x)) = f ( 3 (x + 9)/5)
(3
'#
x+9
= 5 3 (x + 9)/5 − 9 = 5 ×
− 9 = x.
5
#
Answer: f −1 (x) = 3 (x + 9)/5.
38
CHAPTER 4. ADVANCED ALGEBRA
4.3. INVERSE FUNCTION
4. f (x) = 5(x − 9)3 :
Step 1:
y = 5(x − 9)3
x = 5(y − 9)3 ,
→
Step 2:
x = 5(y − 9)3 ,
x
(y − 9)3 = ,
5
[(y − 9)3 ]1/3 = y − 9 =
y=
Step 3:
f
−1
) x *1/3
5
=
&
3
Check:
f (f
=5
Answer: f −1 (x) =
−1
+&
3
(x)) = f
) x *1/3
5
,
+ 9.
x
+ 9.
5
$&
3
%
x
+9
5
,3
x
x
+ 9 − 9 = 5 = x.
5
5
#
3
x/5 + 9.
Note that the functions in #3 and #4 are completely different even that they look
very similarly. Their inverses look similar too, but these are different functions as
well.
5. f (x) = (x − 9)2 for x ≤ 9 :
NOTE: The condition x ≤ 9 is necessary because without it, the function is not oneto-one and therefore will not have an inverse.
Step 1:
y = (x − 9)2
→
x = (y − 9)2 ,
Step 2:
x = (y − 9)2 ,
[(y − 9)2 ]1/2 = x1/2 ,
39
4.3. INVERSE FUNCTION
CHAPTER 4. ADVANCED ALGEBRA
|y − 9| =
√
x.
We need absolute value on the left side because the right side is always positive.
The domain of the original function is the interval (−∞, 9], so the number |y − 9| is
negative. Its absolute value is |y − 9| = −(y − 9) = 9 − y, so
√
|y − 9| = 9 − y = x,
√
y = 9 − x.
Step 3:
f −1 = 9 −
We check:
Answer: f −1 (x) =
√
√
x.
√
f (f −1 (x)) = f (9 − x)
√
√
= [9 − x − 9]2 = (− x)2 = x.
x + 9 for x ≤ 9.
40
CHAPTER 4. ADVANCED ALGEBRA
4.4
4.4. EXP AND LOG FUNCTIONS
Exponential and Logarithmic Functions
4.4.1 CBi1: Recognize graphs of exponential and logarithmic functions
and their inverse relations .
Recognize the inverse function relation between exponential and logarithmic functions
graphically. For example the inverse of exp(x) is log(x).
10
y = exp(x)
8
y=x
6
y
4
2
–10 –8 –6 –4 –2 0
–2
–4
–6
–8
–10
41
y = ln(x)
2
4x6
8 10
4.4. EXP AND LOG FUNCTIONS
CHAPTER 4. ADVANCED ALGEBRA
Match the graph of the function on the left with the graph of its inverse on the right.
1.
a.
10
5
8
4
6
y
4
3
y
2
2
–10 –8
–6
–4
–2 0
–2
1
2
4 x 6
8
10
–5
–4
–3
–2
–4
–1 0
–1
1
2 x 3
4
5
2
4 x 6
8
10
2
4 x 6
8
10
2
4 x 6
8
10
–2
–6
–3
–8
–4
–10
–5
2.
b.
10
10
8
8
6
y
4
6
y
4
2
–10 –8
–6
–4
–2 0
–2
2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
3.
c.
5
10
4
8
3
y
2
6
y
4
1
–5
–4
–3
–2
–1 0
–1
2
1
2 x 3
4
5
–10 –8
–6
–4
–2 0
–2
–2
–4
–3
–6
–4
–8
–5
–10
4.
d.
10
10
8
8
6
y
4
6
y
4
2
–10 –8
–6
–4
–2 0
–2
2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
42
CHAPTER 4. ADVANCED ALGEBRA
4.4. EXP AND LOG FUNCTIONS
Recall that the graphs of the inverses are reflections of each other. The exponential
functions and their inverses are no exceptions.
1.
b.
10
–10 –8
–6
–4
10
8
8
6
y
4
6
y
4
2
2
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
2
4 x 6
8
10
The domain of #1 is (−∞, ∞) and the range is (0, ∞). For b), the range is (−∞, ∞), the
same as the domain of #1, and the domain is (0, ∞). With other words,
The domain of #1 is the same as range of b), and
the range of #1 is the same as the domain of b).
These are reflections of each other with respect to the line y = x.
d.
2.
10
10
8
8
6
y
4
6
y
4
2
–10 –8
–6
–4
–2 0
–2
2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
2
4 x 6
8
10
The domain of #2 is the same as range of d): (−∞, +∞), and the range of #2 is the
same as the domain of d): (2, +∞). These are reflections of each other with respect to the
line y = x.
3.
a.
5
5
4
4
3
y
2
3
y
2
1
–5
–4
–3
–2
–1 0
–1
1
1
2 x 3
4
5
–5
–2
–4
–3
–2
–1 0
–1
1
2 x 3
4
5
–2
–3
–3
–4
–4
–5
–5
The domain of #3 is the same as range of a): (−∞, +∞), and the range of #3 is the
same as the domain of a): (0, +∞). These are reflections of each other with respect to the
line y = x.
43
4.4. EXP AND LOG FUNCTIONS
CHAPTER 4. ADVANCED ALGEBRA
4.
c.
10
10
8
8
6
y
4
6
y
4
2
–10 –8
–6
–4
–2 0
–2
2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–6
–6
–8
–8
–10
–10
2
4 x 6
8
10
The domain of #4 is the same as range of c): (−∞, +∞), and the range of #4 is the same
as the domain of c): (−∞, 0). These are reflections of each other with respect to the line
y = x.
44
CHAPTER 4. ADVANCED ALGEBRA
4.4. EXP AND LOG FUNCTIONS
4.4.2 CBi2: Manipulate expressions using logarithmic functions: log(ab),
log(ap ).
Simplify expressions using the multiplicative and exponent properties of logarithms.
Example: If log(a) = 2 and log(b) = 7 then log(ab) = 9 and log(a3 ) = 6.
In the following a > 0 and b > 0. Do not assume that the base of the logarithm is 10.
Recall the properties of the logarithms that the logarithm of a product of two numbers is
the sum of the logarithms of the numbers. We can represent a2 as a × a, and, using the
property mentioned earlier, log(a2 ) = 2 log(a).
1. If log(a) = 3 and log(b) = −2, then
log(ab)
= log(a) + log(b)
= 3 + (−2) = 1,
and
log(a4 )
= 4 log(a) = 4 × 3 = 12.
2. If log(a) = 2 and log(b) = −2, then
log(ab)
= log(a) + log(b)
= 2 + (−2) = 0,
and
log(a4 b2 )
= log(a4 ) + log(b2 )
= 4 log(a) + 2 log(b)
= 4 × 2 + 2 × (−2) = 4.
3. If log(a) = 1.2 and log(b) = −2.3, then
log(ab)
= log(a) + log(b)
and
= 1.2 + (−2.3) = −1.1,
log(a−2 )
= (−2) log(a) = −2 × 1.2 = −2.4.
45
4.4. EXP AND LOG FUNCTIONS
CHAPTER 4. ADVANCED ALGEBRA
4. If log(a) = −1.3 and log(b) = −2.9, then
log(ab)
= log(a) + log(b)
= 1.3 + (−2.9) = −4.2,
and
log(a1.1 )
= 1.1 × log(a)
= 1.1 × (−1.3) = −1, 43.
5. If log(a2 ) = 3 and log(b3 ) = −2, then
Recall that log(a2 ) = 2 log(a). Therefore, log(a) = 1/2 log(a2 ). We compute the same
way: log(b) = 1/3 log(b3 ).
log(ab)
= log(a) + log(b)
=
1
1
log(a2 ) + log(b3 )
2
3
5
3 2
= − = ,
2 3
6
and
log(a3 )
1
log(a2 )
2
1
9
1
= 3 × log(a2 ) = 3 × × 3 = .
2
2
2
= 3 log(a) = 3 ×
46
CHAPTER 4. ADVANCED ALGEBRA
4.4. EXP AND LOG FUNCTIONS
4.4.3 CBi3: Write logarithmic and exponential functions in equivalent
forms: y = bx ≡ logb (y) = x.
Convert each equation to an equivalent form.
Example: 10x = 4 is the same as log10 (4) = x.
In above example, 10 is the base of the logarithm, x is the function value, and 4 is the
argument of the logarithm function.
1. We use the same method as shown above: 5 is the base of the logarithm, 7 is the
argument of the logarithm function and x is the function value:
5x = 7 is the same as
log5 (7) = x.
2. The base now is a, the logarithm argument is 7, and the function value is 3:
a3 = 7 is the same as
loga (7) = 3.
3. The base is 5, the logarithm argument is y and the function value is 4:
54 = y is the same as
log5 (y) = 4.
4. We reverse the process now - the base of the logarithm is the base of the exponent 7,
the logarithm function value is the exponent y and the argument is the result of the
exponentiation, 9:
log7 (9) = y is the same as
7y = 9.
5. The base is b, the exponent is 19, the result of the exponentiation is 12:
logb (12) = 19 is the same as
b19 = 12.
47
4.4. EXP AND LOG FUNCTIONS
CHAPTER 4. ADVANCED ALGEBRA
6. The base is 3, the exponent is −1.1 and the result of the exponentiation is x:
log3 (x) = −1.1 is the same as
3−1.1 = x.
48
CHAPTER 4. ADVANCED ALGEBRA
4.4. EXP AND LOG FUNCTIONS
4.4.4 CBi4: Solve literal equations using properties of exponents or
logs.
Solve the equation for the indicated variable.
Example: Solve 10x = y for x. Answer: x = log10 (y).
We use the idea shown in CBi3 to solve above for x — the base of the logarithm is the
base of the exponent, the exponentiation value becomes the argument of the logarithm
and exponent becomes the logarithm.
1. Solve y = 10 + 3t for t
We isolate the exponent on one side of the equation, then express the exponent in
its equivalent logarithmic form:
3t = y − 10,
Answer: t = log3 (y − 10).
log3 (y − 10) = t.
2. Solve P = Aert for t:
In this example, we first divide both sides of the equation by A:
P
= ert .
A
The logarithm base e is called natural logarithm. It is denoted by ’ln’. We use the
property that taking the logarithm of exponent function is the exponent itself:
rt = ln(ert ) = ln(P/A).
We now multiply both sides by 1/r:
t=
1
ln(P/A).
r
Answer: t = 1/r ln(P/A).
3. Solve 2x − 2−x = 2y for x:
This equation needs a little more work to be solved. Let’s make a substitution 2x = z.
Since
2−x = (2x )−1
the equation is
z − z −1 = 2y.
We can multiply the equation by z because z = 2x $= 0. After the multiplication we
find we have a quadratic equation for z:
z 2 − 2yz − 1 = 0.
49
4.4. EXP AND LOG FUNCTIONS
CHAPTER 4. ADVANCED ALGEBRA
The equation has two real roots because the discriminant is always bigger than zero:
D = (−2y)2 − 4 × 1 × (−1) = 4y 2 + 4 > 0.
Using the quadratic formula, we write:
#
#
2y + 4y 2 + 4
z1 =
= y + y 2 + 1.
2
Replacing z with 2x we find
2x = y +
Then,
#
x = log2 (y +
Answer: x = log2 (y +
#
y 2 + 1).
y 2 + 1.
#
y 2 + 1).
Note the equation z 2 − 2yz − 1 = 0 has one positive and one negative root. The
exponentiation cannot give us negative value so we do not consider the negative
solution of the quadratic equation.
4. Solve logb (x) = y for b:
We express the logarithm into exponential form:
by = x.
Now we raise both sides to power 1/y:
(by )1/y = x1/y ,
b = x1/y .
Answer: b = x1/y .
5. Solve logb (xp ) = y for p:
We use the property of logarithms log(ap ) = p × log(a) and write:
logb (xp ) = p × logb (x) = y.
We can multiply both sides of this equality by 1/(logb (x)) assuming x $= 1:
p=
Answer: p = y/ logb (x).
50
y
.
logb (x)
CHAPTER 4. ADVANCED ALGEBRA4.5. MANIPULATIONS OF ABSOLUTE VALUES
4.5
Manipulations of Absolute Values
4.5.1 CBc1: Solve |ax + b| < c.
Solve simple inequalities involving absolute value.
Example: If |2x − 4| < 10 then −3 < x < 7.
Recall the absolute value principle:
1. The equation with absolute value may have two solutions because there are two numbers with one and the same absolute value (for example | − 3.1| = |3.1| = 3.1). We see that
|P | = a is equivalent to P = a and P = −a.
2. The inequality
is equivalent to the system
This system can be written as
|P | < a
−a < P < a.
−a < P
AND
P < a.
The solution set is all values which place the expression P in the interval (-a,a).
3. The inequality
|P | > a
is equivalent to the union of the inequalities
P >a
OR
P < −a.
Note that if we have greater than or equal sign, it will give us closed intervals while
strict inequalities give open intervals.
1. Solve for x: |3x + 9| < 15.
We note that we are in case #2 therefore we need to solve
−15 < 3x + 9 < 15.
To avoid confusion, we will solve the left part first: We subtract (9) from both sides
of inequality
−15 − 9 < 3x + 9 − 9,
51
4.5. MANIPULATIONS OF ABSOLUTE VALUESCHAPTER 4. ADVANCED ALGEBRA
We divide by 3 the inequality.
−24 < 3x.
24 ÷ 3 < 3x ÷ 3,
−8 < x.
The solution set is contained in all real numbers bigger than -8. We now turn to the
right side. We subtract (9) from both sides of the inequality:
3x + 9 − 9 < 15 − 9,
3x < 6.
We divide now both sides by 3:
3x ÷ 3 < 6 ÷ 3,
x < 2.
The solution set is all real numbers smaller than 2. What the solution of the inequality |3x + 9| < 15 is the intersection of the two solution sets found above:
x > −8
and
x < 2.
This is the interval (-8,2).
Check: we cannot check all the numbers in the interval (−8, 2). We can, however,
choose any number in the interval and check for it. One good choice is zero if it is
in the interval. So we choose zero now:
|3 × 0 + 9| < 15,
9 < 15 TRUE!
Answer: the solution of |3x + 9| < 15 is the interval (−8, 2).
2. Solve for x: |x − 9| ≥ 4.
We are now in the last case of described above principle. Then, the absolute value
inequality |x − 9| ≥ 4 is equivalent to
x−9≥4
or
x − 9 ≤ −4.
The first one has solution x ≥ 13 and-second one x ≤ 5. The solution of the problem
is the union of the intervals (−∞, 5] [13, ∞).
52
CHAPTER 4. ADVANCED ALGEBRA4.5. MANIPULATIONS OF ABSOLUTE VALUES
Check: now we have two intervals so that the check will consist of two separate
cases. We can choose x = 0 for x ∈ (−∞, 5] and x = 20 for x ∈ [13, ∞).
|0 − 9| ≥ 4,
9 ≥ 4 TRUE!
|20 − 9| ≥ 4,
11 ≥ 4 TRUE!
Answer: the solution of |x − 9| ≥ 4 is (−∞, 5] [13, ∞).
3. Solve for x: |2x − 9| ≥ 5.
We have the same kind of inequality as above. We have two inequalities:
2x − 9 ≥ 5
or
2x − 9 ≤ −5.
We solve these and have from first
- one x ≥ 7 and from second x ≤ 2. The solution
is the union of the sets (−∞, 2] [7, ∞).
Check: we choose the numbers 0 and 8.
|2 × 0 − 9| ≥ 5,
9 ≥ 5 TRUE!
|2 × 8 − 9| ≥ 5,
7 ≥ 5 TRUE!
Be careful! We can choose the endpoint of the interval as well if we have equality
included. Note, in this check we get the equality (x = 2):
|2 × 2 − 9 ≥ 5,
5 ≥ 5 TRUE!
Answer: the solution of |2x − 9| ≥ 5 is the union (−∞, 2]
4. Solve for x: |7x + 4| ≤ 3.
-
[7, ∞).
The inequality is less or equal that means we need to solve
−3 ≤ 7x + 4 ≤ 3.
We can solve these inequalities simultaneously. We subtract 4 to all sides of the
inequality:
−3 − 4 ≤ 7x + 4 − 4 ≤ 3 − 4,
53
4.5. MANIPULATIONS OF ABSOLUTE VALUESCHAPTER 4. ADVANCED ALGEBRA
−7 ≤ 7x ≤ −1.
We divide now the inequality by 7.
−7 ÷ 7 ≤ 7x ÷ 7 ≤ −1 ÷ 7,
1
−1 ≤ x ≤ − .
7
The solution is the closed interval [−1, −1/7].
Check: the number zero is not in the interval so we cannot choose it. We choose the
number x = −1/2.
1
|7 × (− + 4| ≤ 3,
2
7
| − + 4| ≤ 3,
2
1
≤ 3 TRUE!
2
Answer: the solution of |7x + 4| ≤ 3 is the interval [−1, −1/7].
5. Solve for x: | − x + 4| > 3.
We have again absolute value greater than a number. The inequalities are
−x + 4 > 3
or
−x + 4 < −3.
We solve the first one. First, we multiply the inequality with (−1). This will change
the direction of the inequality.
−x + 4 > 3 ⇔ (−x + 4)(−1) > 3(−1),
x − 4 < −3.
We add now 4 to both sides of the inequality
x − 4 + 4 < −3 + 4,
x < 1.
We solve the same way the other inequality and find x >-7. Combining all the
results we write the answer as an union of intervals (−∞, 1] [7, ∞).
In the last inequality I recommend different techniques for transforming the negative sign
into positive. That is adding the variable term (+x) to both sides of the inequality. It will
send the variable term on the other side with positive sign. Then we do not have to
change the direction of the inequality.
Pay attention to the way of the inequality whatever way you use to solve it. Otherwise,
you may have strange answers that make no sense.
54
CHAPTER 4. ADVANCED ALGEBRA4.5. MANIPULATIONS OF ABSOLUTE VALUES
4.5.2 CBc2: Transform a < x < b into |cx + d| < e.
Describe an open interval in terms of absolute value using only integers.
Example: −1 < x < 4 is the same as |2x − 3| < 5.
Example explanation: The solution of the above inequality is an interval. To understand
the solution, we use the geometric representation of the interval on the number line. It is
a line segment. The line segment has lenght and middle point. In a calculus approach, the
representation of the solution ot the above inequality is a difference smaller than a half
of the lenght of the interval from the middle point of the interval. Therefore, we need to
find (1) the lenght of the interval and (2) the middle of the interval. Then, the inequality is
|x − a| < b, where a is the center of the interval and b is half of the lenght of the interval. If
the lenght and/or the coordinate of the center is/are not a whole number(s), we can find
LCD of all the denominators and multiply the inequality by it. This way, we will solve
the problem.
C1
-1
3/2
0
d=60
0
d=5
4
x
C2
43
13
x
73
The lenght of the interval is the difference of coordinates
of the right and the left endpoints: d=4-(-1)=4+1=5. This
is true regardless of the sign of the endpoints:
d=73-13=60.
To find the center of the interval, we add the
coordinates of the endpoints and divide by 2.
For C1: [4+(-1)]x1/2=3/2.
For C2: [13+73]x1/2=43.
The lenght of the interval is the difference between its right and left endpoints. On the
graph, there are two cases: positive endpoints; one positive and one negative endpoint.
However, in both cases, the lenght is the difference between the coordinates of the endpoints. Ovserve in case of negative endpoint that we actually add the distance from the
origin to the left endpoint. The case of two negative endpoint is the same as with two
positive endpoints.
We need the center of the interval as well. The center is equadistance point from both
ends of the interval. We find it by subtracting half of the lenght of the interval from right
endpoint, or adding the same amount to the left one.
In the solved example, the lenght of the interval is 5 = 4 − (−1), and half of it is 5/2. The
center of the interval is 3/2, which is 4 − 5/2 = −1 + 5/2 = 3/2. These are not whole
numvbers so we multiply the inequality by 2 to get the desired result: |2x − 3| < 5.
55
4.5. MANIPULATIONS OF ABSOLUTE VALUESCHAPTER 4. ADVANCED ALGEBRA
1. −5 < x < 12.
We find the lenght of the interval: 12 − (−5) = 17. Then, we find the center of the
interval [(12 + (−5)] ÷ 2 = 7/2. Then, the inequality is
|x − 7/2| < 17/2.
There are still fractions so we multiply the inequality by 2:
.
.
.
.
7
17
2 ..x − .. = |2x − 7| < 2 ×
= 17,
2
2
|2x − 7| < 17.
2. −9 < x < 12
The lenght is 12 − (−9) = 21. Half of the lenght is 21/2. The center of the interval is
[12 + (−9)]/2 = 3/2. The inequaility is:
.
.
.
.
.x − 3 . < 21 .
.
2.
2
We multiply the inequality by 2 andfind the answer:
|2x − 3| < 21.
3. −9 < x < 13.
The interval lenght is 13 − (−9) = 22 and half of it is 11. The center of the interval is
[13 + (−9)]/2 = 2. The inequality is
|x − 2| < 11.
4. 9 < x < 13.
The lenght of the interval is 13 − 9 = 4 and half of it is 2. The center of the interval
is (13 + 9)/2 = 11. The inequality is:
|x − 11| < 2.
5. 9 < x < 23.
We find the lenght of the interval 23 − 9 = 14 and half of it is 7. The center of the
interval is (23 + 9)/2 = 16. The inequality is
|x − 17| < 7.
56
CHAPTER 4. ADVANCED ALGEBRA
4.6
4.6. QUADRATIC INEQUALITIES
Quadratic Inequalities
4.6.1 CBe1: Solve simple quadratic inequalities (e.g., ax2 + bx + c > 0).
Solve quadratic inequalities.
Example: If x2 + 4x + 3 > 0 then x < −3 or x > −1.
1. Solve for x: (3x + 9)2 > 25.
The inequality is the same as (3x + 9)2 − 25 > 0. We consider the function f (x) =
(3x + 9)2 − 25 and find where it crosses the x axis. We include the graph of it as well.
4/3
14/3
The function (3x-9) -25 is
positive here
(3x − 9)2 − 25 = 0,
(3x − 9 − 5)(3x − 9 + 5) = 0,
(3x − 14)(3x − 4) = 0.
We found equivalent form of the function f (x), in which we compute the value of
it by multiplying numbers. The points, where the function obtains value of zero are
57
4.6. QUADRATIC INEQUALITIES
CHAPTER 4. ADVANCED ALGEBRA
x = 14/3 and x = 4/3. If we choose any number smaller than 4/3 then we have the
function value as a product of two negative numbers and it is positive. If we choose
a number bigger than 14/3, the function value can be computed by multiplying two
positive numbers and is therefore positive.
The solution set of (3x − 9)2 > 25 is then
the union of the intervals (−∞, 4/3) (14/3, ∞). Observe between the points where
f (x) = 0 we have negative function values so these points are not solutions of the
interval because we multiply one positive and one negative numbers.
Answer: the solution set is (−∞, 4/3)
2. Solve for x: x2 + 5x + 4 < 0.
-
(14/3, ∞).
We look at the function f (x) = x2 + 5x + 4 = (x + 4)(x + 1). We see that in last
expression of it that it is a product of (x+4) and (x+1). We know that the product of
two numbers is negative (i.e. less than zero) if one number is positive and the other
number is negative. This may happen only the number say, a is bigger than (−4)
and smaller than (−1). This place is exactly between the solutions of the equation
x2 + 5x + 4 = 0.
solution is
(-4,-1)
Answer: the solution set is the interval (-4,-1).
3. Solve for x: (2x + 7)(3x − 4) ≤ 0.
The solutions of (2x + 7)(3x − 4) = 0 are x1 = −7/2 and x2 = 4/3. We need again one
positive factor and one negative factor, so that the solution set is between the two
solutions — the interval [-7/2, 4/3]. The interval is closed because we may have
equality.
Answer: the solution set is [-7/2, 4/3].
4. Solve for x: x2 + 2x + 3 > 0.
58
CHAPTER 4. ADVANCED ALGEBRA
4.6. QUADRATIC INEQUALITIES
We complete the square for the left side of the inequality:
x2 + 2x + 3 = x2 + 2x + 1 + 2 = (x2 + 2x + 1) + 2 = (x + 1)2 + 2.
The left side of the inequality is a sum of two quantities: (x + 1)2 and 2. The first
part (x + 1)2 is nonnegative and the second one is always positive. Therefore, the
answer is all real numbers because the sum of two quantities, one of which is always
positive and the other nonnegative is always positive. The graph below confirms the
result.
There are no intersections
with x axis
Answer: the solution set is all real numbers.
5. Solve for x: x2 + 5x + 1 ≤ 25.
We subtract from both sides of the inequality
x2 + 5x − 24 ≤ 0.
We factor the above expression and have f (x) = x2 + 5x − 24 = (x + 8)(x − 3).
Since we require the expression less or equal to zero, the solution set is between the
solutions of the equation f (x) = 0. They are x1 = −8 and x2 = 3.
Answer: the solution set is the interval [-8, 3].
The method described above can work for any number of factors. Suppose a < b < c
and we look at the polynomial P (x) = (x − a)(x − b)(x − c). We want to find out where
P (x) > 0. We divide the number line into intervals using numbers a, b, c. The polynomial
P (x) is zero when x = a, b, c and it is not zero for any other value of x. The chat below
describes the signs of the polynomial factors and the polynomial itself when x is in any
of the intervals (−∞, a), (a, b), (b, c) and (c, ∞).
59
4.6. QUADRATIC INEQUALITIES
Sign
x−a
x−b
x−c
P (x)
x ∈ (−∞, a)
Minus
Minus
Minus
Minus
x ∈ (a, b)
Plus
Minus
Minus
Plus
x ∈ (b, c)
Plus
Plus
Minus
Minus
+
x ∈ (c, +∞)
Plus
Plus
Plus
Plus
c
b
a
-
CHAPTER 4. ADVANCED ALGEBRA
The sign of product
(x - a )(x - b )(x - c )
changes alternatively
x
+
The above method can be used for any number of different factors. If a factor is raised
to an even power, it does not change the sign at all. It only adds an extra zero to the
polynomial. If a factor is raised to an odd power, it is equivalent to a single factor (the
rest of the factors are always nonnegative quantity) so this is the most general form for
this algorithm.
60
CHAPTER 4. ADVANCED ALGEBRA
4.6. QUADRATIC INEQUALITIES
4.6.2 CBe2: Solve inequalities that lead to simple quadratic (e.g., (ax +
b)/(cx + d) < ex + f .
Solve quadratic inequalities that arise from rational inequalities:
Example: Solve for x: (x2 + 2x + 12)/(x2 + 3x + 6) < 2.
Answer: x > 0 or x < −4.
1. Solve for x:
x2 − 2x + 6
< 2.
x2 − 3x + 3
We subtact 2 from both sides of the inequality
x2 − 2x + 6
− 2 < 0.
x2 − 3x + 3
We simplify it by subtracting 2
LCD = x2 − 3x + 3,
x2 − 2x + 6 − 2(x2 − 3x + 3)
< 0.
x2 − 3x + 3
We open the parentheses and simplify:
x2 − 2x + 6 − 2x2 + 6x − 6
< 0,
x2 − 3x + 3
−x2 + 4x
< 0.
x2 − 3x + 3
Now, we factor the numerator and the denominator:
−x(x − 4)
< 0.
x2 − 3x + 3
The denominator can be expressed as
32 32
3
x2 − 3x + 3 = x2 − 2 × x + 2 − 2 + 3
2
2
2
%2
$
3
3
+ > 0.
= x−
2
4
We see that the sign of the denominator never changes because it is sum of 2 terms,
one nonnegative and the other strictly positive.The numerator is a quadratic function, which is factored. So, the
- solution of the inequality is the set where [−x(x − 4)]
is negative. That is (−∞, 0) (4, ∞).
Answer: the solution is the union of the intervals (−∞, 0)
61
-
(4, ∞).
4.6. QUADRATIC INEQUALITIES
2. Solve for x:
CHAPTER 4. ADVANCED ALGEBRA
x2 − 2x + 6
< 2.
x2 − 4x + 3
As before, we simplify the inequality:
x2 − 2x + 6
− 2 < 0,
x2 − 4x + 3
x2 − 2x + 6 − 2(x2 − 4x + 3)
< 0,
x2 − 4x + 3
−x2 − 6x
< 0.
x2 − 4x + 3
We factor the numerator and the denominator:
−x(x + 6)
< 0.
(x − 1)(x − 3)
We need to make another observation. The sign of the product of two numbers is the
same as the sign of their quotient, therefore we can use the algorithm described at
the end of the previous secton (CBe1). We check the sign when x = −7. It is minus.
Applying the principle, we writhe the signs in all the intervals (see the graph). We
see that the function value is negative when x is less than (−6), between 0 and 1,
and when is bigger than 3.
-6
0
3
1
Answer: the solution is the union of the intervals (−∞, −6)
3. Solve for x:
We simplify the same way:
21x2 − 10x + 12
< 10.
x2 − x + 6
x
-
21x2 − 10x + 12
− 10 < 0,
x2 − x + 6
21x2 − 10x + 12 − 10(x2 − x + 6)
< 0,
x2 − x + 6
11x2 − 48
< 0.
x2 − x + 6
62
(0, 1)
-
(3, ∞).
CHAPTER 4. ADVANCED ALGEBRA
4.6. QUADRATIC INEQUALITIES
We complete the square for the denominator as in #2:
12 12
1
x2 − x + 6 = x2 − 2 × x + 2 − 2 + 6
2
2
2
%2
$
1
12
23
1
23
2
= (x − 2 × x + 2 ) +
= x−
+ .
2
2
4
2
4
We make the same conclussion as above: the denominator never changes its sign,
so the sign of the inequality is the same as the sign of the numerator. We factor it
and observe:
#
#
11(x − 48/11)(x + 48/11)
< 0.
x2 − x + 6
#
#
The solution is between the numbers − 48/11 and 48/11.
#
#
Answer: the solution set is the interval (− 48/11, 48/11).
4. Solve for x:
2x2 − 4x + 4
< 2/3.
x2 − 6x + 9
We simplify the same way
LCD = 3(x2 − 6x + 9),
2x2 − 4x + 4 2
− < 0,
x2 − 6x + 9
3
3(2x2 − 4x + 4) − 2(x2 − 6x + 9)
< 0,
3(x2 − 6x + 9)
4x2 − 6
< 0.
3(x2 − 6x + 9)
We factor the numerator and the denominator:
#
#
4(x − 3/2)(x + 3/2)
< 0.
3(x − 3)2
We see that the denominator never changes its sign but is zero at x =
#3. The solution
#
is where the numerator is negative, that is between the numbers − 3/2 and 3/2.
) #
# *
Answer: the solution is the interval − 3/2, 3/2 .
5. Solve for x:
−3x2 − 2x + 12
> 0.
x2 − 3x + 6
63
4.6. QUADRATIC INEQUALITIES
CHAPTER 4. ADVANCED ALGEBRA
We need to factor the numerator and the denominator. Completing the square for
the denominator we see that it is always positive:
2
x − 3x + 6 =
$
3
x−
2
%2
+
15
.
4
The numerator has two real irrational roots. We write:
√
√
−3[x + 1/3(1 − 37)][x + 1/3(1 + 37)]
> 0.
x2 − 3x + 6
We need to have bigger than zero so we check where it is positive, that is between
the roots.
Answer: the solution is the interval (1/3(−1 −
64
√
√
37), 1/3(−1 + 37)).
CHAPTER 4. ADVANCED ALGEBRA
4.7
4.7. NONLINEAR SYSTEMS
Nonlinear Systems of Equations
4.7.1 CBk1: Solve nonlinear systems of equations .
Solve systems of linear and non-linear inequalities.
Example: √
Solve for √
(x, y): {x2√
+ y 2 = 4, x2√+ 4y 2 = 13}.
Answer: {(1, 3), (1, − 3), (−1, 3), (−1, − 3)}.
Each system must be solved individually. We need to express one variable in terms of the
other. Sometimes it is easy, sometimes it is more difficult.
1. Solve for (x, y): {x2 − y 2 = 4, x2 + 4y = 13}.
We observe that the second equation is linear in y but quadratic in x and the first
equation does not contain linear term of x. So, we express x2 from the second equation and substitute it in the first equation:
x2 + 4y = 13,
x2 = 13 − 4y.
x2 − y 2 = (13 − 4y) − y 2 = 4,
13 − 4y − y 2 = 4.
We have simple quadratic equation and we solve it by using the quadratic formula.
The equation has two real irrational solutions:
√
y1 = −2 − 13
and
y2 = −2 +
√
13.
Now, we return to the equation expressing x. The substitution with y1 will give us
two solutions and substitution with y2 will show other two.
x2 = 13 − 4y1 ,
√
x2 = 13 − 4(−2 − 13),
/
√
x1,1 = + 21 + 4 13,
/
√
x1,2 = − 21 + 4 13.
#
#
√
√
√
√
The two solutions are ( 21 + 4 13, −2 − 13) and (− 21 + 4 13, −2 − 13).
We compute the other solutions:
x2 = 13 − 4y2 ,
65
4.7. NONLINEAR SYSTEMS
CHAPTER 4. ADVANCED ALGEBRA
√
x2 = 13 − 4(−2 + 13),
/
√
x2,1 = + 21 − 4 13,
/
√
x2,2 = − 21 − 4 13.
#
#
√
√
√
The
other
two
solutions
for
the
system
are
(
21
−
4
13,
−2+
13)
and
(−
21 − 4 13, −2+
√
13).
#
√
√
We check our solutions. For x = + 21 + 4 13 and y = −2 − 13:
/
√
√
2
x + 4y = (+ 21 + 4 13)2 + 4(−2 − 13)
√
√
= 21 + 4 13 − 8 − 4 13 = 13 TRUE!
2
2
/
√
√
= (+ 21 + 4 13)2 − (−2 − 13)2
x −y
√
√
= 21 + 4 13 − (4 + 4 13 + 13) = 4 TRUE!
All other solutions are checked the same way:
The
#system√solutions are:
√
(+#21 + 4 13, −2 − 13);
√
√
(−#21 + 4 13, −2 − 13);
√
√
(+#21 − 4 13, −2 + 13);
√
√
(− 21 − 4 13, −2 + 13).
2. Solve for (x, y): {xy = 4, x + 4y = 5}
One way to attack this equation is to express x from the first equation and substitute
it in the second one:
xy = 4,
4
x= .
y
Note, that neither x nor y could be zero because their product is 4.
x + 4y = 5,
4
+ 4y = 5.
y
We clear fractions by multiplying the equation with y and have
4 + 4y 2 = 5y,
66
CHAPTER 4. ADVANCED ALGEBRA
4.7. NONLINEAR SYSTEMS
4y 2 − 5y + 4 = 0.
We solve this quadratic equation for y and find that it has two imaginary solutions
√
5 ± i 39
y1,2 =
8
We compute the
for x:
√ corresponding solutions √
for y1 = (5 + i√39)/8 we have x1 = (5 − i√39)/2;
for y2 = (5 − i 39)/8 we have x2 = (5 + i 39)/2.
We need to check the solutions. We need to show xy = 4 and x + 4y = 5. For (x1 , y1 ):
√
√
5 + i 39
5 − i 39
+4×
x1 + 4y1 =
2
8
√
√
5 − i 39 + 5 + i 39
10
=
=
= 5;
2
2
√
√
5 − i 39 5 + i 39
x1 y1 =
×
2
8
√
25 − i2 ( 39)2
25 + 39
=
=
= 4.
16
16
The same way for (x2 , y2 ):
√
√
5 + i 39
5 − i 39
x2 + 4y2 =
+4×
2
8
√
√
10
5 + i 39 + 5 − i 39
=
= 5.
=
2
2
The product x2 y2 is the same as x1 y1 because we simply interchange the numerators
of the two fractions so their product is 4.
The system
solutions
√
√ are:
([5 − i√39]/2, [5 + i√39]/8);
([5 + i 39]/2, [5 − i 39]/8).
3. Solve for (x, y): {x2 + y 2 = 25, xy = 5}
We do the same as before: express x in one equation and substitute it in the other
one (note, x nor y can be zero):
xy = 5,
5
x= .
y
67
4.7. NONLINEAR SYSTEMS
CHAPTER 4. ADVANCED ALGEBRA
x2 + y 2 = 25,
$ %2
5
+ y 2 = 25,
y
25
+ y 2 = 25.
y2
We clear fractions by multiplying with y 2 . The equation becomes
25 + y 4 = 25y 2 .
If we make the substitution z = y 2 we will have regular quadratic equation we can
solve
z 2 − 25z + 25 = 0.
This equation has two positive real solutions:
25 +
z1 =
25 −
z2 =
√
525
2
√
525
2
,
.
Since z = y 2 we find the values of y:
0
25 +
0
25 +
y1 = +
y2 = −
and
0
y4 = −
525
2
√
525
2
0
y3 = +
√
25 −
25 −
√
,
,
525
2
√
2
525
.
Each value/
of y will give a value of x.
/
√
√
For y1 = + (25 + 525)/2 we find x1 = + (25 − 525)/2;
/
/
√
√
for y2 = − (25 + 525)/2 we find x2 = − (25 − 525)/2;
/
/
√
√
for y3 = + (25 − 525)/2 we find x3 = + (25 + 525)/2;
/
/
√
√
for y4 = − (25 − 525)/2 we find x4 = − (25 + 525)/2.
68
CHAPTER 4. ADVANCED ALGEBRA
4.7. NONLINEAR SYSTEMS
We will show the check of only one of the solutions found above. All others can be
done in a similar manner.
%2 $ /
%2
$ /
√
√
2
2
x1 + y1 = + (25 − 525)/2 + + (25 + 525)/2
25 −
√
√
525
50
=
+
=
= 25,
2
2
2
% $ /
%
$ /
√
√
x1 y1 = + (25 − 525)/2 × + (25 + 525)/2
=
0
252 −
525
1√
4
25 +
22
525
=
&
625 − 525 √
= 25 = 5.
4
The
/system√solutions are:
/
√
(+ (25 − 525)/2, + (25 + 525)/2);
/
/
√
√
(− (25 − 525)/2, − (25 + 525)/2);
/
/
√
√
(+ (25 + 525)/2, + (25 − 525)/2);
/
/
√
√
(− (25 + 525)/2, − (25 − 525)/2).
4. Solve for (x, y): {(x − 1)2 + (y − 2)2 = 4, x − 1 = 4(y − 2)}
We could solve this system directly, but it will give difficult expressions for x and y.
We could make the substitution u = x − 1 and v = y − 2. The system becomes
u2 + v 2 = 4,
u = 4v.
We express u from second equation and find v.
u2 + v 2 = (4v)2 + v 2 = 4,
√
4
2 17
v1 = +
=+
,
17
17
&
√
2 17
4
=−
.
v2 = −
17
17
&
We find corresponding u1 and u2 .
√
8 17
u1 = 4v1 =
,
17
√
8 17
.
u2 = 4v2 = −
17
69
4.7. NONLINEAR SYSTEMS
CHAPTER 4. ADVANCED ALGEBRA
Do not forget that we need x and y but we found u and v! Now, we return to the
substitution and compute x and y
x−1=u⇔x=u+1
and
y = v + 2.
√
We√will have the solutions x1 = u1 + 1 = [(8 17)/17] +√1 and y1 = v1 + 2 =
[(2 17/17] +√
2; the other solution is x2 = u2 + 1 = −[(8 17)/17] + 1 and y2 =
v2 + 2 = −[(2 17/17] + 2
Note the check:
42 3 √
42
√
17
17
+8
+ 1 − 1 + +2
+2−2
17
17
3
(x1 − 1)2 + (y1 − 2)2 =
64 × 17 4 × 17
64 + 4
+
=
= 4,
172
172
17
√
√
17
17
x1 − 1 = +8
7 + 1 − 1 = +8
;
1
17
3 √
4
√
17
17
4(y1 − 2) = 4 +2
+ 2 − 2 = +8
.
17
17
=
Therefore, x1 − 1 = 4(y1 − 2).
The system
solutions are:
√
√
(+[(8√17)/17] + 1, +[(2 √17/17] + 2) and
(−[(8 17)/17] + 1, −[(2 17/17] + 2).
5. Solve for (x, y): {x2 + 4y 2 = 4, y = x2 }
We observe that x2 is expressed in terms of y in the second equation. We substitute
it in the first one and solve the equation for y:
y + 4y 2 = 4,
4y 2 + y − 4 = 0.
This equation has two real roots
√
−1 + 65
y1 =
,
8
√
−1 − 65
.
y2 =
8
70
CHAPTER 4. ADVANCED ALGEBRA
4.7. NONLINEAR SYSTEMS
Since x2 = y, the solution y1 will give us two real solutions for the variable x, the
other two solutions are imaginary.
0
√
−1 + 65
x1 = +
,
8
0
√
−1 + 65
,
x2 = −
8
√
−1 − 65
+1 + 65
x3 = +
= +i
,
8
8
0
0
√
√
−1 − 65
+1 + 65
x4 = −
= −i
.
8
8
0
√
0
We check for the solutions (x1 , y1 ) and (x2 , y2 ). The first equation:
0
x2 + 4y 2 = 
√
2
−1 + 65 
+4×
8
3
√ 42
−1 + 65
8
√
√
−1 + 65
1 − 2 65 + 65
=
+4×
8
64
√
√
√
√
−1 + 65 + 33 − 65
−1 + 65 66 − 2 65
+
=
= 4.
=
8
16
8
It
obvious that the square of x is equal
to y for the solution. System solutions are:
$ is/
%
√
√
+ (−1 + 65)/2, [(−1 + 65)/2] ;
$ /
%
√
√
− (−1 + 65)/2, [(−1 + 65)/2] ;
$ /
%
√
√
+i (+1 + 65)/2, − [(+1 + 65)/2] ;
%
$ /
√
√
−i (+1 + 65)/2, − [(+1 + 65)/2] .
71
4.8. BINOMIAL THEOREM
4.8
CHAPTER 4. ADVANCED ALGEBRA
Binomial Theorem and Factoring
4.8.1 CBg1: Determine coefficients of particular terms in the expansion
of (a + b)n .
Find particular terms in a binomial expansion.
Example: The term in (a + b)9 involving a2 b7 is 36a2 b7 .
We need here to recall the formulas involving (a + b)n :
n
(a + b) =
n $ %
9
n
ak bn−k =
k
$ %
$ %
$ %
$ %
n n 0
n n−1 1
n k n−k
n 0 n
a b +
a b + ... +
a b
+ ... +
ab .
0
1
k
n
k=0
Here,
1n2
k
stands for the number of ways of choosing k items out of set of n items:
$ %
n(n − 1) . . . (n − k + 1)
n
n!
=
=
k
k!
k!(n − k)!
and k! means the product of the first k natural numbers
k! = 1 × 2 × 3 . . . × k.
We define 0! = 1.
1. The term in (a + b)9 involving a7 b2 .
We apply the general formula with
n = 9 and
k = 2.
The binomial coefficient of the term is
$ %
9!
9
9 × 8 × 7!
=
=
2
2!7!
1 × 2 × 7!
We remove factor equal to one (7!/7!):
=
72
= 36.
2
The term is 36a7 b2 .
Answer: 36a7 b2 .
72
CHAPTER 4. ADVANCED ALGEBRA
4.8. BINOMIAL THEOREM
2. The term in (a − b)7 involving a5 b2 .
We can always write
(a − b)n = [a + (−b)]n .
We have now
n = 7 and
k = 5.
The binomial coefficient of the term is
$ %
7
7!
7×6
=
=
= 21.
5
5!2!
1×2
The term is
$ %
7 5
a (−b)2 = 21a5 b2 .
5
The minus sign disappears because there are even number of negative terms, i.e.
(−b)2 = b2 .
Answer: 21a5 b2 .
3. The term in (a + 2b)6 involving a3 b3 .
n = 6 and
k = 3.
The binomial coefficient is
$ %
6!
6
=
= 20.
3
3!3!
The second term of the binomial sum now is 2b, so we look at powers of 2b:
(2b)3 = 8b3 .
$ %
6 3
a (2b)3 = 20a3 8b3 = 160a3 b3 .
3
Answer: 160a3 b3 .
4. The term in (3a + b)10 involving a3 b7 .
n = 10 and
k = 3.
The term:
$ %
10!
10
=
= 120.
3
3!7!
$ %
10
(3a)3 b7 = 120 × 27a3 b6 = 3240a3 b7 .
3
Answer: 3240a3 b7 .
73
4.8. BINOMIAL THEOREM
CHAPTER 4. ADVANCED ALGEBRA
5. The term in (2a + 3b)5 involving a2 b3 .
n = 5 and
k = 2.
$ %
5!
5
=
= 10.
3
2!3!
The term is
$ %
5
(2a)2 (3b)3 = 1080a2 b3 .
3
Answer: 1080a2 b3 .
74
CHAPTER 4. ADVANCED ALGEBRA
4.8. BINOMIAL THEOREM
4.8.2 CBg2: Factor difference of two cubes.
Example: x3 − 8y 3 = (x − 2y)(x2 + 2xy + 4y 2 ).
Recall the formula for sum of cubes:
a3 + b3 = (a + b)(a2 − ab + b2 ).
Observe if we replace b with (−b), we get a formula for difference of cubes:
a3 − b3 = a3 + (−b)3
= [a + (−b)][a2 − a(−b) + (−b)2 ] = (a − b)(a2 + ab + b2 ).
1. Factor as a product of linear and quadratic factors: x3 + 8y 3 .
We can write 8y 3 = (2y)3 and then the above sum is
x3 + (2y)3
= (x + 2y)(x2 − 2xy + 4y 2 ).
We check our work by multiplication:
(x + 2y)(x2 − 2xy + 4y 2 )
= x3 − 2x2 y + 4xy + 2x2 y − 4xy 2 + 8y 3 = x3 + y 3 .
Note, in this check when we multiplied the binomial by the trinomial, the intermediate result contained six terms, and the final result only two. The number of
terms before simplifying is always the product of the number of terms in the first
parentheses and the number of terms in the second pair. In this example, we have
2 × 3 = 6 terms!
2. Factor as a product of linear and quadratic factors: 125x3 − 27y 3
Now, 125x3 = (5x)3 and 27y 3 = (3y)3 . We write
125x3 − 27y 3 =
(5x)3 − (3y)3
= (5x − 3y) [(5x)2 + 5x3y + (3y)2 ]
= (5x − 3y)(25x2 + 15xy + 9y 2 ).
We check:
(5x − 3y)(25x2 + 15xy + 9y 2 )
= 125x3 + 75x2 y + 45xy 2 − 75x2 y − 45xy 2 − 27y 3
=
25x3 − 27y 3 .
75
4.8. BINOMIAL THEOREM
CHAPTER 4. ADVANCED ALGEBRA
3. Factor as a product of linear and quadratic factors: t3 x3 + y 3 .
The terms are t3 x3 = (tx)3 and the other cube is simply y 3 .
t3 x3 + y 3 =
(tx)3 + y 3
= (tx + y) [(tx)2 − txy + y 2 ]
= (tx + y)(t2 x2 − txy + y 2 ).
Check:
(tx + y)(t2 x2 − txy + y 2 )
= t3 x3 − t2 x2 y + txy 2 + t2 x2 y − txy 2 + y 3
=
t3 x3 + y 3 .
4. Factor as a product of linear and quadratic factors: (x3 /8) + y 3 .
We see that x3 /8 = (x/2)3 .
x3 /8 + y 3 =
(x/2)3 + y 3
= (x/2 + y) [(x/2)2 − (x/2)y + y 2 ]
=
(x/2 + y)(x2 /4 − xy/2 + y 2 ).
We check:
(x/2 + y)(x2 /4 − xy/2 + y 2 )
= x3 /8 − x2 y/4 + xy 2 /2 + x2 y/4 − xy 2 /2 + y 3
=
x3 /8 + y 3 .
5. Factor as a product of linear and quadratic factors: 64x3 − 216y 3 .
We have now 64x3 = (4x)3 and 216y 3 = 6y 3 .
64x3 − 216y 3 =
(4x)3 − (6y)3
= (4x − 6y) [(4x)2 + (4x)(6y) + (6y)2 ]
=
(4x − 6y)(16x2 + 24xy + 36y 2 ).
Check:
(4x − 6y)(16x2 + 24xy + 36y 2 )
= 64x3 + 96x2 y + 144xy 2 − 96x2 y − 144xy 2 − 216y 3
=
64x3 − 216y 3 .
76
CHAPTER 4. ADVANCED ALGEBRA
4.8. BINOMIAL THEOREM
4.8.3 CBg3: Factor an ± bn when n > 3.
Factor each of the following using differences of squares and cubes.
1. a6 − b6
= (a3 )2 − (b3 )2
= (a3 − b3 )(a3 + b3 )
= (a − b)(a2 + ab + b2 )(a + b)(a2 − ab + b2 ).
2. a8 − b8
= (a4 )2 − (b4 )2
!
"
= (a4 − b4 )(a4 + b4 ) = (a2 )2 − (b2 )2 (a4 + b4 )
= (a2 − b2 )(a2 + b2 )(a4 + b4 )
= (a − b)(a + b)(a2 + b2 )(a4 + b4 ).
3. a6 + b6
= (a2 )3 + (b2 )3
"
!
= (a2 + b2 ) (a2 )2 − a2 b2 + (b2 )2
= (a2 + b2 )(a4 − a2 b2 + b4 ).
4. a6 − 64b6
1 22 !
"2
= a3 − (2b)3
1
21
2
= a3 − (2b)3 a3 + (2b)3
= (a − 2b)(a2 + 2ab + 4b2 )(a + 2b)(a2 − 2ab + 4b4 ).
5. a9 − b9
= (a3 )3 − (b3 )3
!
"
= (a3 − b3 ) (a3 )2 + a3 b3 + (b3 )2
= (a − b)(a2 + ab + b2 )(a6 + a3 b3 + b6 ).
77
4.9. SUMMATION NOTATION
4.9
CHAPTER 4. ADVANCED ALGEBRA
Summation notation
4.9.1 CBj1: Recognize and find sums of arithmetic and geometric series.
Sum the indicated series.
Example: 1 + 2 + 3 + · · · + 100 = 5050.
1. 2 + 4 + 6 + · · · + 200 =
We observe that the difference betweeen the second and the first term of the sequence is 2, the difference between the third and the second term is the same and so
on. We see, therefore, an arithmetic sequence. The first term is a1 = 2, last term (we
call it n-th term) is an = 200. The common difference, denoted with d, is 2 (recall
that d = a2 − a1 = 4 − 2 = 2).
We know that a1 = 2, an = 200 and d = 2. In the arithmetic sequence, the second
therm is the first term plus the difference d, i.e. a2 = a1 + d. The third therm
is the second therm plus the differnce d, or the first therm plus two differences:
a3 = a2 + d = (a1 + d) + d. Generalizing this, the formula for the n term is
an = a1 + (n − 1)d.
Recall that the sum of the first n terms of the arithmetic sequence is
S n = a 1 + a2 + a 3 + · · · + an =
n
× (a1 + an ).
2
To see this, we observe the sum of the first and the last term is a1 + an = a1 + (a1 +
(n − 1)d) = 2a1 + (n − 1)d. The sum of the second and the term before the last one
is a2 + an−1 = a1 + d + a1 + (n − 2)d = 2a1 + (n − 1)d. We can continue the same
way comparing the sums of the terms equally distant from the beginning and the
end and see they are equal to (2a1 + (n − 1)d)! At each step, we add two terms, so
the number of times we add the sum is n/2. The formula above follows.
We can evaluate this expression using the data given above and find the number of
terms in the sequence, i.e. n:
200 = 2 + (n − 1)2,
n = 100.
Now we use the formula for the sum and compute:
S100 =
100
× (2 + 200) = 10100.
2
The sum of this series is 10100.
78
CHAPTER 4. ADVANCED ALGEBRA
4.9. SUMMATION NOTATION
2. 1 + 2 + 4 + 8 + · · · + 256 =
One way of computing the sum (easy method) is to denote the sum by x. We then
multiply the sum by 2 and subtract it:
1 + 2 + 4 + 8 + · · · + 256 = x,
2(1 + 2 + 4 + 8 + · · · + 256) = 2x,
x = 2x − x = 512 + 256 + · · · + 2 − 1 − 2 − 4 − · · · − 256
We rearrange the terms of the two series so we see that the second term in the first
sequence, 256 is equal to the last term form the other sequence and so the other parts
of the sum. At the end, only the term 1 is not paired with term of the multiplied
sequence. Therefore, the final result is the difference between the last term, 512 and
the first term of the other sequence, 1:
= 512 + 256 − 256 + 128 − 128 + · · · + 2 − 2 − 1 = 512 − 1 = 511.
The other way: We investigate the relation between the terms in the sequence. We
see that the ratio of the consecutive terms is always the same
2/1 = 4/2 = · · · = 2
so the sequence is a geometric sequence. Recall the formulas for computing the nth
term
an = a1 × rn−1
where a1 is the first term of the sequence, n is the number of the term and r is the
ratio.
The sum of the first n terms is
Sn =
a1 (1 − rn )
.
1−r
We have ratio 2 and know that a1 is 1. We need to find n. We use the formula stated
above:
an = a1 × rn−1 .
an = 256;
a1 = 1;
r = 2.
256 = 1 × 2n−1 ,
256 = 28 = 2n−1 ,
79
4.9. SUMMATION NOTATION
CHAPTER 4. ADVANCED ALGEBRA
We use the principle of the exponential equality, that is, if the basis of the exponents
are the same and not equal to one, and bx = by then x = y. This means
8 = n − 1,
n = 9.
We use the sum formula now:
Sn =
a1 (1 − rn )
1(1 − 29 )
=
= 511.
1−r
1−2
The sum of the first nine members of this geometric series is 511.
3. 1 + (1/2) + (1/4) + · · · + (1/2)100 =
We can employ the same idea to the above sum. After denoting the sum by x, we
multiply it by 1/2:
1 + (1/2) + (1/4) + · · · + (1/2)100 = x,
1
(1 + (1/2) + (1/4) + · · · + (1/2)100 )
2
x
1 1
= + + · · · + (1/2)101 = .
2 4
2
We subtract above two sums:
x−
x
1 1
= 1 + (1/2) + (1/4) + · · · + (1/2)100 − ( + + · · · + (1/2)101 )
2
2 4
= 1 + 1/2 − 1/2 + 1/4 − 1/4 + · · · + (1/2)100 − (1/2)100 − (1/2)101 ,
x
= 1 − (1/2)101 ,
2
x = 2 − (1/2)100 .
The method with the number of terms. This is again geometric series with
a1 = 1;
r = a2 /a1 = 12 /1 = 12 ;
an = (1/2)100 .
Using the formula for finding n (as in #2) we find n = 101.
We compute the sum now:
S101
1
=1+ +
2
$ %2
$ %100
1
1
+ ··· +
=
2
2
80
CHAPTER 4. ADVANCED ALGEBRA
4.9. SUMMATION NOTATION
:
$ %101 ;
$ %100
1
1
1[1 − (1/2)101 ]
=2 1−
.
=2−
=
1 − 1/2
2
2
The sum of the first 101 terms is 2 − (1/2)100 .
4. 1 + 4 + 16 + · · · + 425 =
We denote the sum by x and multiply by 4.
1 + 4 + 16 + · · · + 425 = x
4(1 + 4 + 16 + · · · + 425 )
4 + 16 + · · · + 426 = 4x
Subtracting the equations, we obtain
4x − x = 4 + 16 + · · · + 426 − (1 + 4 + 16 + · · · + 425 ),
3x = 426 − 1,
426 − 1
.
3
The other way of solving the problem. We observe:
a1 = 1;
x=
r = a2 /a1 = 4/1 = 4;
The number of the terms is n = 26.
The sum is
S26 = 1 + 4 + 42 + · · · + 425 =
1(1 − 426 )
426 − 1
=
.
1−4
3
The sum of the first 26 terms is (426 − 1)/3.
5. 1 + 3 + 5 + · · · + 2001 =
This is an arithmetic sequence. In it, a1 = 1, an = 2001 and the common difference is
d = 3−1 = 2. To find the sum, we compute the number of the terms in the sequence:
an = a1 + (n − 1)d,
2001 = 1 + (n − 1)2,
2000 = 2(n − 1),
81
4.9. SUMMATION NOTATION
CHAPTER 4. ADVANCED ALGEBRA
1000 = n − 1 ⇔ n = 1001.
We use the sum formula with a1 = 1, a1001 = 2001, n = 1001:
S1001 =
1001
(2001 + 1) = 1, 002, 001
2
Note: Observe that when find the sum of the geometric series, often your intermediate
results could be negative, but your answer is still positive.
82
CHAPTER 4. ADVANCED ALGEBRA
4.9. SUMMATION NOTATION
4.9.2 CBj2: Understand and use summation notation.
Example:
4
9
k 2 = 30.
k=1
Solution of the example:
We could use summation notation when we know the general term of a sequence. In this
example, the general term is k 2 . So, we read the expression
4
9
k2
k=1
as follows: the sum as k goes from 1 to 4 of k 2 . The number k is called index of the summation. The meaning, in mathematical notation, is
4
9
k 2 = 12 + 22 + 32 + 42 .
k=1
We can create the same way the sum of the first n squares (sum as k goes from one to n):
n
9
k=1
1.
5
9
k 2 = 12 + 22 + 32 + · · · + n2 .
k2.
k=1
We have now the
sum as k goes from 1 to 5 of k 2 .
We evaluate
5
9
k 2 = 12 + 22 + 32 + 42 + 52 = 55.
k=1
2.
7
9
(2k − 3).
k=1
We read the sum as k goes from 1 to 7 of (2k − 3). We look at the numbers (2 × 1 − 3),
(2 × 2 − 3), . . . , (2 × 7 − 3). These are the numbers (−1), 1, 3,. . . ,11. We observe an
arithmetic sequence with first member (−1) and last member 11. There are two ways
to evaluate the sum — direct summing and using the formula for first n members
of arithmetic sequence. We use the second one. First, we find how many members
there are (recall previous section formulas):
an = 11;
83
4.9. SUMMATION NOTATION
CHAPTER 4. ADVANCED ALGEBRA
a1 = −1;
d = 1 − (−1) = 2.
an = a1 + (n − 1)d.
We solve for n and find n = 7. We note that this is the same number as the end index
of summation notation. Therefore, the sum is
7
9
k=1
(2k − 3) = (a1 + an ) ×
n
2
(−1 + 11)
× 7 = 35.
2
The second method is simply adding the numbers for which the sumation notation
is a shortcut.
=
7
9
(2k−3) = (2×1−3)+(2×2−3)+(2×3−3)+(2×4−3)+(2×5−3)+(2×−3)+(2×7−3)
k=1
= −1 + 1 + 3 + 5 + 7 + 9 + 11 = 35.
3.
6
9
2k .
k=2
We read sum as k goes from 2 to 6 of 2k . We see that this is a geometric series with
a1 = 22 and
r = a2 /a1 = 23 /22 = 2.
We need to know how many terms are there. There are five terms (count on your
fingers or see the other way at the end of this section, if the terms are many). We
therefore write:
6
9
2(1 − 25 )
2k =
=
1−2
k=2
2(1 − 32)
= 62.
−1
So the summation is a short way of writing the following
6
9
2k = 22 + 23 + 24 + 25 + 26
k=2
= 4 + 8 + 16 + 32 + 64 = 62.
8
9
4.
(k − 4)2 = We read sum as k goes from 3 to 8 of (k − 4)2 We evaluate the sum
k=3
8
9
k=3
(k − 4)2 = (−1)2 + 02 + 12 + · · · + 52 + 62 .
84
CHAPTER 4. ADVANCED ALGEBRA
4.9. SUMMATION NOTATION
We know that 12 + · · · + 52 = 55 (see #1 in this section). We compute
8
9
k=3
(k − 4)2 = (−1)2 + 02 + (12 + · · · + 52 ) + 62
= 1 + 55 + 36 = 92.
9
9
5.
(1/2)k =
k=6
We read sum as k goes from 6 to 9 of (1/2)k . This is another geometric series with
a1 = (1/2)6 ;
an = (1/2)9 ;
r = (1/2).
The number of terms is n = 4. Using the sum of geometric series we compute
9
9
(1/2)k =
k=6
(1/2)6 [1 − (1/2)4 ]
=
1 − (1/2)
(1/5)5 × [1 − (1/2)4 ] =
15
.
512
Or, equivalently, we write
9
9
(1/2)k = (1/2)6 + (1/2)7 + (1/2)8 + (1/2)9
k=6
1
= 6
2
$
1 1 1
1+ + +
2 4 8
%
=
15
.
512
We can see that, for example in #5, we can write
9
9
k=6
k
(1/2) =
4
9
(1/2)(l+5) .
l=1
We simply changed k making it starting at k = 1. To do it, we simply subtracted 5 (that
is in general terms 1 = 6 − 5). We must do so with start and end index number, in this
example 9 − 5 = 4. As observed in #2, the end index value will show how many terms
there are. Check for yourself all the examples.
This is a substitution. We need to express in the new variable all the values of the old
variable values in the formula. The exponent is k = l + 5 because l = k − 5.
85
4.10. COMPLEX NUMBERS
4.10
CHAPTER 4. ADVANCED ALGEBRA
Complex Numbers
4.10.1 CBd1: Perform arithmetic operations.
Perform arithmetic operations with complex numbers.
Example: Add, subtract, multiply and divide (2 + 3i) and (4 − i). Answer:
(2 + 3i) + (4 − i) = 6 + 2i,
(2 + 3i) − (4 − i) = −2 + 4i,
(4 − i) − (2 + 3i) = 2 − 4i,
(2 + 3i) × (4 − i) = 11 + 10i,
(2 + 3i) ÷ (4 − i) = (5/17) + (14/17)i, (4 − i) ÷ (2 + 3i) = (5/13) − (10/13)i.
Explanation of the Example. Recall that we add complex numbers by adding their real parts,
that is the real part of the sum, and (the same for imaginary part) the answer is the sum of
imaginary parts of the members of the additives. The subtraction is the same, only now
we subtract the real parts and imaginary parts.
Multiplication is more difficult. We need to use the rule of multiplying binomials and the
fact that i2 = −1.
The division presents additional difficulties. We can divide ty a real number but we cannot directly divide by a complex number. Therefore, we need to make the denominator to
be a real number. Then, we can divide the real part by the denominator and the imaginary
part by the denominator. To do so, we make the denominator real by multiplying it with
the expression of one containing the complex conjugate of the denominator — that is, for
example,(4 + i) for (4 − i); (2 − 3i) for (2 + 3i).
(4 + i)(4 − i) = 42 − i2 = 16 − (−1) = 17.
The product of a complex number and its conjugate is a real number. To see this we can
use FOIL or simply multiply using the formula sum times difference, and use the fact that
i2 = −1.
FOIL for the above example: (4 + i)(4 − i) = 16 − 4i + 4i − i2 = 16 − (−1) = 17.
1. Perform the indicated operations.
(2 − 3i) + (4 + i) = (2 + 4) + [(−3) + 1]i
= 6 − 2i.
(2 − 3i) − (4 + i) = (2 − 4) + [(−3) − 1]i
= −2 − 4i.
(4 + i) − (2 − 3i) = (4 − 2) + [1 − (−3)]i
= 2 + 4i.
86
CHAPTER 4. ADVANCED ALGEBRA
4.10. COMPLEX NUMBERS
(2 − 3i) × (4 + i) = [2 × 4 − (−3) × 1] + [2 × 1 + (−3) × 4]i
= (8 + 3) + (2 − 12)i = 11 − 10i.
(2 − 3i) ÷ (4 + i)
=
=
(2 − 3i)(4 − i)
2 − 3i
=
4+i
(4 + i)(4 − i)
5 − 14i
(8 + 3i2 ) + (−2i − 12i)
=
= 5/17 − 14/17i.
2
2
4 −i
17
4+i
(4 + i)(2 + 3i)
=
2 − 3i
(2 + 3i)(2 − 3i)
2
(8 + 3i ) + (2i + 12i)
5 + 14i
=
=
= 5/13 + 14/13i.
22 − (3i)2
13
(4 + i) ÷ (2 − 3i)
=
2. Perform the indicated operations.
(12 − 13i) + (4 + i) = (12 + 4) + [(−13) + 1]i
= 16 − 12i.
(12 − 13i) − (4 + i) = (12 − 4) + [(−13) − 1]i
= 8 − 14i.
(4 + i) − (12 − 13i) = (4 − 12) + [1 − (−13)]i
= −8 + 14i.
(12 − 13i) × (4 + i) = [12 × 4 − (−13) × 1] + [12 × 1 + (−13) × 4]
(48 + 13) + (12 − 52)i = 61 − 40i.
(12 − 13i) ÷ (4 + i)
=
(12 − 13i)(4 − i)
12 − 13i
=
4+i
(4 + i)(4 − i)
(48 + 13i2 ) + (−52i − 12i)
35 − 64i
=
= 35/17 − 64/17i.
=
42 − i2
17
87
4.10. COMPLEX NUMBERS
(4 + i) ÷ (12 − 13i)
=
CHAPTER 4. ADVANCED ALGEBRA
(4 + i)(12 + 13i)
4+i
=
12 − 13i
(−12 − 13i)(12 + 13i)
=
(48 + 13i2 ) + (12i + 52i)
35 + 64i
=
= 35/313 + 64/313i.
2
2
12 − (13i)
313
3. Perform the indicated operations.
(2 − 3i) + (14 + 2i) = (2 + 14) + [(−3) + 2]i
= 16 − i.
(2 − 3i) − (14 + 2i) = (2 − 14) + [(−3) − 2]i
= −12 − 5i.
(14 + 2i) − (2 − 3i) = (14 − 2) + [2 − (−3)]
= 12 + 5i.
(2 − 3i) × (14 + 2i) = [2 × 14 − (−3) × 2] + [2 × 2 + (−3) × 14]
= (28 + 6) + (4 − 42)i = 34 − 38i.
(2 − 3i) ÷ (14 + 2i)
=
=
(2 − 3i)(14 − 2i)
2 − 3i
=
14 + 2i
(14 + 2i)(14 − 2i)
(28 + 6i2 ) + (−42i − 4i)
22 − 46i
=
= 11/98 − 23/98i.
2
2
14 − (2i)
196
(14 + 2i) ÷ (2 − 3i)
=
=
(14 + 2i)(2 + 3i)
14 + 2i
=
2 − 3i
(2 − 3i)(2 + 3i)
(28 + 6i2 ) + (4i + 42i)
34 − 38i
= 34/13 − 38/13i.
=
2
2
2 − (3i)
13
4. Perform the indicated operations.
(2 − 3i) + (4 + 5i) = (2 + 4) + [(−3) + 5]i
= 6 + 2i.
88
CHAPTER 4. ADVANCED ALGEBRA
4.10. COMPLEX NUMBERS
(2 − 3i) − (4 + 5i) = (2 − 4) + [(−3) − 5]i
= −2 − 8i.
(4 + 5i) − (2 − 3i) = (4 − 2) + [5 − (−3)]i
= 2 + 8i.
(2 − 3i) × (4 + 5i) = [2 × 4 − (−3) × 5] + [2 × 5 + (−3) × 4]i
(8 + 15) + (10 − 12)i = 23 − 2i.
(2 − 3i) ÷ (4 + 5i)
=
=
2 − 3i
(2 − 3i)(4 − 5i)
=
4 + 5i
(4 + 5i)(4 − 5i)
(8 + 15i2 ) + (−12i − 10i)
−7 − 22i
=
= −7/41 − 22/41i.
2
2
4 − (5i)
41
(4 + 5i) ÷ (2 − 3i)
=
=
(4 + 5i)(2 + 3i)
4 + 5i
=
2 − 3i
(2 − 3i)(2 + 3i)
(8 + 15i2 ) + (10i + 12i)
23 − 2i
=
= 23/13 − 2/13i.
2
2
2 − (3i)
13
5. Perform the indicated operations.
(2 − 3i) + (2 + 3i) = (2 + 2) + [(−3) + 3]i
= 4.
(2 − 3i) − (2 + 3i) = (2 − 2) + [(−3) − 3]i
= −6i.
(2 + 3i) − (2 − 3i) = (2 − 2) + [3 − (−3)]i
= 6i.
89
4.10. COMPLEX NUMBERS
CHAPTER 4. ADVANCED ALGEBRA
We note that these numbers are complex conjugates so their sum and product are
real numbers.
(2 − 3i) × (2 + 3i) = 22 − (3i)2 = 13.
(2 − 3i) ÷ (2 + 3i)
=
=
(2 − 3i)(2 − 3i)
2 − 3i
=
2 + 3i
(2 + 3i)(2 − 3i)
−5 − 6i
(4 + 9i2 ) + (−6i − 6i)
=
= −5/13 − 6/13i.
2
2
2 − (3i)
13
(2 + 3i) ÷ (2 − 3i)
=
=
2 + 3i
(2 + 3i)(2 + 3i)
=
2 − 3i
(2 − 3i)(2 + 3i)
−5 + 6i
(4 + 9i2 ) + (6i + 6i)
=
= −5/13 + 6/13i.
22 − (3i)2
13
90
CHAPTER 4. ADVANCED ALGEBRA
4.10. COMPLEX NUMBERS
4.10.2 CBd2: Recognize and use the graphical representation of complex numbers and their sums.
Graph each pair of complex numbers and their sum in the complex plane.
Example: (2 + i), (3 − 4i).
Solution: (2 + i) + (3 − 4i) = (5 − 3i).
When we have only real numbers, geometrically they can be drawn using an axis with
positive direction and unit section. To geometrically represent complex numbers, we
draw a second axis called imaginary axis. Then to each complex number we draw the
real part on the real axis and the imaginary part on the imaginary axis. We then find the
point in the plain having corresponding real and imaginary part (see the graph).
5
4
3
iR
2
1
±5 ±4 ±3 ±2
0
±1
±2
±3
±4
±5
2+i
1 2 3 4 5 6 7 8 9 10
R
(2+i)+(3±4i)=(5±3i)
3±4i
1. (3 − 2i), (5 + 6i).
Im
(5+6i)
(8+4i)=
(3-2i)+(5+6i)
Re
(3-2i)
91
4.10. COMPLEX NUMBERS
CHAPTER 4. ADVANCED ALGEBRA
2. (3 + 2i), (5 − 6i).
Im
(3+2i)
Re
(8-4i)=
(3+2i)+(5-6i)
(5-6i)
3. (−3 + 2i), (−5 + 6i).
Im
(-8+8i)=
(-3+2i)+(-5+6i)
(-5+6i)
(-3+2i)
- Re
4. (4 − 7i), (8 − 6i).
Re
(8-6i)
(4-7i)
-Im
(12-13i)=
(4-7i)+(8-6i)
92
CHAPTER 4. ADVANCED ALGEBRA
4.10. COMPLEX NUMBERS
4.10.3 CBd3: Manipulate expressions involving conjugates or magnitude.
Give the conjugate and magnitude of each complex number. z denotes the conjugate of z,
and |z| denotes the magnitude of z.
Example: 3 − 4i = 3 + 4i, |3 − 4i| = 5.
1. 3 + 4i.
Complex conjugates are numbers for which two the sum and the product are real
numbers. We produce the conjugate of a complex number by reversing the sign of
the imaginary part, as shown in all the examples.
3 + 4i = 3 − 4i.
The magnitude of a complex number shows how far it is from the origin. We look
at the triangle OHA, O(0, 0), H(3, 0), A(3, 4). It is a right triangle and the magnitude
is the hypotenuse OA.
OA - magnitude
A
(3,4)
H
(3,0)
O
(0,0)
In all cases, the magnitude is the square root of the sum of the squares of the coordinates.
√
|3 + 4i| = |OA| = OH 2 + HA2
√
√
= 32 + 42 = 25 = 5.
2. 2 + 4i
= 2 − 4i.
93
4.10. COMPLEX NUMBERS
|2 + 4i|
CHAPTER 4. ADVANCED ALGEBRA
=
√
22 + 42 =
√
√
20 = 2 5.
3. 5 + 7i
5 − 7i.
|5 + 7i|
=
4. 5 − 12i
√
52 + 72 =
√
74.
= 5 + 12i.
|5 − 12i|
=
√
52 + 122 =
√
169 = 13.
√
5. (1/2) − ( 3/2)i
√
= (1/2) + ( 3)/2i.
√
|(1/2) − ( 3/2)i|
=
/
#
√
√
(1/2)2 + ( 3/2)2 = (1/4) + (3/4) = 1 = 1.
94
CHAPTER 4. ADVANCED ALGEBRA
4.11
4.11. THEORY OF EQUATIONS
Theory of Equations
4.11.1 CBf1: Recognize and use relationship between roots and factors.
Recognize and use the relationship between roots and factors to find the zeros of polynomials of degree higher than 2.
2
Example: x√= 1 is root of p(x) = 3x3 +
√ 2x + x − 6. Find the other roots. Answer:
x = (−5/6) + ( 47/6)i and x = (−5/6) − ( 47/6)i.
You are given one root of each polynomial. Find the remaining roots.
1. x = 1 is root of p(x) = 3x3 + 2x2 + 3x − 8.
Any method for factoring polynomials is good for solving the problem. For polynomials with four of more terms, the general method for factoring is grouping. We
need to have a factor of (x−1) in the factored polynomial. One way of obtaining this
factor is representing one of the terms in the polynomial p(x) as a sum of two terms
so that we can group the polynomial terms in two polynomials which are easier to
factor.These polynomials must have a common factor of (x − 1). Since the needed
factor is (x − 1), we note that if we represent the second term 2x2 in a form of
2x2 = −6x2 + 8x2
and group the polynomial terms, here is one solution:
p(x)
3
2
2
= 3x − 6x + 3x + 8x − 8
= 3x(x − 1)2 + 8(x − 1)(x + 1)
= (x − 1)(3x2 − 3x + 8x + 8)
=
=
=
=
3x3 + (−6x2 + 8x2 ) + 3x − 8
[3x(x2 − 2x + 1)] + 8(x2 − 1)
(x − 1)[3x(x − 1) + 8(x + 1)]
(x − 1)(3x2 + 5x + 8).
Note, that the first three terms form a perfect square after factoring 3x and the remaining terms form a difference of squares after factoring out 8. The rest is factoring
out (x − 1), opening the parenthesis and combining like terms.
To find the remaining roots, we need to solve
3x2 + 5x + 8 = 0.
Recall the quadratic formula
x1,2 =
−b ±
√
b2 − 4ac
.
2a
In our case a = 3, b = 5 and c = 8.
The roots of the equation (and these are the remaining zeros of the original polynomial) are
√
−5 + 71i
,
x2 =
6
95
4.11. THEORY OF EQUATIONS
CHAPTER 4. ADVANCED ALGEBRA
√
−5 − 71i
.
x3 =
6
√
√
The roots of the polynomial p(x) are 1, (−5 + 71i)/6 and (−5 − 71i)/6.
2. x = −1 is root of p(x) = 3x3 + 2x2 + 3x + 4.
Using the same idea as in #1, we represent 2x2 :
2x2 = 6x2 − 4x2 .
The solution is
p(x)
3
2
2
= (3x + 6x + 3x) + (−4x + 4)
= 3x(x + 1)2 − 4(x + 1)(x − 1)
= (x + 1)[3x2 + 3x − 4x + 4]
=
=
=
=
3x3 + (6x2 − 4x2 ) + 3x + 4
3x(x2 + 2x + 1) − 4(x2 − 1)
(x + 1)[3x(x + 1) − 4(x − 1)]
(x + 1)(3x2 − x + 4).
We need to solve
3x2 − x + 4 = 0.
We use quadratic formula to solve the equation.
The roots of the equation are
√
1 + 47i
x2 =
,
6
√
1 − 47i
.
x3 =
6
√
√
The roots of the polynomial p(x) are (−1), (1 + 47i)/6 and (1 − 47i)/6.
3. x = 2 is root of p(x) = x3 + 2x2 − 16.
There is a missing term in the polynomial. We can represent the missing term
0x = 8x − 8x.
We choose this representation motivated by the last term in the polynomial (−16).
Indeed, we could group 8x − 16 = 8(x − 2). Here is the solution:
p(x) = x3 + 2x2 + (8x − 8x) − 16
= x3 + 2x2 − 8x + 8x − 16 = x(x2 + 2x − 8) + 8(x − 2)
= x(x − 2)(x + 4) + 8(x − 2) = (x − 2)[x(x + 4) + 8]
= (x − 2)(x2 + 4x + 8).
96
CHAPTER 4. ADVANCED ALGEBRA
4.11. THEORY OF EQUATIONS
To factor the trinomial in the above solution, we used FOIL method. The simplified
quadratic trinomial will give the remaining solutions by solving the equation
x2 + 4x + 8 = 0.
Again, we use the quadratic formula and find the solutions
√
−4 + 16i
x2 =
= −2 + 2i,
2
√
−4 − 16i
= −2 − 2i.
x3 =
2
The roots of p(x) are 2, (−2 + 2i) and (−2 − 2i).
4. x = −2 is root of p(x) = x3 + 3x2 − 4.
Motivated by the root x = −2, we write the missing term as
0x = 2x − 2x.
The solution is
p(x) = x3 + 3x2 + (2x − 2x) − 4
= x3 + 3x2 + 2x − 2x − 4 = x(x2 + 3x + 2) − 2(x + 2)
= x(x + 1)(x + 2) − 2(x + 2) = (x + 2)[x(x + 1) − 2]
= (x + 2)(x2 + x − 2).
The quadratic equation we solve is
x2 + x − 2 = 0.
We use factoring for this equation and find x2 = −2 and x3 = 1.
The roots of the polynomial p(x) are x1,2 = −2 and x3 = 1.
5. x = 3 is root of p(x) = x3 + 3x − 36.
The coefficient of x2 in the polynomial is zero. We add and subtract 4x2 representing
the missing term as
0x2 = 4x2 − 4x2 .
The factoring is
p(x)
3
2
2
= x − 4x + 3 + 4x − 36
= x(x − 1)(x − 3) + 4(x − 3)(x + 3)
= (x − 3)[x2 − x + 4x + 12]
=
=
=
=
x3 + (4x2 − 4x2 ) + 3x − 36
x(x2 − 4x + 3) + 4(x2 − 9)
(x − 3)[x(x − 1) + 4(x + 3)]
(x − 3)(x2 + 3x + 12).
We solve the equation
x2 + 3x + 12 = 0
97
4.11. THEORY OF EQUATIONS
CHAPTER 4. ADVANCED ALGEBRA
using the quadratic formula given above:
√
−3 + 39i
,
x2 =
2
√
−3 − 39i
x3 =
.
2
√
√
The roots of p(x) are 3, (−3 + 39i)/2 and (−3 − 39i)/2.
Motivation for representing the missing term. We need a difference of squares containing (x − 3)(x + 3) = x2 − 9. Observing that 36 = 4 × 9 we decide to add and
subtract 4x2 because 4x2 − 36 = 4(x2 − 9) = 4(x − 3)(x + 3).
98
CHAPTER 4. ADVANCED ALGEBRA
4.11. THEORY OF EQUATIONS
We know one root of each polynomial. As an alternative of factoring in such case we
can use polynomial or synthetic division of polynomials. Below are shown the results of
the appropriate synthetic divisions.
1. x = 1 is root of p(x) = 3x3 + 2x2 + 3x − 8.
+3
+2
+3
+5
1|
+3
+3
+5
+8
-8
+8
|0
+3
+1
+4
+4
-4
|0
Thus, we write p(x) = (x − 1)(3x2 + 5x + 8).
2. x = −1 is root of p(x) = 3x3 + 2x2 + 3x + 4.
+3
+2
-3
-1
-1|
+3
p(x) = (x + 1)(3x2 − x + 4).
3. x = 2 is root of p(x) = x3 + 2x2 − 16.
Note, how the missing term is written in the synthetic division.
+1
+2|
+1
+2
+2
+4
+0
+8
+8
-16
+16
|0
+3
-2
+1
+0
-2
-2
-4
+4
|0
+3
+9
+12
-36
+36
|0
p(x) = (x − 2)(x2 + 4x + 8).
4. x = −2 is root of p(x) = x3 + 3x2 − 4.
+1
-2|
+1
p(x) = (x + 2)(x2 + x − 2).
5. x = 3 is root of p(x) = x3 + 3x − 36.
+1
+0
+3
+3
+3|
+1
p(x) = (x − 3)(x2 + 3x + 12).
99
4.11. THEORY OF EQUATIONS
CHAPTER 4. ADVANCED ALGEBRA
4.11.2 CBf2: Recognize relationship between degree of a polynomial
and maximum possible number of roots and conjugate pairs of
complex roots.
Give the maximum number of roots of a polynomial without finding them. Example:
p(x) = x4 − 3x + 7 has at most 4 roots.
Recall that the degree of polynomial is the degree of its leading term, that is the term of
the highest degree. The degree of polynomial is the same as the maximum number of
roots a polynomial may have (the example polynomial is of degree 4).
1. p(x) = 9x5 + 12x3 + 2x − 9 has at most 5 roots.
2. p(x) = −9x7 + 2x3 − 2x − 99 has at most 7 roots.
3. p(x) = 999x51 − 12x3 − 2x − 9 has at most 51 roots.
4. p(x) = x4 + x5 + x3 + 2x − 3 has at most 4 roots.
5. p(x) = x5 + x4 + 12x3 + 2x2 + 2x has at most 5 roots.
100
CHAPTER 4. ADVANCED ALGEBRA
4.11. THEORY OF EQUATIONS
4.11.3 CBf3: Find, characterize, or interpret solutions to quadratic equations with negative discriminant.
Example: A quadratic polynomial has a double root at x = 3. The discriminant of this
polynomial is 0.
We assume all polynomial have whole or rational number coefficients.
1. A quadratic polynomial has roots of x = 3 and x = 5. The discriminant is (positive),
(negative), (zero), (undefined), (15). (Circle one.)
The discriminant is positive and is a perfect square (see explanations below).
The roots of this quadratic polynomial are real numbers so that the discriminant
cannot be negative.
It is not zero since the roots are different.
In all cases, one can find the discriminant by evaluating the expression (x − 5)(x −
3) and compute it (i.e., it is not undefined). Let A be a rational number. For this
example we solve P (x) = 0 where
P (x) = A(x − 5)(x − 3) = Ax2 − 8Ax + 15A = 0.
The discriminant is
D = (−8A)2 − 4 × A × 15A = 64A2 − 60A2 = 4A2 = (2A)2 .
It is perfect square because the roots are rational numbers, so that it is not 15.
2. A quadratic polynomial has roots of x = 3 + i and x = 3 − i. The discriminant is
(positive), (negative), (zero), (undefined), (15). (Circle one.)
The discriminant is negative because only then are the roots of a quadratic polynomial complex conjugates.
3. The discriminant of a quadratic polynomial is 12. How many roots does it have?
The only case when the quadratic polynomial has one double root is when the discriminant is zero.In our case, the discriminant is 12 so that the polynomial has two
roots.
4. The discriminant of a quadratic polynomial is −1 and one root is 3 + i. What is the
other root?
We know that one of the roots is complex number. We also know that the coefficients
101
4.11. THEORY OF EQUATIONS
CHAPTER 4. ADVANCED ALGEBRA
of the polynomial are real numbers, so the root must be complex conjugates. The
only complex conjugate of 3 + i is 3 − i. The other root is 3 − i.
5. Find the discriminant of p(x) = 12x2 − 13x + 2.
We simply need to evaluate the discriminant :
D = (−13)2 − 4 × 12 × 2 = 169 − 96 = 75.
102
CHAPTER 4. ADVANCED ALGEBRA
4.12
4.12. APPLICATIONS
Applications
4.12.1 CBw1: General story/word/application problem involving sub
score 1 objectives.
Write rules for functions based on verbal descriptions. Give the domain for the function.
Example: The initial temperature of a glass of milk is 10o centigrade, and it gains 2o
centigrade per minute for 10 minutes. Express this as a function. Answer: Let M (t) be the
temperature after t minutes, expressed in degrees centigrade. M (t) = (10 + 2t) degrees
centigrade, and 0 ≤ t ≤ 10.
1. A snowball masses 102 grams and it loses 3 grams per minute until it is totally
melted.
After one minute the snowball’s mass will be 102 − (1 × 3) = S(1), in two minutes
S(2) = 102 − (2 × 3) = 96. The conclusion is that in t minutes the mass is
S(t) = 102 − 3t
The domain. We cannot use above expression for all positive values of t because in
120 minutes the mass will be
S(120) = 102 − 120 × 3 = −258 grams,
which is not possible. Therefore we stop the melting once S(t) = 0. We solve for t
102 − 3t = 0
t = 34 min
Answer:
S(t) = 102 − 3t and t ∈ [0, 34] in minutes.
2. A spherical balloon has an initial radius of 10 centimeters and is leaking air at the
rate of 2 cubic centimeters per minute.
Let we denote the initial volume of the balloon with V0 . The rate at which it leaks is
2 cc per minute. In t minutes the volume will be V (t) = V0 − 2t in cubic centimeters.
Let find the initial volume using the formula for volume of sphere with radius r =
10:
4
V0 = πr3 cc.
3
The volume in t minutes is
4
V (t) = πr3 − 2t.
3
103
4.12. APPLICATIONS
CHAPTER 4. ADVANCED ALGEBRA
To determine the domain, as in #1, we set V (t) = 0 and solve it for t.
4 3
πr − 2t = 0,
3
t = 2093.5 min.
Answer: V (t) = 4/3πr3 − 2t and t ∈ [0, 2093.5] in minutes.
3. A square of cardboard with an area of 100 square inches is to be folded into a box
by cutting a square of side length x out of each corner.
If the area of the cardboard is 100 square inches, then the side of the cardboard is 10
inches. From it, we can make box as shown in the picture.
10
10-2x
10
x
x
10-2x
2
V=(10-2x) x
We cannot cut bigger pieces than 5 inches a side because we do not have more
cardboard.
Answer:
V (x) = (10 − 2x)2 x and x ∈ [0, 5] in inches.
4. The volume of a sphere is proportional to the cube of its radius. The volume of
sphere of radius 1 meter is 4π/3 cubic meters.
104
CHAPTER 4. ADVANCED ALGEBRA
4.12. APPLICATIONS
We denote with V (r) the volume of sphere with radius r. It is proportional to the
cube of the radius, that means the ratio of volumes of two spheres with radii r and
1 meter is as r3 /13 . We write the ratios an find the volume of a sphere with radius r:
V (r)
r3
= 3 = r3
V (1)
1
We find V (r) using above
4π 3
r
3
For the domain, r could be any possible length, that is r ≥ 0.
Answer:
V (r) = V (1)r3 =
V (r) =
4π 3
r
3
and r ≥ 0, r in meters.
5. The distance a ball falls in a vacuum is proportional to the square of the time it is
falling. In 1 second the ball falls 5 meters. The ball is dropped from a height of 10
meters, and d(t) is its distance from the ground after t seconds.
We drop the ball at time t = 0. We know that the distance traveled is proportional
to the square of the time, i.e.
d(t) = a × t2 .
We know also that after t = 1 sec, the ball has fallen 5 meters. Thus, we can find the
coefficient a
2
d(1)=a x1 =5
5m
d(tend)=10
10 m
105
4.12. APPLICATIONS
CHAPTER 4. ADVANCED ALGEBRA
d(1) = a × 12 = 5,
a = 5 meters.
But the ball cannot fall indefinitely. After time tend it will hit the ground, and therefore the function is defined only for time less than tend and bigger than zero. We
compute tend setting d(tend ) = 10:
d(tend ) = 10 = 5t2 ,
√
t = 2 sec.
Answer:
√
d(t) = 5t2 and t ∈ [0, 2] in seconds.
106
CHAPTER 4. ADVANCED ALGEBRA
4.12. APPLICATIONS
4.12.2 CBw2: General story/word/application problem involving sub
score 2 objectives.
Solve problems involving exponential and logarithmic functions.
Example: The temperature (measured in degrees centigrade) of a solid at time t (measured in seconds) is given by S(t) = 10 + Aekt . At t = 0 seconds the temperature is 20
degrees centigrade, and at t = 30 minutes the temperature is 31 degrees centigrade. What
is the temperature at t = 60 minutes? Round your answer to the nearest 0.01 degrees.
Answer: 54.10 degrees centigrade.
1. The temperature (measured in degrees centigrade) of a solid at time t (measured in
seconds) is given by S(t) = 10 + Aekt . At t = 0 seconds the temperature is 5 degrees
centigrade, and at t = 30 minutes the temperature is -4 degrees centigrade. What is
the temperature at t = 60 minutes? Round your answer to the nearest 0.01 degrees.
We know the value of the function for time t = 0. We find the value of the parameter
A using this condition:
S(0) = 10 + Aek×0 = 5,
10 + A = 5,
A = −5.
We know the temperature when t = 30. We write the temperature using the found
value of A and solve for k:
S(30) = 10 − 5ek×30 = −4,
−5e30k = −14,
e30k =
14
,
5
1
2
ln e30k = ln(14/5),
k=
1
ln(14/5).
30
2. The mass of bacteria, in grams, in a Petri dish is given by M (t) = Aekt , where t is
measured in days. Initially there are 0.10 grams of bacteria, and after 3 days, there
are 0.75 grams. How many grams of bacteria are there after 10 days?
107
4.12. APPLICATIONS
CHAPTER 4. ADVANCED ALGEBRA
There are two solutions. We know that in the beginning, when t = 0 the mass was
0.1 grams. Thus, the coefficient A is
M (0) = Aek×0 = A,
0.1 = A.
For any amount of time we know the mass of bacteria
M (t) = Aekt = 0.1ekt .
For ten days,
M (10) = 0.1ek×10 .
But for three days we have
M (3) = 0.1 × ek×3 = 0.75.
We find that
0.75
= 7.5.
0.1
We rewrite the expression for the ten days:
ek×3 =
M (10) = 0.1ek×10 = 0.1e(k×3)×10/3 ,
M (10) = 0.1(ek×3 )10/3 = 0.1 × 7.510/3 ≈ 82.6 grams.
Here is another solution.
We need to determine the coefficients A, k in the function M (t). We have two relations: first, we know how much bacteria was there initially and second, after three
days. From the first one we find A, from second k. We write
M (0) = Aek×0 = Ae0 = A.
Since in the beginning we had 0.1 grams bacteria, A = 0.1.
Using the fact that after three days there are 0.75 grams of bacteria, we write
0.75 = M (3) = 0.1ek×3 ,
7.5 = e3k .
We take natural logarithm of both sides
ln(7.5) = ln(e3k ) = 3k,
k=
ln(7.5)
≈ 0.6716.
3
108
CHAPTER 4. ADVANCED ALGEBRA
4.12. APPLICATIONS
The mass of bacteria is given by the formula M (t) = 0.1e0.6716t . We evaluate M (t) at
t = 10:
M (10) = 0.1e0.6716×10 = 82.6 grams.
Answer: 82.6 grams.
3. Water is evaporating from a pond according to W (t) = Aekt where W (t) is the volume of water in gallons, and t is time measured in weeks. Initially the pond contains
100,000 gallons of water, and after 2 weeks it contains 90,000 gallons. How long is it
until the pond is half full?
Using the same strategy as in the second solution in #2 we can find A from the
function’s initial condition
W (0) = Ae0k = A,
A = 100, 000 gallons.
We know also that there were 90,000 gallons of water in two weeks. We put this
information in the function and find k
90, 000 = 100, 000e2k ,
e2k = 0.9.
(4.1)
We take natural logarithm of both sides of the equation and find k
ln(0.9) = ln(e2k ) = 2k,
k=
ln(0.9)
= −0.05268.
2
The evaporation function is
W (t) = 100, 000e−0.5268t
where t is given in weeks. Since we need to know when the pond is half full, we set
the equation W (t) = 50, 000 and solve it for t.
50, 000 = M (t) = 100, 000e−0.05268t ,
e−0.05268t = 0.5,
ln(0.5) = ln(e−0.05268t ) = −0.05268t,
t=
Answer: 13 weeks.
ln(0.5)
= 13.15 ≈ 13 weeks.
−0.05268
109
4.12. APPLICATIONS
CHAPTER 4. ADVANCED ALGEBRA
Second method.
We use the relationship for two weeks period to find e2k = 0.9 (see 4.1):
50, 000 = W (t) = 100, 000 × ekt ,
1
50, 000
= = ekt = e(2k)×t/2 = (0.9)t/2 .
100, 000
2
We take natural logarithm and solve for t.
ln(1/2) = ln(.9(t/2) ) =
t=2×
t
ln(0.9),
2
ln(0.5)
≈ 13 weeks.
ln(0.9)
4. When interest is compounded continuously, A dollars invested for t years at the rate
of 100r% interest is worth Aert dollars. If after 10 years 100 dollars has grown to 175
dollars, what was the interest rate?
Reading the text carefully we can see all the information needed to solve the problem. The value of A is given in the second sentence ”. . . 100 hundred dollars has
grown to. . . ”. The amount of time for growing form 100 to 175 dollars is given there
too. We find r from the formula:
A = A0 ert
where A = 175 and A0 = 100; these are initial amount and the amount in ten years.
175 = 100e10r ,
e10r = 1.75,
ln(1.75) = ln(e10r ) = 10r,
r=
ln(1.75)
= 0.05596 = 5.596%.
10
Answer: 5.596%.
5. After t seconds, the horizontal velocity, H(t), of a cannon ball, measured in meters
per second, is given by H(t) = Aekt . After 1 second the velocity is 150 meters per
second, and after 2 seconds it is 125 meters per second. What was the initial velocity?
We know the velocity after one second 150 = H(1) = Aek and after two seconds
125 = H(2) = Ae2k . If we divide H(2) by H(1) we will have a relation only for k:
125
H(2)
Ae2k
=
=
,
150
H(1)
Aek
110
CHAPTER 4. ADVANCED ALGEBRA
4.12. APPLICATIONS
125
= ek ,
150
k = ln(125/150) = −0.1823.
We can use any of the conditions now to find the initial velocity. For time t = 1
second,
150 = Ae−0.1823 ,
150
A = −0.1823 = 180 m/sec.
e
Answer: 180 m/sec.
111
4.13. QUALITATIVE OBJECTS
4.13
CHAPTER 4. ADVANCED ALGEBRA
Qualitative Objects
4.13.1 CBq1: Graph interpretation.
Read information from a graph.
Example: Use the graphs of y = f (x) to find the value(s) for x that makes f (x) largest.
a.
b.
Solution: x=0 in both cases.
For each of the above graphs answer the following questions:
112
CHAPTER 4. ADVANCED ALGEBRA
4.13. QUALITATIVE OBJECTS
1. Use the graph of y = f (x) to solve f (x) = 8.
We draw a horizontal line y = 8 and tend where it crosses the graph of the function.
a) x1 = 2, x2 = −0.8;
b) x = 0.
a.
b.
2. Use the graph of y = f (x) to solve f (2) = y.
We draw vertical line x = 2 and observe the intersection point between it and the
graph of the function.
a) y = 8;
b) y = 6.
a.
b.
113
4.13. QUALITATIVE OBJECTS
CHAPTER 4. ADVANCED ALGEBRA
3. Use the graph of y = f (x) to find the value(s) of x that make f (x) smallest.
For the graph a) the smallest value is observed for x ≈ −5.2. The other graph
does not have the smallest value. This is because the function value of f (x) is 8 at
x = 0.We can get as close to (−6) as we please but we never get the value of (−6).
a) x = −5.0;
b) No such value.
a.
b.
4. Use the graph of y = f (x) to find the zeros of f (x).
These values are easy to identify. We observe the graph and see where it crosses the
x-axis.
a) x1 = −10, x2 = −3.2, x3 = 10;
b) x1 = −6, x2 = 8.
114
CHAPTER 4. ADVANCED ALGEBRA
4.13. QUALITATIVE OBJECTS
5. Use the graph of y = f (x) to solve 3 < f (x) ≤ 5.
We draw two horizontal lines: y = 3 and y = 5. We then observe where these lines
cross the graph of the function. We draw vertical lines through these intersection
points and cross them with x-axis. The values of x for which f (x) is in desired
interval are the solutions of the problem. The intervals are open at the side where
the function approaches 3. Observe also that the graph of the function b) has value
of 4 when x = −10. The line y = 5 will intersect the continuation of the graph of
the left part of the function. This function is defined when x = −10, therefore the
interval is closed at that point.
There are two intervals in which the values of the function obtain the values between 3 and 5. The notation for-the fact that the solutions are in one or in the other
interval is called ”union sign” ( ).
a) x ∈ (−2.25, −1.6] [5, 7); b) x ∈ [−10, −9) [3, 5).
a.
b.
y=5
y=3
y=5
y=3
115
4.13. QUALITATIVE OBJECTS
CHAPTER 4. ADVANCED ALGEBRA
4.13.2 CBq2: Algebraic Proportionality and Scaling.
Solve problems using proportional reasoning.
Example: A circle has area 4 square meters. If we double its radius, what is the area of the
new circle? Answer: 16 square meters.
Note: We need a specific way to solve these problems. Even one may have easier methods
to do them, it is desirable to solve them this particular way. Please, read the explanations
given carefully.
1. A square has side length 4 meters. What is the side length of a square with 3 times
this area?
The area of a square with side a is
A = a × a = a2 = 42 = 16 sq. m.
Let increase the side x times. Therefore, its area is
A(x) = ax × ax = x2 a2 = x2 A
We want A(x) = 3A. Therefore,
3A = 48 = x2 × 16,
√
Answer: 4 3 meters.
x2 = 48/16 = 3,
√
x = 3.
2. A sphere has volume 4 cubic meters. What is the radius of a sphere with with 1/2
this volume?
The volume of the sphere is proportional to the third power of the radius, as shown
by the formula for the volume:
V (r) = (4π/3)r3 ,
4 = 4π/3r3 ,
#
r = 3 (3/π).
If we take two spheres and compare the
ratio of their radii, it is as one-third power
#
3
of the ratio of their volumes (r1 /r2 = V1 /V2 . If V1 = 2 cubic meters and V2 = 4
cubic meters,
116
CHAPTER 4. ADVANCED ALGEBRA
4.13. QUALITATIVE OBJECTS
r1
V1
= ,
r2
V2
#
#
r1 = r2 3 V1 /V2 = r2 3 1/2
where r2 is the radius of a sphere with volume 4 cubic meters. Using the expression
above we compute it:
#
r2 = 3 3/π ≈ 0.9847.
Then,
r1 = r2 ×
Answer: 0.7816 meters.
#
3
1/2 ≈ 0.7816.
3. A sphere has volume 10 cubic meters. What is the volume of a sphere with 3 times
the radius of this sphere?
In reverse of the above, the ratio of the volumes is as the third power of the ratio
of the radiuses. Therefore, if we increase the radius three times, we increase the
volume 33 = 27 times. The new volume is v2 = 27 × 10 = 270 cubic meters.
Answer: 270 cubic meters.
4. A cube has volume 10 cubic meters. What is the volume of a cube with 1/3 the side
length of this cube?
The reasoning is the same as in #3. However, the side is three times smaller which
means that the ratio is 1/3. The volume of the new cube is (1/3)3 times of the volume
of the original one (we denote V1 the original volume and V2 the new one):
V2 = V1 × (1/3)3 = 10 × (1/27) ≈ 0.370 cubic meters.
Answer: 0.370 cubic meters
5. A rectangle has area 15 square feet. If I double its length and triple its height, what
is the new area?
The area is proportional to the sides of the rectangle (a, b are the sides of the rectangle):
A = ab
If we double one of the sides and keep the other the same, we double the area. If we
double one of the sides and triple the other one, then the area will increase 2 × 3 = 6
times!. Therefore, the new rectangle has area A2 = A1 × 6 = 15 × 6 = 90 square
117
4.13. QUALITATIVE OBJECTS
CHAPTER 4. ADVANCED ALGEBRA
meters.
Answer: 90 square meters.
118
Chapter 5
Analytic Geometry
5.1
Lines
5.1.1 CGa1: Recognize and use relationship between slope and parallel
or perpendicular lines.
In each set of three lines, there is either a pair of parallel lines, or a set of perpendicular
lines. Identify that pair and say if they are parallel or perpendicular.
Example: {2x + 6y = 9; 4x + 12y = −5; 12x + 4y = 7}.
Answer: 2x + 6y = 9 and 4x + 12y = −5 are parallel lines.
Recall that two lines are parallel if they have one and the same slope. Two lines are
perpendicular if the product of their slopes is negative one.
1. {3x + 8y = 9; 4x + 12y = −5; 8x − 3y = 7}.
Recall the intercept-slope form of an equation of a line is y = mx + b. We call the
coefficient m slope and b is y-intercept (that is, the second coordinate in which the
line crosses y axis. We transform all equations in slope-intercept form.
3x + 8y = 9,
8y = −3x + 9,
3
9
y =− x+ .
8
8
The second and third equations are
y=−
5
1
5
4
x−
=− x− ,
12
12
3
12
8
7
y = x− .
3
3
119
5.1. LINES
CHAPTER 5. ANALYTIC GEOMETRY
We look now at the slopes of all lines. They are (−3/8), (−1/3) and (8/3). There are
no parallel lines because these will have one and the same slope. But there are perpendicular lines. These are the first and the third lines. The product of their slopes
is negative one.
Answer: The lines 3x + 8y = 9 and 8x − 3y = 7 are perpendicular.
2. {y = 3x + 9; 2y − 6x = −5; 3x − 3y = 7}.
The slope-intercept form of the equations are
y = 3x + 9,
5
y = 3x − ,
2
7
y =x− .
3
We see that first and second lines have the same slopes. This means that they are
parallel.
Answer: the lines y = 3x + 9 and 2y − 6x = −5 are parallel.
3. {4x + 8y = 9; x = 2y − 5; 2x − y = 7}
The slope-intercept form of the equations of the lines are
9
1
y =− x+ ,
2
8
5
1
y = x+ ,
2
2
y = 2x − 7.
We see that the product of the slopes of the first and the third lines is negative one.
Therefore, these lines are perpendicular.
Answer: the lines 4x + 8y = 9 and 2x − y = 7 are perpendicular.
4. {3x + 8y = 9; x + y = −5; x = y + 7}.
The slope-intercept form of the equations of the lines are
9
3
y =− x+ ,
8
8
120
CHAPTER 5. ANALYTIC GEOMETRY
5.1. LINES
y = −x − 5,
y = x − 7.
Again, there are no parallel lines. But the product of the slopes of the second and
the third lines is minus one. These are perpendicular.
Answer: the lines x + y = −5 and x = y + 7 are perpendicular.
5. {3x + 8y = 9; 4x + 12y = −5; 6x + 16y = 7}
The slope-intercept form of the equations of the lines are
3
9
y =− x+ ,
8
8
5
1
y =− x− ,
3
12
7
3
7
6
=− x+ .
y =− x+
16
16
8
16
The first and the third equations have one and the same slope of −3/8. Therefore
these two lines are parallel.
Answer: the lines 3x + 8y = 9 and 6x + 16y = 7 are parallel.
121
5.1. LINES
CHAPTER 5. ANALYTIC GEOMETRY
5.1.2 CGa2: Find the equation of a line through a point parallel or perpendicular to a given line.
Find an equation of a line given a point on the line and a line that the line is (1) parallel to
and (2) perpendicular to.
Example: An equation of the line parallel to 2x + 3y = 7 and passing through (2, 3) is
2x + 3y = 13 and an equation of the line perpendicular to 2x + 3y = 7 and passing through
(2, 3) is −3x + 2y = 0.
If a and b are nonzero numbers then the lines ax + by = c and −bx + ay = d are always
perpendicular. This is because their slopes are −a/b and b/a. When we multiply these
numbers, their product is minus one (see above example a = 2 and b = 3.
The other fact we’ll take advantage of is that if a point is on the line, when we substitute
the coordinates in the equation, this makes it true sentence.
For each question, part (1) is for parallel line and part (2) is for perpendicular line.
1. Find equations for the lines through (2, −3) that are (1) parallel to, and (2) perpendicular to 4x − 9y = 7.
(1) The parallel line to 4x − 9y = 7 passing through point (2, −3) is
4x − 9y = b.
We find b by substituting x = 2 and y = −3:
4 × 2 − 9 × (−3) = b,
b = 35.
The equation of the line parallel to 4x − 9y = 7 is 4x − 9y = 35.
(2) We construct perpendicular line the way we showed in the beginning: we interchange the coefficients in front of x and y and change the sign of one of them:
9x + 4y = b.
We use b because we need the line to pass through a certain point. We evaluate the
expression 9x + 4y = b for x = 2 and y = −3:
9 × 2 + 4 × (−3) = b,
b = 6.
The line perpendicular to 4x − 9y = 7 and passing through point (2, −3) is 9x + 4y =
6.
122
CHAPTER 5. ANALYTIC GEOMETRY
5.1. LINES
2. Find equations for the lines through (0, −3) that are (1) parallel to, and (2) perpendicular to 4x + 9y = 17.
(1)
x = 0,
y = −3,
4x + 9y = b.
4 × 0 + 9 × (−3) = b,
b = −27.
The equation of the line parallel to 4x − 9y = 7 and passing through point (0, −3) is
4x + 9y = −27.
(2)
x = 0,
y = −3,
−9x + 4y = b.
−9 × 0 + 4 × (−3) = b,
b = −12.
The line perpendicular to 4x − 9y = 7 and passing through point (0, −3) is
−9x + 4y = −12.
3. Find equations for the lines through (12, −3) that are (1) parallel to, and (2) perpendicular to 4x + 19y = 7.
(1)
x = 12,
y = −3,
4x + 19y = b.
4 × 12 + 19 × (−3) = b,
123
5.1. LINES
CHAPTER 5. ANALYTIC GEOMETRY
b = −9.
The equation of the line parallel to 4x + 19y = 7 and passing through point (12, −3)
is 4x + 19y = −9.
(2)
x = 12,
y = −3,
19x − 4y = b.
19 × 12 − 4 × (−3) = b,
b = 276.
The equation of the line perpendicular to 4x + 19y = 7 and passing through point
(12, −3) is 19x − 4y = 276.
4. Find equations for the lines through (5, −3) that are (1) parallel to, and (2) perpendicular to 14x − 9y = 8.
(1)
x = 5,
y = −3,
14x − 9y = b.
14 × 5 − 9 × (−3) = b,
b = 97.
The equation of the line parallel to 14x − 9y = 8 and passing through point (5, −3)
is 14x − 9y = 97.
(2)
x = 5,
y = −3,
124
CHAPTER 5. ANALYTIC GEOMETRY
5.1. LINES
9x + 14y = b.
9 × 5 + 14 × (−3) = b,
b = 3.
The equation of the line perpendicular to 14x − 9y = 8 and passing through point
(5, −3) is 9x + 14y = 3.
5. Find equations for the lines through (22, −33) that are (1) parallel to, and (2) perpendicular to y = 7.
(1) The line y = 7 has values of 7 for all values of x — it is parallel to x axis. Therefore,
the line parallel to it and going through (22, 33) is
y = −33.
(2) The line perpendicular to y = 7 and going through the same point is parallel to
y axis and is
x = 22.
Another observation we can make here is that one cannot apply the slope product
rule for perpendicular lines to be minus one. The line y = 33 has slope zero, and the
line x = 22 has an undefined slope so that the product of them is not defined.
125
5.2. DISTANCE
5.2
CHAPTER 5. ANALYTIC GEOMETRY
Distance
5.2.1 CGb1: Recognize and apply the distance formula (two dimensions).
Use the distance formula or the Pythagorean theorem to solve problems
√
Example: What is the distance from (3, 4) to (5, 11). Answer: 53.
1. How far is it from (3, 5) to (7, 11)?
Let A has coordinates (3, 5) and B — (7, 11). We draw a line parallel to y-axis through
point B. We draw also line parallel to x axis through point A. These two lines intersect each other in point H. The coordinates of all points are given on the drawing.
B (7,11)
e
nc
ta
dis
A
(3,5)
H (7,5)
We look at the triangle AHB. It is a right triangle. The distance AB is a hypotenuse
in it. To find it we use the Pythagorean theorem. The line segment AH has length
AH = 7 − 3 = 4 units.
The line segment HB has length
HB = 11 − 5 = 6 units.
We write the Pytagorean theorem for triangle AHB:
√
AB = AH 2 + HB 2 ,
#
AB = (7 − 3)2 + (11 − 5)2 ,
√
√
AB = 16 + 36 = 52.
In general, finding a distance between two points in the plane we use Pythagorean
theorem. The distance is the hypotenuse in it and the legs are parallel to the axes.
The length of the legs is the difference of the corresponding coordinates of the two
126
CHAPTER 5. ANALYTIC GEOMETRY
5.2. DISTANCE
given points (see AH and HB lengths).
If the line segment is parallel to one of the axes, the two pints have one and the same
(either) first or second coordinate. In this case, the length is simply the difference of
the different coordinates.
Answer: the distance between (3, 5) and (7, 11) is
√
52.
2. How far is it from (3, −5) to (−7, 11)?
Let denote the distance with d. We use the same way:
#
d = [(−7) − 3]2 + [11 − (−5)]2 ,
√
d = 100 + 256,
√
d = 356.
Answer: the distance between points (3, −5) and (−7, 11) is
√
356.
3. The distance from (3, 5) to (x, x + 2) is 8 units. What are the possible values of x?
H(3,x+2)
X(x,x+2)
A(3,5)
We look at the triangle AHX. The lengths of the legs are
AH = (x + 2) − 5 = x − 3
and
HX = x − 3.
We want AX = 8. We now express AX 2 = 64:
AX 2 = AH 2 + HX 2 ,
82 = (x − 3)2 + (x − 3)2 .
We solve this equation for x and find:
127
5.2. DISTANCE
CHAPTER 5. ANALYTIC GEOMETRY
√
32 = 3 + 4 2,
√
√
x2 = 3 − 32 = 3 − 4 2.
x1 = 3 +
√
√
√
Answer:
the
coordinates
of
point
(x,
x+2)
8
units
from
(3,
5)
are
([3+4
2],
[5+4
2])
√
√ 2
1
and [3 − 4 2], [5 − 4 2] .
4. The distance from (3, 5) to (2x+1, x+2) is 8 units. What are the possible values of x?
We denote c1 the leg parallel to x axis and c2 the one parallel to y axis. The distance
we denote with d.
c2
A(3,5)
d
c1
X(2x+1,x+2)
H(3,x+2)
We need d = 8, therefore d2 = 64. We write
c1 = (x + 2) − 5 = x − 3,
c2 = (2x + 1) − 3 = 2x − 2,
d2 = c21 + c22 ,
64 = (x − 3)2 + (2x − 2)2 .
We solve the above equation for x:
x2 − 6x + 9 + 4x2 − 8x + 4 = 64,
5x2 − 8x − 51 = 0,
x1 =
and
14 +
#
√
(−14)2 − 4 × 5 × (−51)
7 + 4 19
=
2×5
5
√
7 − 4 19
.
x2 =
5
128
CHAPTER 5. ANALYTIC GEOMETRY
5.2. DISTANCE
We evaluate the expressions in the point X(2x + 1, x + 2) to find the answer. For the
solution of the equation x1 we find
√
7 + 4 19
+1
2x + 1 = 2 ×
5
√
√
14 + 8 19
19 + 8 19
=
+1=
;
5
5
√
7 + 4 19
+2
x+2=
5
√
√
7 + 4 19 + 10
17 + 4 19
=
=
.
5
5
For the solution x2 the values are the conjugates of the above numbers.
√
√
√
√
Answer: the points ([19 + 8 19]/5, [17 + 4 19]) and ([19 − 8 19]/5, [17 − 4 19]) are
8 units from (3, 5).
129
5.2. DISTANCE
CHAPTER 5. ANALYTIC GEOMETRY
5. The distance from (3, 5) to (x, x + 2) is 8 units more than the distance from (0, 0) to
(x, x + 2). What are the possible values of x?
We denote d1 the distance between points A(3, 5) and X(x, x + 2) and d2 between
O(0, 0) and X(x, x + 2).
A(3,5)
d1
H1(3,x+2)
X(x,x+2)
d2
H2(x,0)
O(0,0)
We write
d1 = d2 + 8
We express d2 form triangle OH2 X
d2 =
and d1 from triangle AH1 X:
d1 =
#
x2 + (x + 2)2
#
(x − 3)2 + (x − 3)2
√
√
When x ≥ 3 then d1 = 2(x − 3), and when x < 3 then d1 = 2(3 − x). We use the
relation between d1 and d2 :
√
d1 = d2 + 8
#
2(x − 3) = x2 + (x + 2)2 + 8
The equation d1 = d2 + 2 leads to radical equation. This equation has two solutions
which are
√
√
24 2 + 39
24 2 − 39
x1 = √
x2 = √
.
8 2+8
8 2−8
The details of solving the equation are given on the next page.
130
CHAPTER 5. ANALYTIC GEOMETRY
5.2. DISTANCE
SOLUTION OF THE EQUATION
d1 = d2 + 8
We use the absolute value principle: d1 =
The equation
√
2(x − 3) then x ≥ 3 and d1 =
√
2(3 − x) when x < 3.
d1 = d2 + 8
is equivalent to the union of the equations
√
#
2(x − 3) = x2 + (x + 2)2 + 8
#
√
2(3 − x) = x2 + (x + 2)2 + 8
when x ≥ 3
when x < 3.
We need to solve both of them and the solutions will make the solution of the original one.
We start with
√
2(x − 3) =
#
x2 + (x + 2)2 + 8.
We take 8 on the other side and square both sides:
#
√
[ 2(x − 3) − 8]2 = [ x2 + (x + 2)2 ]2 ,
√
2(x − 3)2 − 16 2(x − 3) + 64 = x2 + (x + 2)2 .
We evaluate both sides and simplify
√
√
2(x2 − 6x + 9) − 16 2x + 48 2 + 64 = x2 + x2 + 4x + 4,
√
√
2x2 − 12x + 18 − 16 2x + 48 2 + 64 = 2x2 + 4x + 4.
We take all terms on one side and simplify combining like terms
√
√
2x2 − 12x + 18 − 16 2x + 48 2 + 64 − 2x2 − 4x − 4 = 0,
√
√
(2 − 2)x2 + (−12 − 16 2 − 4)x + (18 + 48 2 + 64 − 4) = 0,
√
√
(−16 − 16 2)x + (78 + 48 2) = 0.
This is a linear equation for x. The solution is
√
78 + 48 2
√
x=
16 + 16 2
We see that the numerator and the denominator have common factor 2 so we remove it:
√
2(39 + 24 2)
√ ,
x=
2(8 + 8 2)
131
5.2. DISTANCE
CHAPTER 5. ANALYTIC GEOMETRY
√
39 + 24 2
√ .
x=
8+8 2
We simply need to verify that this number is bigger than 3 (this was the condition under which
we solve the equation). We express the solution in the following form
√
39 + 24 2
√
x=
8+8 2
√
15
24 + 24 2
√ +
√
=
8+8 2
8+8 2
15
√ .
=3+
8+8 2
The solution is in the domain because it is a sum of 3 and a number bigger than zero.
We solve the same way the other equation
√
2(3 − x) =
√
#
x2 + (x + 2)2 + 8
2(3 − x) − 8 =
when x < 3,
#
x2 + (x + 2)2 ,
√
2(9 − 6x + x2 ) − 16 2(3 − x) + 64 = x2 + x2 + 4x + 4,
√
√
18 − 12x + 2x2 − 48 2 + 16 2x + 64 − 2x2 − 4x − 4 = 0,
√
√
(−12 − 4 + 16 2)x + (18 − 48 2 + 64 − 4) = 0,
√
√
(16 2 − 16)x = 48 2 − 78,
√
24 2 − 39
x= √
.
8 2−8
Here is the the proof that this solution is less than 3:
√
24 2 − 39
x= √
8 2−8
√
24 2 − 24
15
= √
− √
8 2−8
8 2−8
15
.
=3− √
8 2−8
We subtract a positive number from 3 so that the solution we found is smaller than 3.
132
CHAPTER 5. ANALYTIC GEOMETRY
5.3
5.3. CIRCLES
Circles
5.3.1 CGc1: Recognize and apply the definition of a circle as locus of
points equidistant from a fixed point.
Find equations of circles from geometric information.
Example: The point (3, 4) is center of a circle of radius 5. Give the coordinates of four
points on the circle. Answer: (3, 9), (8, 4), (3, −1) and (−2, 4).
1. The point (−3, 2) is center of a circle of radius 7. Give the coordinates of four points
on the circle.
We can choose points on a circle as shown on the figure below. To find more points
on the circle, systematic way will be shown at the end of the section.
(a,b+r)
(a-r,b)
r
O(a,b)
(a+r,b)
(a,b-r)
Four points on the circle are (a,b+r),
(a,b-r),(a+r,b), (a-r,b).
The values of the parameters for this problem are a = −3, b = +2 and r = 7.
Therefore, the points are:
(a + r, b) = (−3 + 7, 2) = (+4, +2),
(a − 4, b) = (−3 − 7, 2) = (−10, 2),
(a, b + r) = (−3, 2 + 7) = (−3, +9),
(a, b − r) = (−3, 2 − 7) = (−3, −5).
Four points on the circle are (4, 2), (−10, 2), (−3, 9) and (−3, −5).
2. The point (−1, 2) is center of a circle of diameter 4. Give the coordinates of four
points on the circle.
133
5.3. CIRCLES
CHAPTER 5. ANALYTIC GEOMETRY
We know the diameter of the circle is twice the radius, so we find the radius of the
circle: r = d/2 = 4/2 = 2. For this circle a = −1, b = +2 and r = 2. Answer: four
points on a circle with center (−1, 2) and diameter 2 are
(−1, −4), (−1, 0), (1, 2) and (−3, 2).
3. The point (3, 21) is center of a circle of radius 9. Give the coordinates of four points
on the circle.
Now a = 3, b = 21 and r = 9.
Answer: four points on a circle with center (3, 21) and radius 9 are
(3, 30), (3, 12), (12, 21) and (−6, 21).
4. The point (−13, 2) is center of a circle of diameter 10. Give the coordinates of four
points on the circle.
The values of the parameters are a = −13, b = 2 and r = 5. Answer: four points on
a circle with center (−13, 2) and diameter 10 are
(−13, 7), (−13, −3), (−8, 2) and (−18, 2).
5. The point (−3, 22) is center of a circle of circumference 14π. Give the coordinates of
four points on the circle.
The radius and the circumference are connected by the formula
c = 2πr
where the circumference is c and the radius is r. Thus the radius in the problem is
r = 14π/(2π) = 7.
The value of the parameters are a = −3, b = 22 and r = 7. Answer: four points on
a circle with center (−3, 22) and of circumference 14π are (−3, 15), (−3, 29), (10, 22)
and (−4, 22).
Finding other points
For a circle with center O(a, b and radius r the formula determining all the points X(x, y)
on the circle is
(x − a)2 + (y − b)2 = r2 .
134
CHAPTER 5. ANALYTIC GEOMETRY
5.3. CIRCLES
O(a,b)
r
X(x,y)
The distance between X and O is
XO= (x-a) 2 + (y-b) 2 =r
Note that we have infinitely many choices for one of the coordinates so that the equation
(x − a)2 + (y − b)2 = r2 has real solutions. For example, in #1, the coordinates of the
center are a = −3 and b = +2. We can choose y = 0. Then the equation is
[x − (−3)]2 + (0 − 2)2 = 72 .
We solve it for x:
(x + 3)2 = 45,
The solutions of the equation are
#
√
(x + 3)2 = ± 45.
x1 = −3 +
and
x2 = −3 −
√
45
√
45.
√
We thus found two points (−3 + 45, 0) and (−3 − 45, 0).
We make now x = 0 and solve for y.
√
[0 − (−3)]2 + (y − 2)2 = 49,
y1,2 = 2 ±
√
40.
√
We found two more points (0, 2 + 40) and (0, 2 − 40).
Four other
circle with center√(−3, 2) and radius
√ points on a√
√7 are
(−3 + 45, 0), (−3 − 45, 0), (0, 2 + 40) and (0, 2 − 40).
We show this way that there are infinitely many solutions of the above problems.
√
135
5.3. CIRCLES
CHAPTER 5. ANALYTIC GEOMETRY
5.3.2 CGc2: Find center and radius of a circle from its equation or vice
versa.
Give the center and radius of a circle given its equation, and give an equation of the circle
given the radius and center of the circle.
Example: An equation of a circle is x2 + 2x + y 2 + 4y = 4. Answer: The center of this
circle is (−1, −2) and radius of this circle is 3.
Example: The center of a circle is at (2, 3) and the radius of this circle is 9. Answer: An
equation of this circle is (x − 2)2 + (y − 3)2 = 81.
1. The center of a circle is at (5, −6) and the radius of the circle is 4.
We use the same formula as in previous section. We have
a = 5;
b = −6;
r = 4.
We write
(x − a)2 + (y − b)2 = r2 ,
(x − 5)2 + [y − (−6)]2 = 42 .
Answer: the equation of a circle with radius 4 and center (5, −6) is
(x − 5)2 + (y + 6)2 = 16.
2. The center of a circle is at (−3, −6) and the radius of the circle is 2.
The same way:
a = −3;
b = −6;
r = 2.
We write the equation
[x − (−3)]2 + [y − (−6)]2 = 22 .
Answer: the equation of a circle with center (−3, 2) and radius 2 is
(x + 3)2 + (y − 6)2 = 4.
3. An equation of a circle is x2 +6x+y 2 −8y = 5. Give the center and radius of this circle.
We need to have an equation in a form (x − a)2 + (y − b)2 = r2 . To do so we use a
process known as ”completing a square”. Let look at the terms containing x. They
are x2 + 6x. We wish to have a binomial (x + a)2 .
x2 + 6x + a2 = (x + a)2 ,
x2 + 6x + a2 = x2 + 2ax + a2 .
136
CHAPTER 5. ANALYTIC GEOMETRY
5.3. CIRCLES
Therefore,
6x = 2ax
and
a = 3.
We need to add 9 to the left side. We must not change the equality so we add 9 to
the right side 9 too.
x2 + 6x + y 2 − 8y = 5,
x2 + 6x + 9 + y 2 − 8y = 5 + 9,
(x + 3)2 + y 2 − 8y = 14.
We complete the same way the square for y
(x + 3)2 + (y 2 − 8y + 16) = 14 + 16,
(x + 3)2 + (y − 4)2 = 30.
We can write now
The right side is r2 .
√
[x − (−3)]2 + (y − 4)2 = ( 30)2 .
Answer: the circle x2 + 6x + y 2 − 8y = 5 has radius of
√
30 and center (−3, 4).
4. An equation of a circle is x2 − 16x + y 2 + 14y = 5. Give the center and radius of this
circle.
The terms involve x are x2 and 16x. The crossproduct shows the second term must
be 8. For the term y it is 7. To complete the square we need to add to both sides
(82 + 72 ) = 113.
x2 − 16x + y 2 + 14y = 5,
[x2 − 16x + 64] + [y 2 + 14y + 49] = 5 + 113,
√
(x − 8)2 + (y + 7)2 = ( 118)2 ,
√
Answer: The center of the circle is (8, −7) and the radiusl is 118)2 .
5. The center of a circle is at (5, 12) and the circle passes through the origin.
We know the center of the circle and a point of the circle. Part of the circle is shown
on the graph.
137
5.3. CIRCLES
CHAPTER 5. ANALYTIC GEOMETRY
A(5,12)
r
r
O(0,0)
H(5,0)
We use triangle OHA to find the radius of the circle. In it, the legs are OH = 5 and
HA = 12.
r2 = OH 2 + HA2 ,
√
r = OH 2 + HA2 ,
√
r = 52 + 122 = 13.
We are ready to write the equation. Its center is (5, 12) and the radius is 13. The
equation is
(x − 5)2 + (y − 12)2 = 132 .
Answer: the equation of a circle with center (5, 12) and passes through the origin is
(x − 5)2 + (y − 12)2 = 132 .
138
CHAPTER 5. ANALYTIC GEOMETRY
5.4
5.4. CONICS
Conics
5.4.1 CGd1: Given an equation of the form Ax2 +By 2 +Cx+Dy +E = 0,
be able to identify the graph or name of the conic or vice versa.
Example: The equation 4x2 + 16x + y 2 − 2y = 0 describes an ellipse.
6
5
4
3y
2
1
–6 –5 –4 –3 –2 –1
x
1 2
–1
–2
–3
–4
(1) Circle the equation that best describes this graph.
2
–6
–4
x
1
–2
2
4
0
–1
–2 y
–3
–4
(a)
x2 + 4y 2 + 2x + 8y = 4,
(b)
x2 − 4y 2 + 2x + 8y = 4,
(c)
4x2 + y 2 + 2x + 8y = 4,
(d) −x2 + 4y 2 + 2x + 8y = 4,
(e)
x2 + 2x + 8y = 4.
This is a graph of an ellipse. The equation of an ellipse is
(x − a)2 (y − c)2
+
= 1.
b2
d2
The center of the elipse is the point (a, c). In case b > d, the major axes is the horizontal
diameter of the elipce with lenght of 2b. The minor axes is perpendicular to the major
axes and has lenght of 2d. Equations (a) and (c) could be transformed into this form. All
139
5.4. CONICS
CHAPTER 5. ANALYTIC GEOMETRY
other ones are different curves than ellipse. We’ll show the process using (a). We complete
squares on the left side by adding (12 + 12 ) = 2 to the equation.
x2 + 4y 2 + 2x + 8y = 4,
(x2 + 2x + 12 ) + (4y 2 + 8y + 12 ) = 4 + 2,
(x + 1)2 + (2y + 1)2 = 6,
(x + 1)2 + [2(y + 1/2)]2 = 6.
√
We divide both sides with 6 and express it as 6 = ( 6)2 . The term containing y has
in addition multiplier of 4, so we represent it as 4 = 1/(1/4) so it will be a part of the
denominator. We thus obtain
(x + 1)2 (y + 1/2)2
√
+ √
= 1.
( 6)2
( 3/2)2
The other ellipse is
(x + 1/4)2
(y + 4)2
#
+#
= 1.
( 101/16)2
101/4
We choose any point from the curve. We evaluate the the equation at that point and the
one which makes the equation true sentence is the equation for the curve. The point
(−4, −1) is on the curve. To evaluate, we can use the original equations of the curves. We
needed to show simply that these are ellipses.
For
x = −4 and
y = −1 we evaluate (a) and (c):
(a)
x2 + 4y 2 + 2x + 8y = 4,
(−4)2 + 4(−1)2 + 2(−4) + 8(−1) = 4,
16 + 4 − 8 − 8 = 4,
4 = 4 TRUE!
We will evaluate (c) too:
4x2 + y 2 + 2x + 8y = 4,
4(−4)2 + (−1)2 + 2(−4) + 8(−1) = 4,
64 + 1 − 8 − 8 = 4,
49 = 4 FALSE!
The elipse has an equation
a) x2 + 4y 2 + 2x + 8y = 4.
(2) Circle the equation that best describes this graph.
140
CHAPTER 5. ANALYTIC GEOMETRY
5.4. CONICS
8
6
y4
–8 –6 –4 –2
2
2
4x6
8 10
–2
–4
–6
–8
–10
(a)
x2 + 4y 2 + 6x + 8y = 12
(c)
−4x2 + y 2 − 8x + 2y = 4 (d)
(b)
x2 − 4y 2 − 2x + 8y = 4
−x2 + 4y 2 + 2x + 8y = 4
(e) y 2 + 2x + 8y = 4
This is a graph of a hyperbola. The equation of a hyperbola is
(x − a)2 (y − c)2
−
= ±1
b2
d2
Looking at the equations, only (b), (c) and (d) can be put in this form. Note, that (a) is an
equation of an ellipse and in (e) there is no term containing x2 . We evaluate (b), (c) and
(d) at point x = 0 and y = 1.
(b)
x2 − 4y 2 − 2x + 8y = 4,
02 − 4(12 ) − 2 × 0 + 8 × 1 = 4,
0 − 4 − 0 + 8 = 4,
4 = 4 TRUE!
We evaluate the other two the same way (details not shown):
(c)
−4x2 + y 2 − 8x + 2y = 4,
3 = 4 FALSE!
(d)
−x2 + 4y 2 + 2x + 8y = 4,
12 = 4 FALSE!
Answer is (b) x2 − 4y 2 − 2x + 8y = 4.
141
5.4. CONICS
CHAPTER 5. ANALYTIC GEOMETRY
(3) Match the function on the left with its graph on the right.
1.
x2 + 2x + 2y 2 + 3y + 1 = 0.
a.
8
6
4y
–10 –8 –6 –4 –2
2
2
x
4
6
8
–2
–4
–6
–8
–10
2. x2 + 2x − 2y 2 + 3y + 1 = 0.
b.
2
–6
–4
2x
–2
4
–2
–4
y
–6
–8
–10
3.
x2 + 2x + 3y + 1 = 0.
c.
1
0.5
–3
–2
x
–1
1
0
–0.5
–1 y
–1.5
–2
We have one hyperbola, one ellipse and one parabola. The equation of a hyperbola has
coefficients in front of x2 and y 2 with different signs, the ellipse has these with positive
sign and parabola has either of the terms x2 and y 2 , but not both. We have then 2 →
hyperbola, 1 → ellipse and 3 → parabola.
142
CHAPTER 5. ANALYTIC GEOMETRY
1.
5.4. CONICS
x2 + 2x + 2y 2 + 3y + 1 = 0.
c.
1
0.5
–3
x
–2
–1
1
0
–0.5
–1 y
–1.5
–2
2. x2 + 2x − 2y 2 + 3y + 1 = 0.
a.
8
6
4y
–10 –8 –6 –4 –2
2
2
x
4
6
–2
–4
–6
–8
–10
3.
x2 + 2x + 3y + 1 = 0.
b.
2
–6
–4
2x
–2
–2
–4
y
–6
–8
–10
143
4
8
5.4. CONICS
CHAPTER 5. ANALYTIC GEOMETRY
5.4.2 CGd2: Given an equation of a parabola in the form Ax2 + Bx +
Cy +D = 0 or Ay 2 +By +Cx+D = 0, describe the parabola, giving
its vertex, axis, and the direction it opens, and vice versa.
Example: The equation 4x2 + 16x − 2y + 8 = 0 describes a parabola that opens upward
with vertex located at (−2, −4) and axis of symmetry x = −2.
1. Describe the parabola described by the equation x2 + 2x + 3y + 1 = 0.
In general, the variable which does not have a second power shows the direction
(axis) in which the parabola will open. The sign in front of the second power of the
other variable will show the positive or negative direction of opening the parabola.
We rearrange the equation in a form of a function with independent variable x. For
this, we take y on the other side of the equation and divide by (−3).
x2 + 2x + 3y + 1 = 0,
−3y = x2 + 2x + 1,
1
2
1
y = − x2 − x − .
3
3
3
Therefore, the direction of this parabola is in y axis and it is open downward. We
complete the square on the right side:
1
y = − (x2 + 2x + 1).
3
In the parenthesis we have perfect square of (x + 1). The equation is
1
y = − (x + 1)2 + 0.
3
We add 0 to underline the process of finding the vertex. This is the value of the
other coordinate (in this case — y) when the complete square is zero. The coefficient
of the squared binomial is negative, so that the parabola is open downward. When
x = −1, the parabola takes its highest point and this is the axes of symmetry.
Answer: This is a parabola open downward. The axis of symmetry is x = −1, and
the vertex is (−1, 0).
144
CHAPTER 5. ANALYTIC GEOMETRY
5.4. CONICS
x= -1 is
axis of symm
vertex is
(-1,0)
The parabola opens downward.
2. Describe the parabola described by the equation y 2 + 2x + 3y + 1 = 0.
We use the same method but this time we express x in terms of y.
y 2 + 2x + 3y + 1 = 0,
−2x = y 2 + 3y + 1,
$ %2 $ %2
3
3
3
2
−2x = y + 2 × x +
−
+ 1,
2
2
2
%2
$
3
5
−2x =
y+
− ,
2
4
$
%2
3
5
1
y+
+ .
x = −
2
2
8
We complete the square in variable y leaving the coefficient (−2) in front of x. Expressing the term 3y in a form of double crossproduct, we note 3y = 2 × (3/2) × y.
Therefore, the second term for the completing the square is (3/2)2 . We add and subtract that term and complete the square. On the last step, we divide the identity by
(−2).
vertex is
(5/8, -3/2)
y= - 3/2
axis of symm
The parabola opens leftward.
145
5.4. CONICS
CHAPTER 5. ANALYTIC GEOMETRY
Answer: this is a parabola open leftward. The axis of symmetry is y = −3/2 and the
vertex is (5/8, −3/2). For all other parabolas, only the equations and the plot will
be shown.
3. Describe the parabola described by the equation −2y 2 + 2x + 3y + 1 = 0.
−2y 2 + 2x + 3y + 1 = 0,
2x = 2y 2 + 3y − 1,
1
1
x = (2y 2 − 3y) − 1 × ,
2
2
3
9
1
9
2
x = (y − y + ) − − ,
2
16
2 16
$
%2
3
17
x= y+
− .
4
16
Answer: The parabola opens to the right. The axis of symmetry is y = −3/4 and the
vertex is (−3/4, −17/16).
vertex is
(-17/16, -3/4)
axis of symm
y= - 3/4
The parabola opens to the right.
4. Describe the parabola described by the equation x2 + 2x − 3y + 2 = 0.
x2 + 2x − 3y + 2 = 0,
x2 + 2x + 2 = +3y,
1
2
y = (x2 + 2x) + ,
3
3
1
1
2
y = (x + 1) + .
3
3
Answer: The parabola opens up. The axis of symmetry is x = −1 and the vertex is
(−1, 1/3).
146
CHAPTER 5. ANALYTIC GEOMETRY
5.4. CONICS
axis of symm
x= -1
vertex is
(-1,1/3)
The parabola opens up
5. Describe the parabola described by the equation x2 + 4x + 3y + 4 = 0.
x2 + 4x + 3y + 4 = 0,
−3y = x2 + 4y + 4,
−3y = (x + 2)2 ,
1
y = − (x + 2)2 .
3
Answer: the parabola opens down. The axis of symmetry is x = −2 and the vertex
is (−2, 0).
axis of symm
x=-2
vertex is
(-2, 0)
The parabola opens down.
147
5.4. CONICS
CHAPTER 5. ANALYTIC GEOMETRY
5.4.3 CGd3: Recognize the graph of y = (ax + b)/(cx + d) from equation
and vice versa.
Recognize the graph of y = (ax + b)/(cx + d) from its equation and vice versa.
Example: The graph of y = (x + 2)/(2x + 11) is
10
8
6
y
4
2
–15
–10
–5
x
0
–2
5
–4
–6
–8
–10
(1) Circle the equation that best describes this graph.
(a) y = (x + 1)/(x − 7) (b) y = (x − 1)/(x + 7)
(c)
y = (1 − x)/(x + 7)
(e)
y = (2 − x)/(x − 9)
(d) y = (1 − x)/(x − 9)
10
8
6
y
4
2
–20
–16
–12 x –8 –6 –4 –2
02
–2
4
6
8
–4
–6
–8
–10
The graph has two asymptotes: horizontal at y = −1 and vertical at x = −7. The horizontal asymptote for this kind of rational functions is the ratio of the coefficients infront
of the terms containing xis the ratio of the coefficients in front of x in the numerator and
the denominator. Example: for the function (a) it is
y=
1
= 1.
1
The vertical asymptote shows where the function is not continuous, in other words,
where the denominator becomes zero. In our case it is at x = −7.
148
CHAPTER 5. ANALYTIC GEOMETRY
5.4. CONICS
We look for an expression which has
(1) zero denominator at x = −7 and
(2) gest close to (−1) as x increases without bound, or decreases without bound.
The horizontal asymptotes is found in the following way: when x $= 0, we can factor out
x and reduce the fraction. When x increases or decreases without bound a fraction of a
type number/x gets very small. For example, (e):
2−x
x(2/x − 1)
−1 + 2/x
−1
=
=
≈
= −1 when x increases/decreases without bound.
x−9
x(1 − 9/x)
1 − 9/x
1
For all the functions, the horizontal and vertical asymptotes are shown in the following
table:
function
(a)
(b)
(c)
(d)
(e)
horizontal
y = +1
y = +1
y = −1
y = −1
y = −1
vertical
x = +7
x = −7
x = −7
x = +9
x = +9
The functions with property (1) are (b) and (c). For the property (2) these are (c), (d) and
(e). Therefore, the answer is (c) because it is the only function with both properties.
Answer: the function that best describes the graph is y = (1 − x)/(x + 7).
(2) Circle the equation that best describes this graph.
(a)
y = (2x + 1)/(3x − 7)
(b)
y = (2x − 1)/(3x + 7)
(c)
y = (2 − x)/(3x + 7)
(d)
y = (1 − x)/(x − 9)
(e)
y = (2 − x)/(x − 9)
20
16
12
y
8
4
–20
–12 –8 –4 0
–4
4
–8
–12
–16
–20
149
8 x12 16 20
5.4. CONICS
CHAPTER 5. ANALYTIC GEOMETRY
Looking at the graph, the first expression has horizontal asymptote y = 1 and vertical
asymptote about x = 2. The actual asymptotes of the functions (a) - (e) are given in the
table below.
function
(a)
(b)
(c)
(d)
(e)
horizontal
y = 2/3
y = 2/3
y = −1/3
y = −1
y = −1
vertical
x = 7/3
x = −1/7
x = −3/7
x=9
x=9
Looking at the table we see that the actual asymptotes are: horizontal y = 2/3 and
vertical x = 7/3. Therefore, the answer is function (a).
Answer: the function that best describes the graph isy = (2x + 1)/(3x − 7).
(3) Match the function on the left with its graph on the right.
1.
y = (x + 1)/(x + 2).
a.
20
16
12
y
8
4
–12 –8 –4 0
–4
4
8 12 16 20 24 28 32 36 40
x
–8
–12
–16
–20
2.
y = (2x + 6)/(x − 12).
b.
30
25
20
15y
10
5
–50
3.
y = (−2x + 6)/(x + 12).
–40
–30 x –20
0 10 15 20 25 30
–5
–10
–15
–20
–25
–30
–10
c.
10
8
6
y
4
2
–12
–8 –6 –4 –2
x
02
–2
–4
–6
–8
–10
150
4
6
8
CHAPTER 5. ANALYTIC GEOMETRY
5.4. CONICS
An easy way to find out is to spot the vertical asymptote and match it with the function
which has it. The functions have vertical asymptotes :
(1) x = 2,
(2) x = 12,
(3) x = −12.
For the graphs vertical asymptotes are as follows:
(a) x = 12,
(b) x = −12,
(c) x = 2.
The answer is shown below.
1.
y = (x + 1)/(x + 2).
c.
10
8
6
y
4
2
–12
–8 –6 –4 –2
x
02
–2
4
6
8
–4
–6
–8
–10
2.
y = (2x + 6)/(x − 12).
a.
20
16
12
y
8
4
–12 –8 –4 0
–4
4
8 12 16 20 24 28 32 36 40
x
–8
–12
–16
–20
3.
y = (−2x + 6)/(x + 12).
b.
30
25
20
15y
10
5
–50
151
–40
–30 x –20
–10
0 10 15 20 25 30
–5
–10
–15
–20
–25
–30
5.4. CONICS
CHAPTER 5. ANALYTIC GEOMETRY
5.4.4 CGd4: Solve verbal problems.
Give an equation for each conic. Put your equation in a standard form.
Example: Give the standard equation for a parabola with focus at (0, 1) and with directrix at y = −1. Answer: y = x2 /4.
1. Give the standard equation for a parabola with focus at (1, 2) and with directrix
y = −7.
Recall that the parabola is the set of all points which are at equal distance form a
fixed point, called focus, and a line called directrix. Let P (x, y) is a point on the
parabola, F (1, 2) is the focus and H(x, −7) is the point on the directrix, corresponding to the distance for point P .
.F(1,2)
FP= FA2 + AP 2
= (x-1)2 + (y-2) 2 .
PH=y-(-7).
. A(1,y) . P(x,y)
. H(x,-7)
We express the distances F P from triangle F P A and P H:
#
√
F P = P A2 + AF 2 =
(x − 1)2 + (y − 2)2 ,
P H = y − (−7) = y + 7.
We know that F P = P H so we equal the right sides of the above equations. To
reveal the function, we solve for y:
#
(x − 1)2 + (y − 2)2
#
( (x − 1)2 + (y − 2)2 )2
(x − 1)2 + y 2 − 4y + 4
(x − 1)2 + 4 − 49
= y + 7,
= (y + 7)2 ,
= y 2 + 14y + 49,
= 14y + 4y,
5
1
(x − 1)2 − .
y =
18
2
Answer: the standard form of the parabola with focus (1, 2) and directrix y = −7 is
y = 1/18(x − 1)2 − 5/2.
152
CHAPTER 5. ANALYTIC GEOMETRY
5.4. CONICS
2. Give the standard equation for an ellipse with foci at (1, 2) and (3, 2) and major axis
of length 10.
B
e
b a
F2
c F1
a
C
F1 (3,2)
F2 (1,2)
A
C(2,2)
F1 ,F2 - foci
C -center of the ellipse
The standard equation of the elipse with center C(c, d) and halfaxis a and b is
(x − c)2 (y − d)2
+
= 1.
a2
b2
We know that for an ellipse with major axes 2a, minor axes 2b and the distnace
between the foci 2c, the excentricity of the ellipse, denoted by e is
c
e= .
a
The relationship between the lenght of the axis and the distance between the center
and one of the foci and the excentricity is
&
b2
e = 1 − 2.
a
We equal the right sides of the above identities:
&
c
b2
= 1 − 2,
a
a
b2
c2
=
1
−
,
a2
a2
c 2 = a2 − b 2 ,
a2 = c 2 + b 2 .
153
5.4. CONICS
CHAPTER 5. ANALYTIC GEOMETRY
For an ellipse, the center is the midpoint of the line segment with endpoints foci.
Knowing the foci, we found the center of the ellipse C(2, 2). We know that the
major axes is 10, thus a = 10/2 = 5 units. The distance c between the center of the
ellipse and one of the foci is
c = 1 unit.
We find the minor axis b:
b=
√
a2 − c 2 =
√
52 − 12 =
√
24.
The major axis has half-lenght 5, minor axis has half-length
the ellipse is C(2, 2).
√
24 and the center of
Answer: the ellipse with foci (1, 2) and (3, 2) and major axis of length 10 is
2
(x−2)2
√
+ ((y−2)
= 1.
52
24)2
3. Give the standard equation for a hyperbola with foci at (0, 2) and (4, 2) and vertices
at (1, 2) and (3, 2).
The standard equation of a hyperbola is
x2 y 2
− 2 = 1,
a2
b
where a < b and the axes of symmetry are coordinate axes. To shift these axes the
equation is
(x − h)2 (y − k)2
−
= 1.
a2
b2
The axes of symmetry are x = h and y = k. The vertices and foci lye on the horizontal axis of symmetry. The vertical axis of symmetry is the line cutting in half
the distance between foci and the distance between vertices. Thus, in this particular
case, x = 2 is vertical axis of symmetry and y = 2 is the horizontal one.
154
CHAPTER 5. ANALYTIC GEOMETRY
5.4. CONICS
F1
F2
Answer: the standard equation for a hyperbola with foci at (0, 2) and (4, 2) and
vertices at (1,
√2) and (3, 2) is
x2 /12 − y 2 /( 3)2 = 1.
4. Give the standard equation for an ellipse with foci at (1, 2) and (3, 2) and minor axis
of length 10.
We use the same equation as in #2. This time we have minor axis of length 10. Thus,
we know c = 1 and b = 5. The foci are at the same place so that the center is still
C(2, 2). Using the same relation c2 + b2 = a2 , we find a
√
a = c 2 + b2 ,
√
√
a = 12 + 52 = 26.
Answer: the equation of an ellipse with foci (1, 2) and (3, 2) and minor axis of length
10 is
√
(x − 2)2 /( 26)2 + (y − 2)2 /52 = 1.
5. Give the standard equation for a parabola with focus at (1, 2) and directrix x = 12.
We reason as in #1. Let F (1, 2) is the focus of the parabola, P (x, y) is a point on the
parabola and H is the point giving the distance from the point P to the line x = 12.
155
5.4. CONICS
CHAPTER 5. ANALYTIC GEOMETRY
P(x,y)
H(12,y)
F(1,2)
We write
#
(x − 1)2 + (y − 2)2
(x − 1)2 + (y − 2)2
x2 − 2x + 1 + (y − 2)2
−22x
=
=
=
=
|x − 12|,
(x − 12)2 ,
x2 − 24x + 144,
(y − 2)2 − 143,
1
13
x = − (y − 2)2 + .
22
2
Answer: the standard equation for a parabola with focus at (1, 2) and directrix x =
12 is
x = −1/22(y − 2)2 + 13/2.
156