Download Thinking Mathematically by Robert Blitzer

Complex Numbers
Complex Numbers
The imaginary unit i is defined as
1  i
i  1
2
Example
 81   1 81
 i  9  9i
Complex Numbers
• The set of all numbers in the form a+bi with
real numbers a and b, and i, the imaginary
unit, is called the set of complex numbers.
The real number a is called the real part,
and the real number b is called the
imaginary part, of the complex number
a+bi.
Equality of Complex Numbers
• a+bi = c+di if and only if a = c and b = d
Adding and Subtracting Complex Numbers
• (a+bi) + (c+di) = (a+b) + (c+d)i
• (a+bi) - (c+di) = (a-c) + (b-d)i
Text Example
Perform the indicated operation, writing the
result in standard form.
(-5 + 7i) - (-11 - 6i)
Combine the real and imaginary parts:
(5-(-11)) + (7-(-6))i = (-5+11) + (7+6)i =
6+13i
Example
3  2i  6i  8
 5  4i
Example
(2  i )(1  3i )
 2  6i  i  3i
 2  5i  3
 5  5i
2
Conjugate of a Complex Number
• The complex conjugate of the number a+bi
is a-bi, and the conjugate of a-bi is a+bi.
Example
2
2 1 i


1 i 1 i 1 i
2  2i 2  2i


2
1 i
11
2  2i

 1 i
2
Principal Square Root of a
Negative Number
For any positive real number b, the principal
square root of the negative number -b is
defined by
(-b) = ib
Example
 16   9
 4i  3i
 12i  12
2
Quadratic Functions
Graphs of Quadratic Functions
The graph of any quadratic function is called a parabola.
Parabolas are shaped like cups, as shown in the graph below. If
the coefficient of x2 is positive, the parabola opens upward;
otherwise, the parabola opens downward. The vertex (or turning
point) is the minimum or maximum point.
The Standard Form of a
Quadratic Function
• The quadratic function
• f (x)  a(x  h)2  k,
a0
• is in standard form. The graph of f is a parabola
whose vertex is the point (h, k). The parabola is
symmetric to the line x  h. If a > 0, the parabola
opens upward; if a < 0, the parabola opens
downward.
Graphing Parabolas With
Equations in Standard Form
•
1.
2.
3.
4.
5.
To graph f (x)  a(x  h)2  k:
Determine whether the parabola opens upward or
downward. If a > 0, it opens upward. If a < 0, it opens
downward.
Determine the vertex of the parabola. The vertex is (h,
k).
Find any x-intercepts by replacing f (x) with 0. Solve
the resulting quadratic equation for x.
Find the y-intercept by replacing x with zero.
Plot the intercepts and vertex. Connect these points with
a smooth curve that is shaped like a cup.
Text Example
• Graph the quadratic function f (x)  2(x  3)2  8.
Solution We can graph this function by following the steps in the preceding
box. We begin by identifying values for a, h, and k.
Standard form
f (x)  a(x  h)2  k
a  -2
h3
k8
Given equation f (x)  2(x  3)2  8
Step 1 Determine how the parabola opens. Note that a, the coefficient of
x 2, is -2. Thus, a < 0; this negative value tells us that the parabola opens
downward.
Text Example cont.
Step 2 Find the vertex. The vertex of the parabola is at (h, k). Because h 
3 and k  8, the parabola has its vertex at (3, 8).
Step 3
Find the x-intercepts. Replace f(x) with 0 in f(x)  2(x  3)2  8.
0  2(x  3)2  8
Solve for x. Add 2(x  3)2 to both sides of
the equation.
2(x  3)  8
2
(x  3)2  4
(x  3)  2
x  3  2
x1
Find x-intercepts, setting f (x) equal to zero.
or
or
Divide both sides by 2.
Apply the square root method.
x32
x5
Express as two separate equations.
Add 3 to both sides in each equation.
The x-intercepts are 1 and 5. The parabola passes through (1, 0) and (5, 0).
Text Example cont.
Step 4
Find the y-intercept. Replace x with 0 in f(x)  2(x  3)2  8.
f(0)  2(0  3)2  8  2(3)2  8  2(9)  8  10
The y-intercept is –10. The parabola passes through (0, 10).
Step 5 Graph the parabola. With a vertex at (3, 8), x-intercepts at 1 and 5,
and a y-intercept at –10, the axis of symmetry is the vertical line whose
equation is x  3.
The Vertex of a Parabola Whose
Equation Is f (x)  ax 2  bx  c
• Consider the parabola defined by the
quadratic function
• f (x)  ax 2  bx  c. The parabola's vertex is
at
 b  b 
 , f  
 2a  2a 
Example
Graph the quadratic function f (x)  x  6x .
2
Solution:
Step 1 Determine how the parabola opens. Note that a, the
coefficient of x 2, is -1. Thus, a < 0; this negative value tells us that the
parabola opens downward.
Step 2 Find the vertex. We know the x-coordinate of the vertex is x =
-b/(2a). We identify a, b, and c to substitute the values into the equation
for the x-coordinate:
x = -b/(2a) = -6/2(-1)=3.
The x-coordinate of the vertex is 3. We substitute 3 for x in the equation of
the function to find the y-coordinate:
y=f(3) = -(3)^2+6(3)-2=-9+18-2=7, the parabola has its vertex at (3,7).
Example
Graph the quadratic function f (x)  x2  6x .
• Step 3 Find the x-intercepts. Replace f(x) with 0 in f(x)  x2  6x
 2.
a  1,b  6,c  2
• 0 = x2  6x  2
b  b 2  4ac
x
2a
6  6 2  4(1)(2)

2(1)
6  36  8

2
6  28 6  2 7

2
2
 3 7

Example
Graph the quadratic function f (x)  x  6x .
2
• Step 4 Find the y-intercept. Replace x with 0 in f(x)  x2  6x 
2.
•
f(0)  02  6 • 0  2  
 The y-intercept is –2. The parabola passes through
(0, 2).
10
8
 Step 5 Graph the parabola.
6
4
2
-10 -8 -6 -4 -2
2
-2
-4
-6
-8
-10
4
6
8 10
Minimum and Maximum:
Quadratic Functions
•
Consider f(x) = ax2 + bx +c.
1. If a > 0, then f has a minimum that occurs at
x = -b/(2a). This minimum value is f(-b/(2a)).
2. If a < 0, the f has a maximum that occurs at
x = -b/(2a). This maximum value is f(-b/(2a)).
Strategy for Solving Problems Involving
Maximizing or Minimizing Quadratic
Functions
1.
2.
3.
4.
5.
Read the problem carefully and decide which quantity is
to be maximized or minimized.
Use the conditions of the problem to express the quantity
as a function in one variable.
Rewrite the function in the form f(x) = ax2 + bx +c.
Calculate -b/(2a). If a > 0, then f has a minimum that
occurs at x = -b/(2a). This minimum value is f(-b/(2a)).
If a < 0, the f has a maximum that occurs at x = -b/(2a).
This maximum value is f(-b/(2a)).
Answer the question posed in the problem.
Polynomial
Functions and Their
Graphs
Definition of a Polynomial Function
Let n be a nonnegative integer and let an, an1,…, a2, a1, a0, be real numbers with an  0.
The function defined by
f (x)  anxn  an-1xn-1 … a2x2  a1x  a0
is called a polynomial function of x of
degree n. The number an, the coefficient of
the variable to the highest power, is called
the leading coefficient.
Smooth, Continuous Graphs
Two important features of the graphs of polynomial functions are that they are
smooth and continuous. By smooth, we mean that the graph contains only
rounded curves with no
y
y
Smooth
sharp corners. By
Smooth
rounded
rounded
corner
corner
continuous, we mean
that the graph has no
breaks and can be
drawn without lifting
x
x
your pencil from the
rectangular coordinate
system. These ideas are
Smooth
Smooth
rounded
rounded
illustrated in the figure.
corner
corner
The Leading Coefficient Test
As x increases or decreases without bound, the graph of the polynomial
function
f (x)  anxn  an-1xn-1  an-2xn-2 … a1x  a0 (an  0)
eventually rises or falls. In particular,
1. For n odd:
an > 0
an < 0
If the
leading
coefficient is
positive, the
graph falls to
the left and
rises to the
right.
Rises right
Falls left
If the leading
coefficient is
negative, the
graph rises
to the left
and falls to
the right.
Rises left
Falls right
The Leading Coefficient Test
As x increases or decreases without bound, the graph of the polynomial
function
f (x)  anxn  an-1xn-1  an-2xn-2 … a1x  a0 (an  0)
eventually rises or falls. In particular,
1. For n even:
an > 0
an < 0
If the
leading
coefficient is
positive, the
graph rises
to the left
and to the
right.
Rises right
Rises left
If the leading
coefficient is
negative, the
graph falls to
the left and
to the right.
Falls left
Falls right
Text Example
Use the Leading Coefficient Test to determine the end behavior of the graph of
Graph the quadratic function f(x)  x3  3x2  x  3.
y
Solution Because the degree is odd
(n  3) and the leading coefficient, 1,
is positive, the graph falls to the left
and rises to the right, as shown in the
figure.
Rises right
x
Falls left
Text Example
Find all zeros of f(x)  x4  4x3  4x2.
Solution
We find the zeros of f by setting f(x) equal to 0.
x4  4x3  4x2  0
x4  4x3  4x2  0
x2(x2  4x  4)  0
x2(x  2)2  0
x2  0
x0
or
(x  2)2  0
x2
We now have a polynomial equation.
Multiply both sides by 1. (optional step)
Factor out x2.
Factor completely.
Set each factor equal to zero.
Solve for x.
Multiplicity and x-Intercepts
If r is a zero of even multiplicity, then the
graph touches the x-axis and turns around at
r. If r is a zero of odd multiplicity, then the
graph crosses the x-axis at r. Regardless of
whether a zero is even or odd, graphs tend
to flatten out at zeros with multiplicity
greater than one.
Example
• Find the x-intercepts and multiplicity of
f(x) = 2(x+2)2(x-3)
Solution:
• x=-2 is a zero of multiplicity 2 or even
• x=3 is a zero of multiplicity 1 or odd
Graphing a Polynomial Function
f (x)  anxn  an-1xn-1  an-2xn-2    a1x  a0 (an  0)
1. Use the Leading Coefficient Test to determine the
graph's end behavior.
2. Find x-intercepts by setting f (x)  0 and solving the
resulting polynomial equation. If there is an x-intercept
at r as a result of (x  r)k in the complete factorization of
f (x), then:
a. If k is even, the graph touches the x-axis at r and turns
around.
b. If k is odd, the graph crosses the x-axis at r.
c. If k > 1, the graph flattens out at (r, 0).
3.
Find the y-intercept by setting x equal to 0 and
computing f (0).
Graphing a Polynomial Function
f (x)  anxn  an-1xn-1  an-2xn-2    a1x  a0 (an  0)
4.
5.
Use symmetry, if applicable, to help draw the graph:
a. y-axis symmetry: f (x)  f (x)
b. Origin symmetry: f (x)   f (x).
Use the fact that the maximum number of turning points
of the graph is n  1 to check whether it is drawn
correctly.
Text Example
Graph: f(x)  x4  2x2  1.
Solution
Step 1 Determine end behavior. Because the degree is even (n  4) and
the leading coefficient, 1, is positive, the graph rises to the left and the right:
y
Rises
left
Rises
right
x
Text Example cont.
Graph: f(x)  x4  2x2  1.
Solution
Step 2 Find the x-intercepts (zeros of the function) by setting f(x)  0.
x4  2x2  1  0
(x2  1)(x2  1)  0
(x  1)(x  1)(x  1)(x  1)  0
(x  1)2(x  1)2  0
(x  1)2  0
x  1
or
(x  1)2  0
x1
Factor.
Factor completely.
Express the factoring in more compact notation.
Set each factor equal to zero.
Solve for x.
Text Example cont.
Graph: f(x)  x4  2x2  1.
Solution
Step 2 We see that -1 and 1 are both repeated zeros with multiplicity 2.
Because of the even multiplicity, the graph touches the x-axis at 1 and 1 and
turns around. Furthermore, the graph tends to flatten out at these zeros with
multiplicity greater than one:
y
Rises
right
Rises
left
x
1
1
Text Example cont.
Graph: f(x)  x4  2x2  1.
Solution
Step 3 Find the y-intercept. Replace x with 0 in f(x)  x  4x  1.
f(0)  04  2 • 02  1  1
There is a y-intercept at 1, so the graph passes through (0, 1).
y
Rises
left
Rises
right
1
x
1
1
Text Example cont.
Graph: f(x)  x4  2x2  1.
Solution
Step 4 Use possible symmetry to help draw the graph. Our partial graph
suggests y-axis symmetry. Let's verify this by finding f(x).
f(x)  x4  2x2  1
Replace x with x.
f(x)  (x)4  2(x)2  1  x4  2x2  1
Because f(x)  f(x), the graph of f is symmetric with respect to the y-axis. The
following figure shows the graph.
Text Example cont.
Graph: f(x)  x4  2x2  1.
Solution
Step 5 Use the fact that the maximum
number of turning points of the graph
is n  1 to check whether it is drawn
correctly. Because n  4, the maximum
number of turning points is 4 - 1, or 3.
Because our graph has three turning
points, we have not violated the maximum
number possible.
y
x
Dividing
Polynomials
Long Division of Polynomials
• Arrange the terms of both the dividend and the divisor in
descending powers of any variable.
• Divide the first term in the dividend by the first term in the
divisor. The result is the first term of the quotient.
• Multiply every term in the divisor by the first term in the
quotient. Write the resulting product beneath the dividend
with like terms lined up.
• Subtract the product from the dividend.
• Bring down the next term in the original dividend and
write it next to the remainder to form a new dividend.
• Use this new expression as the dividend and repeat this
process until the remainder can no longer be divided. This
will occur when the degree of the remainder (the highest
exponent on a variable in the remainder) is less than the
degree of the divisor.
Text Example
Divide 4 – 5x – x2 + 6x3 by 3x – 2.
Solution We begin by writing the divisor and dividend in descending
powers of x. Next, we consider how many times 3x divides into 6x3.
Multiply.
3x – 2
2x2
6x3 – x2 – 5x + 4
6x3 – 4x2
3x2 – 5x
Divide: 6x3/3x = 2x2.
Multiply: 2x2(3x – 2) = 6x3 – 4x2.
Subtract 6x3 – 4x2 from 6x3 – x2
and bring down –5x.
Now we divide 3x2 by 3x to obtain x, multiply x and the divisor, and subtract.
Multiply.
3x – 2
2x2 + x
6x3 – x2 – 5x + 4
6x3 – 4x2
3x2 – 5x
3x2 – 2x
-3x + 4
Divide: 3x2/3x = x.
Multiply: x(3x – 2) = 3x2 – 2x.
Subtract 3x2 – 2x from 3x2 – 5x
and bring down 4.
Text Example cont.
Divide 4 – 5x – x2 + 6x3 by 3x – 2.
Solution Now we divide –3x by 3x to obtain –1, multiply –1 and the
divisor, and subtract.
Multiply.
3x – 2
2x2 + x
6x3 – x2 – 5x
6x3 – 4x2
3x2 – 5x
3x2 – 2x
-3x
-3x
– 1
+ 4
+ 4
+ 2
2
Divide: -3x/3x = -1.
Multiply: -1(3x – 2) = -3x + 2.
Subtract -3x + 2 from -3x + 4,
leaving a remainder of 2.
The Division Algorithm
If f (x) and d(x) are polynomials, with d(x) = 0, and the degree of d(x) is less
than or equal to the degree of f (x), then there exist unique polynomials q(x)
and r(x) such that
f (x)
=
d(x)
•
q(x)
+
r(x).
Dividend
Divisor
Quotient
Remainder
The remainder, r(x), equals 0 or its is of degree less than the degree of d(x).
If r(x) = 0, we say that d(x) divides evenly in to f (x) and that d(x) and q(x)
are factors of f (x).
Example
• Divide:
3x  2 x  4 x  3
2
x  3x  3
3
2
Example cont.
x  3x  3 3 x  2 x  4 x  3
2
3
2
Example cont.
3x
x  3x  3 3 x  2 x  4 x  3
2
3
2
Example cont.
3x
x  3x  3 3 x  2 x  4 x  3
2
3
2
3x3 + 9x2 + 9x
Example cont.
3x
x  3x  3 3 x  2 x  4 x  3
2
3
3
3x +
2
2
9x +
9x
-11x2 - 5x - 3
Example cont.
3x -11
x  3x  3 3 x  2 x  4 x  3
2
3
2
3
-(3x +
2
9x +
9x)
-11x2 - 5x - 3
Example cont.
3x -11
x  3x  3 3 x  2 x  4 x  3
2
3
3
3x +
2
2
9x +
9x
-11x2 - 5x - 3
-11x2 - 33x - 33
Example cont.
3x -11
x  3x  3 3 x  2 x  4 x  3
2
3
3
3x +
2
2
9x +
9x
-11x2 - 5x - 3
-11x2 - 33x - 33
28x+30
Example cont.
3x  2 x  4 x  3

2
x  3x  3
3
2
28 x  30
(3x  11)  2
x  3x  3
Synthetic Division
To divide a polynomial by x – c
Example
1. Arrange polynomials in descending powers,
with a 0 coefficient for any missing terms.
x – 3 x3 + 4x2 – 5x + 5
2. Write c for the divisor, x – c. To the right,
write the coefficients of the dividend.
3
1 4 -5 5
3. Write the leading coefficient of the dividend
on the bottom row.
3
1 4 -5 5
Bring down 1.
1
4. Multiply c (in this case, 3) times the value
just written on the bottom row. Write the
product in the next column in the 2nd row.
3
1 4 -5 5
3
1
Multiply by 3.
Synthetic Division
5. Add the values in this new column, writing
the sum in the bottom row.
6. Repeat this series of multiplications and
additions until all columns are filled in.
3
1
1
3
1
1
4
3
7
-5
Add.
4
3
7
-5
21
16
5
5
Add.
Multiply by 3.
3
7. Use the numbers in the last row to write the
quotient and remainder in fractional form.
The degree of the first term of the quotient is
one less than the degree of the first term of
the dividend. The final value in the row is the
remainder.
1
1
4
3
7
-5
21
16
5
48 Add.
53
Multiply by 3.
53
1x2 + 7x + 16 + x – 3
Text Example
Use synthetic division to divide 5x3 + 6x + 8 by x + 2.
Solution The divisor must be in the form x – c. Thus, we write x + 2 as x
– (-2). This means that c = -2. Writing a 0 coefficient for the missing
x2term in the dividend, we can express the division as follows:
x – (-2) 5x3 + 0x2 + 6x + 8 .
Now we are ready to set up the problem so that we can use synthetic division.
Use the coefficients of the dividend
in descending powers of x.
This is c in x-(-2).
-2
5
0
6
8
Text Example cont.
Solution We begin the synthetic division process by bringing down 5. This
is following by a series of multiplications and additions.
2. Multiply: -2(5) = -10.
1. Bring down 5.
-2
5
0
6
8
5
-2
5
0
6
-10
3. Add: 0 + (-10) = -10.
8
-2
5
5
4. Multiply: -2(-10) = 20.
5. Add: 6 + 20 = 26.
-2
-2
5
5
0
6
-10 20
-10
8
5
5
0
6
-10 20
-10 26
6. Multiply: -2(26) = -52.
7. Add: 8 + (-52) = -44.
-2
-2
5
5
0
6
-10 20
-10 26
8
-52
5
5
5
0
6
-10 20
-10 26
8
Add.
8
-52 Add.
-44
0
6
8
-10 Add.
-10
Text Example cont.
Solution
The numbers in the last row represent the coefficients of the quotient and the
remainder. The degree of the first term of the quotient is one less than that of
the dividend. Because the degree of the dividend is 3, the degree of the
quotient is 2. This means that the 5 in the last row represents 5x2.
-2
5
5
0
6
-10 20
-10 26
8
-52
-44
Thus,
44
5x2 – 10x + 26 –
x+2
x + 2 5x3 + 0x2 + 6x + 8
The Remainder Theorem
• If the polynomial f (x) is divided by x – c,
then the remainder is f (c).
The Factor Theorem
• Let f (x) be a polynomial.
• If f (c ) = 0, then x – c is a factor of f (x).
• If x – c is a factor of f (x), then f ( c) = 0.
Text Example
Solve the equation 2x3 – 3x2 – 11x + 6 = 0 given that
3 is a zero of f (x) = 2x3 – 3x2 – 11x + 6.
Solution We are given that f (3) = 0. The Factor Theorem tells us that x – 3
is a factor of f (x). We’ll use synthetic division to divide f (x) by x – 3.
3
2
2
-3
6
3
-11 6
9
-6
-2 0
2x2 + 3x – 2
x – 3 2x3 – 3x2 – 11x + 6
Equivalently,
2x3 – 3x2 – 11x + 6 = (x – 3)(2x2 + 3x – 2)
Text Example cont.
Solution
Now we can solve the polynomial equation.
2x3 – 3x2 – 11x + 6 = 0
This is the given equation.
(x – 3)(2x2 + 3x – 2) = 0
Factor using the result from the synthetic
division.
(x – 3)(2x – 1)(x + 2) = 0
Factor the trinomial.
x – 3 = 0 or 2x – 1 = 0 or x + 2 = 0
Set each factor equal to 0.
x=3
x = 1/2
The solution set is {-2, 1/2 , 3}.
x = -2
Solve for x.
Zeros of Polynomial
Functions
The Rational Zero Theorem
• If f (x)  anxn  an-1xn-1 … a1x  a0 has
integer coefficients and p/q (where p/q is
reduced) is a rational zero, then p is a factor
of the constant term a0 and q is a factor of
the leading coefficient an.
Example
Find all of the possible real, rational roots of
f(x) = 2x3-3x2+5
Solution:
p is a factor of 5 = 1, 5
q is a factor of 2 = 1, 2
p/q = 1, 1/2, 5, 5/2
Properties of Polynomial
Equations
• If a polynomial equation is of degree n, then
counting multiple roots separately, the
equation has n roots
• If a+bi is a root of the equation, then a-bi is
also a root.
Example
Find all zeros of f(x) = x3+12x2+21x+10
p/q = 1, 2, 5, 10
f(1) = 44
f(-1) = 0
Divide out -1 to get x2+11x-10
Use the quadratic formula to find the last 2
zeros. x=-11.844 and .844
The solutions are -1, -11.844, and .844
Text Example
Solve:
x4  6x2  8x + 24  0.
Solution The graph of f(x)  x4  6x2  8x + 24 is shown the figure below.
Because the x-intercept is 2, we will test 2 by synthetic division and show that
it is a root of the given equation.
2
•
1
0 6 8 24
2
4 4 24
2 2 12 0
The zero remainder
indicates that 2 is a root
of x4  6x2  8x + 24  0.
x-intercept: 2
Text Example cont.
Solve:
Solution
x4  6x2  8x + 24  0.
Now we can rewrite the given equation in factored form.
x4  6x2  8x + 24  0
(x – 2)(x3  2x2  2x  12)  0
x–20
or
x3  2x2  2x  12  0
This is the given equation.
This is the result obtained from the
synthetic division.
Set each factor equal to zero.
Text Example cont.
Solve:
x4  6x2  8x + 24  0.
Solution We can use the same approach to look for rational roots of the
polynomial equation x3  2x2  2x  12 = 0, listing all possible rational roots.
However, take a second look at the figure of the graph of x4  6x2  8x + 24 
0. Because the graph turns around at 2, this means that 2 is a root of even
multiplicity. Thus, 2 must also be a root of x3  2x2  2x  12 = 0, confirmed
by the following synthetic division.
These are the coefficients
of x3  2x2  2x  12 = 0.
•
1
1
x-intercept: 2
2
2
4
2 12
8 12
6
0
The zero remainder
indicates that 2 is a root
of x3  2x2  2x  12 = 0.
Text Example cont.
Solve:
Solution
x4  6x2  8x + 24  0.
Now we can solve the original equation as follows.
x4  6x2  8x + 24  0
(x – 2)(x3  2x2  2x  12)  0
(x – 2)(x – 2)(x2  4x  6)  0
x – 2  0 or x – 2  0 or x2  4x  6  0
x2
x2
x2  4x  6  0
This is the given equation.
This was obtained from the first
synthetic division.
This was obtained from the second
synthetic division.
Set each factor equal to zero.
Solve.
Text Example cont.
Solve:
Solution
x4  6x2  8x + 24  0.
We can use the quadratic formula to solve x2  4x  6  0.
The solution set of the original equation is: {2, -2 - i2, -2+i2}
Descartes’s Rule of Signs
• If f (x)  anxn  an1xn1  …  a2x2  a1x  a0 be a
polynomial with real coefficients.
• 1. The number of positive real zeros of f is either equal to
the number of sign changes of f (x) or is less than that
number by an even integer. If there is only one variation in
sign, there is exactly one positive real zero.
• 2. The number of negative real zeros of f is either equal
to the number
of sign changes of f (x) or is less than
that number by an even integer. If f (x) has only one
variation in sign, then f has exactly one negative real zero.
Text Example
Determine the possible number of positive and negative real zeros of
f(x)  x3  2x2  5x + 4.
Solution
1. To find possibilities for positive real zeros, count the number of sign
changes in the equation for f(x). Because all the terms are positive, there
are no variations in sign. Thus, there are no positive real zeros.
2. To find possibilities for negative real zeros, count the number of sign
changes in the equation for f(x). We obtain this equation by replacing x
with x in the given function.
f(x)  x3  2x2  5x + 4
Replace x with x.
f(x)  (x)3  2(x)2  5(x)  4
 x3  2x2  5x + 4
This is the given polynomial function.
•Text Example cont.
Determine the possible number of positive and negative real zeros of
f(x)  x3  2x2  5x + 4.
Solution
Now count the sign changes.
f(x)  x3  2x2  5x + 4
1
2
3
There are three variations in sign. The number of negative real zeros of f is
either equal to the number of sign changes, 3, or is less than this number by an
even integer. This means that there are either 3 negative real zeros or 3  2  1
negative real zero.
Rational Functions
and Their Graphs
Example
• Find the Domain of
this Function.
• Solution:
• The domain of this
function is the set of
all real numbers not
equal to 3.
x7
f ( x) 
x 3
Arrow Notation
Symbol
Meaning
xa
xa
x
x approaches a from the right.
x approaches a from the left.
x approaches infinity; that is, x
increases without bound.
x
x approaches negative infinity; that is, x
decreases without bound.
Definition of a Vertical Asymptote
The line x  a is a vertical asymptote of the graph of a function f if f(x)
increases or decreases without bound as x approaches a.
f (x)   as x  a 
f (x)   as x  a 
y
y
f
f
x
a
x=a
a
x
x=a
Thus, f (x)   or f(x)    as x approaches a from either the left or the
right.
Definition of a Vertical Asymptote
The line x  a is a vertical asymptote of the graph of a function f if f(x)
increases or decreases without bound as x approaches a.
f (x)    as x  a 
f (x)   as x  a 
y
y
x=a
x=a
x
a
f
a
x
f
Thus, f(x)   or f(x)    as x approaches a from either the left or the
right.
Locating Vertical Asymptotes
• If f(x) = p(x) / q(x) is a rational function in
which p(x) and q(x) have no common
factors and a is a zero of q(x), the
denominator, then x = a is a vertical a
vertical asymptote of the graph of f.
Definition of a Horizontal
Asymptote
The line y = b is a horizontal asymptote of the graph of a function f if f(x)
approaches b as x increases or decreases without bound.
y
y
y
y=b
y=b
f
f
x
f
y=b
x
x
f(x)  b as x  
f(x)  b as x  
f (x)  b as x 
Locating Horizontal Asymptotes
Let f be then rationaln1function given by
an x  an 1 x  ...  a1 x  a0
f (x) 
, an  0,bm  0
m
m 1
bm x  bm1 x  ...  b1 x  b0
The degree of the numerator is n. The degree of the
denominator is m.
1. If n<m, the x-axis, or y=0, is the horizontal
asymptote of the graph of f.
2. If n=m, the line y = an/bm is the horizontal
asymptote of the graph of f.
3. If n>m,t he graph of f has no horizontal
asymptote.
Strategy for Graphing a Rational Function
•
Suppose that f(x) = p(x) / q(x),
where p(x) and q(x) are polynomial functions with no common factors.
1. Determine whether the graph of f has symmetry.
f (x)  f (x): y-axis symmetry
f (x)  f (x): origin symmetry
2. Find the y-intercept (if there is one) by evaluating f (0).
3. Find the x-intercepts (if there are any) by solving the equation p(x)  0.
4.
Find any vertical asymptote(s) by solving the equation q (x)  0.
5.
Find the horizontal asymptote (if there is one) using the rule for
determining the horizontal asymptote of a rational function.
6. Plot at least one point between and beyond each x-intercept and
vertical asymptote.
7. Use the information obtained previously to graph the function between
and beyond the vertical asymptotes.
Sketch the graph of
2x  3
f ( x) 
5 x  10
2x  3
f ( x) 
5 x  10
• the graph has no symmetry
2x  3
f ( x) 
5 x  10
• the graph has no symmetry
• The y-intercept is (0,-3/10)
2x  3
f ( x) 
5 x  10
• the graph has no symmetry
• The y-intercept is (0,-3/10)
• The x-intercept is (3/2, 0)
2x  3
f ( x) 
5 x  10
•
•
•
•
the graph has no symmetry
The y-intercept is (0,-3/10)
The x-intercept is (3/2, 0)
The vertical asymptote is x = -2
2x  3
f ( x) 
5 x  10
•
•
•
•
•
the graph has no symmetry
The y-intercept is (0,-3/10)
The x-intercept is (3/2, 0)
The vertical asymptote is x = -2
The horizontal asymptote is y = 2/5
2x  3
f ( x) 
5 x  10
•
•
•
•
•
•
the graph has no symmetry
The y-intercept is (0,-3/10)
The x-intercept is (3/2, 0)
The vertical asymptote is x = -2
The horizontal asymptote is y = 2/5
Test points include (-3, 9/5), (0, -3/10), (2,
1/20)
2x  3
f ( x) 
5 x  10
10
8
6
4
2
-10 -8 -6 -4 -2
2
-2
-4
-6
-8
-10
4
6
8 10
Text Example
Find the slant asymptote of
x 2  4x  5
f (x) 
x 3
Solution Because the degree of the numerator, 2, is exactly one more
than the degree of the denominator, 1, the graph of f has a slant
asymptote. To find the equation of the slant asymptote, divide x  3
into x2  4x  5.
The equation of the slant asymptote is y  x  1. Using our strategy for
7
graphing rational
functions, the graph is shown.
6
5
Slant asymptote:
y=x-1
4
3
2
1
-2
-1
1
-1
-2
-3
2
3
4
5
6
7
8
Vertical asymptote:
x=3
Polynomial and
Rational Inequalities
Definition of a Polynomial
Inequality
A polynomial inequality is any inequality
that can be put in one of the forms
f(x) < 0, f(x) > 0, f(x) < 0, or f(x)> 0
where f is a polynomial function.
Procedure for Solving Polynomial Inequalities
• Express the inequality in the standard form
f(x) < 0 or
f(x) > 0.
• Find the zeros of f. The real zeros are the boundary points.
• Locate these boundary points on a number line, thereby
dividing the number line into intervals.
• Choose one representative number within each interval and
evaluate f at that number. If the value of f is positive, then
f(x)>0 for all numbers, x, in the interval. If the value of f is
negative, then f(x)<0 for all numbers, x, in the interval.
• Write the solution set; selecting the interval(s) that satisfy the
given inequality.
Example
Solve and graph the solution set on a real number line:
2x2 – 3x > 2.
Solution
Step 1
Write the inequality in standard form. We can write by
subtracting 2 from both sides to get zero on the right.
2x2 – 3x – 2 > 2 – 2
2x2 – 3x – 2 > 0
Step 2
Solve the related quadratic equation. Replace the inequality sign
with an equal sign. Thus, we will solve.
2x2 – 3x – 2 =
(2x + 1)(x – 2) =
2x + 1 = 0 or x – 2 =
x = -1/2
or
x =
The boundary points are –1/2 and 2.
0
0
0
2
This is the related quadratic equation.
Factor.
Set each factor equal to 0.
Solve for x.
Example cont.
Solve and graph the solution set on a real number line:
2x2 – 3x > 2.
Solution
Locate the boundary points on a number line. The
number line with the boundary points is shown as follows:
Step 3
-1/2
2
x
-5 -4 -3 -2 -1 0
1 2
3 4 5
The boundary points divide the number line into three test
intervals. Including the boundary points (because of the given
greater than or equal to sign), the intervals are (-ºº, -1/2], [-1/2, 2],
[2, -ºº).
Example cont.
Solve and graph the solution set on a real number line:
2x2 – 3x > 2.
Solution
Step 4
Take one representative number within each test interval and
substitute that number into the original inequality.
Test Interval
Representative
Number
Substitute into
2x∆ – 3x > 2
Conclusion
(-ºº, -1/2]
-1
2(-1) 2 – 3(-1) > 2
5 > 2, True
(-ºº, -1/2] belongs
to the solution set.
[-1/2, 2]
0
2(0) 2 – 3(0) > 2
0 > 2, False
[2, ºº)
3
2(3)2 – 3(3) > 2
9 > 2, True
[-1/2, 2] does not
belong to the
solution set.
[2, ºº) belongs to
the solution set.
Example cont.
Solve and graph the solution set on a real number line:
2x2 – 3x > 2.
Solution
Step 5
The solution set are the intervals that produced a true
statement. Our analysis shows that the solution set is
(-ºº, -1/2] or [2, ºº).
The graph of the solution set on a number line is shown as follows:
-1/2
2
)
(
-5 -4 -3 -2 -1 0
1 2
x
3 4 5
Text Example
Solve and graph the solution set:
x 1
2
x3
Solution
Step 1
Express the inequality so that one side is zero and the other side
is a single quotient. We subtract 2 from both sides to obtain zero on the right.
x 1
 2 This is the given inequality.
x3
x 1
2 0
x3
Subtract 2 from both sides, obtaining 0 on the right.
x 1 2(x  3)

 0 The least common denominator is x + 3. Express 2
x3
x3
in terms of this denominator.
x  1  2(x  3)
0
x3
x  1  2x  6
0
x3
x  5
0
x3
Subtract rational expressions.
Apply the distributive property.
Simpli fy.
Text Example cont.
Solve and graph the solution set:
x 1
2
x3
Solution
Step 2
Find boundary points by setting the numerator and the
denominator equal to zero.
-x - 5
x = -5
= 0
x+3 = 0
x = -3
Set the numerator and denominator equal to 0.
These are the values that make the previous
quotient zero or undefined.
Solve for x.
The boundary points are -5 and -3. Because equality is included in the given
less-than-or-equal-to symbol, we include the value of x that causes the
quotient to be zero. Thus, -5 is included in the solution set. By contrast, we do
not include 1 in the solution set because -3 makes the denominator zero.
Text Example cont.
Solve and graph the solution set:
Step 3
x 1
2
x3
Locate boundary points on a number line.
The boundary points divide the number line into three test intervals, namely
(-ºº, -5], [-5, -3), (-3, ºº).
Text Example cont.
x 1
2
x3
Solve and graph the solution set:
Step 4
Take one representative number within each test interval and
substitute that number into the original equality.
Test
Interval
Representative
Number
(-ºº, -5) -6
(-5,-3)
-4
f (x) 
Substitute into
f (6) 
(6)  5
6  3
 1/ 3
f (4) 
(4)  5
4  3
1
(-3, ºº)
0
f (0) 
 5/ 3
0  5
03
x  5
x  3 Conclusion
f(x )< 0 for all x in
(-ºº, -5)
f(x) > 0 for all x in
(-5,-3)
f(x) < 0 for all x in
(-3, ºº)
Text Example cont.
Solve and graph the solution set:
x 1
2
x3
Step 5
The solution set are the intervals that produced a true
statement. Our analysis shows that the solution set is [-5, -3).
The Position Formula for a Free-Falling
Object Near Earth’s Surface
• An object that is falling or vertically projected into
the air has its height in feet above the ground
given by
•
s = -16 t 2 + v0 t + s0
• where s is the height in feet, v0 is the original
velocity (initial velocity) of the object in feet per
second, t is the time that the object is in motion in
seconds, and s0 is the original height (initial
height) of the object in feet.
Example
An object is propelled straight up from ground level with an initial velocity of
80 fps. Its height at time t is described by s = -16 t 2 + 80 t where the height, s,
is measured in feet and the time, t, is measured in seconds. In which time
interval will the object be more than 64 feet above the ground?
Solution
-16 t 2 + 80 t > 64
-16 t 2 + 80 t – 64 > 0
-16 t 2 + 80 t – 64 = 0
-16 (t – 1)(t – 4) = 0
(t – 1)(t – 4) = 0
t – 1 = 0 or t – 4 = 0
t=1
t=4
This is the inequality implied by the problem’s
question. We must find t.
Subtract 64 from both sides.
Solve the related quadratic equation.
Factor.
Divide each side by -16.
Set each factor equal to 0.
Solve for t. The boundary points are 1 and 4.
Example cont.
An object is propelled straight up from ground level with an initial velocity of
80 fps. Its height at time t is described by s = -16 t ∆ + 80 t where the height, s,
is measured in feet and the time, t, is measured in seconds. In which time
interval will the object be more than 64 feet above the ground?
Solution
-16 t 2 + 80 t > 64
t=1
t=4
This is the inequality implied by the problem’s
question. We must find t.
The boundary points are 1 and 4.
Since neither boundary point satisfy the
inequality, 1 and 4 are not part of the solution.
1
4
x
-5 -4 -3 -2 -1 0
1 2
3 4 5
With test intervals (-ºº, 1), (1, 4), and (4, ºº), we could use 0, 2, and 5 as test
points for our analysis.
Example cont.
An object is propelled straight up from ground level with an initial velocity of
80 fps. Its height at time t is described by s = -16 t 2 + 80 t where the height, s,
is measured in feet and the time, t, is measured in seconds. In which time
interval will the object be more than 64 feet above the ground?
Solution
Substitute into
(x – 1)(x – 4) < 0
Conclusion
(-ºº, 1) 0
(0 – 1)(0 – 4) < 0
4 < 0, False
(-ºº, 1) does not belong to
the solution set.
(1, 4)
2
(2 – 1)(2 – 4) < 0
-2 < 0, True
(1, 4) belongs to the
solution set.
(4, ºº)
5
(5 – 1)(5 – 4) < 0
4 < 0, False
(4, ºº) does not belong to
the solution set.
Test
Interval
Representative
Number
The object will be above 64 feet between 1 and 4 seconds.
Modeling Using
Variation
Direct Variation
• If a situation is described by an equation in
the form
• y  kx
• where k is a constant, we say that y varies
directly as x. The number k is called the
constant of variation.
Solving Variation Problems
1. Write an equation that describes the given
English statement.
2. Substitute the given pair of values into the
equation in step 1 and find the value of k.
3. Substitute the value of k into the equation
in step 1.
4. Use the equation from step 3 to answer the
problem's question.
Text Example
The amount of garbage, G, varies directly as the population, P.
Allegheny County, Pennsylvania, has a population of 1.3 million
and creates 26 million pounds of garbage each week. Find the
weekly garbage produced by New York City with a population of
7.3 million.
Solution
Step 1
Write an equation.
expressed as
We know that y varies directly as x is
y  kx.
By changing letters, we can write an equation that describes the following
English statement: Garbage production, G, varies directly as the population, P.
G  kP
Text Example cont.
The amount of garbage, G, varies directly as the population, P. Allegheny
County, Pennsylvania, has a population of 1.3 million and creates 26 million
pounds of garbage each week. Find the weekly garbage produced by New
York City with a population of 7.3 million.
Solution
Step 2 Use the given values to find k. Allegheny County has a population
of 1.3 million and creates 26 million pounds of garbage weekly. Substitute 26
for G and 1.3 for P in the direct variation equation. Then solve for k.
26 = 1.3k
26/1.3 = k
20 = k
Text Example cont.
The amount of garbage, G, varies directly as the population, P. Allegheny
County, Pennsylvania, has a population of 1.3 million and creates 26 million
pounds of garbage each week. Find the weekly garbage produced by New
York City with a population of 7.3 million.
Solution
Step 3 Substitute the value of k into the equation.
G  kP
Use the equation from step 1.
G  20P
Replace k, the constant of variation, with 20.
Text Example cont.
The amount of garbage, G, varies directly as the population, P. Allegheny
County, Pennsylvania, has a population of 1.3 million and creates 26 million
pounds of garbage each week. Find the weekly garbage produced by New
York City with a population of 7.3 million.
Solution
Step 4 Answer the problem's question. New York City has a population
of 7.3 million. To find its weekly garbage production, substitute 7.3 for P in
G  20P and solve for G.
G = 20P
Use the equation from step 3.
G = 20(7.3) Substitute 7.3 for P.
G = 146
The weekly garbage produced by New York City weighs approximately 146
million pounds.
Direct Variation with Powers
• y varies directly as the nth power of x if
there exists some nonzero constant k such
that
• y  kx n.
• We also say that y is directly proportional
to the nth power of x.
Inverse Variation
• If a situation is described by an equation in
the form
•
y  k/x
• where k is a constant, we say that y varies
inversely as x. The number k is called the
constant of variation.
Example
• Describe in words the variation shown by
the given equation:
kT
H
Q
Solution:
H varies directly as T and inversely as Q
Joint Variation
• Joint variation is a variation in which a
variable varies directly as the product of
two or more other variables. Thus, the
equation y  kxz is read "y varies jointly as x
and z."
Example
An object’s weight on the moon, M, varies directly as its
weight on Earth, E. A person who weighs 75 kilograms
on Earth weighs 12 kilograms on the moon. What is
the moon weight of a person who weighs 80 kilograms
on Earth?
Solution
M = kE
12 = k*75
k = .16
M = .16E
M = .16 * 80
M = 12.8 kilograms