Transcript Thinking Mathematically by Robert Blitzer
Complex Numbers
Complex Numbers
The imaginary unit i is defined as 1
i i
2 1
Example
81
i
9 9
i
1 81
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 2
i
6
i
8 5 4
i
Example
( 2
i
)( 1 3
i
) 2 6
i
i
3
i
2 2 5
i
3 5 5
i
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
1 2 2
i
1
i
2
i
2 1 2 2 2
i
2
i
2 1 1 2
i i i
1 1 1
i
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 4
i
3
i
9 12
i
2 12
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
x
2
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.
• 1.
2.
3.
4.
5.
Graphing Parabolas With Equations in Standard Form
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
Given equation
f
(
x
)
a
-2 2(
h
3
x
3)
2
k
8 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.
Step 2
3 and
k
Text Example cont.
Find the vertex.
The vertex of the parabola is at (
h, k
)
.
Because
h
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
.
2(
x
0 3)
2
8 2(
x
3)
2
8
Find x-intercepts, setting f (x) equal to zero.
Solve for x. Add 2(x the equation.
3) 2 to both sides of
(
x
3)
2
4 (
x
3) 2
Divide both sides by 2.
Apply the square root method.
x
3
x
2 or
x
3 2
Express as two separate equations.
1 or
x
5
Add 3 to both sides in each equation.
The
x
-intercepts are 1 and 5. The parabola passes through (1, 0) and (5, 0).
Step 4
Text Example cont.
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) The
y
-intercept is –10. The parabola passes through (0, 10). 8 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
2
a
,
f
b
2
a
Example
Graph the quadratic function
f
(
x
)
x
2
6
x
.
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
)
x
2
6
x
.
•
Step 3
2
.
• 0 =
x
2 Find the x-intercepts.
6
x
2
a
1,
b
6,
c
Replace
f
(
x
) with 0 in
f
(
x
) 2
x
b
b
2 4
ac
2
a
x
2
6
x
6 6 2 4( 1)( 2) 2( 1) 6 36 2 8 6 2 3 7 28 6 2 7 2
Example
• • Graph the quadratic function
f
(
x
)
Step 4
x
2
6
x
.
Find the y-intercept.
Replace
x
with 0 in
f
(
x
)
x
2
6
x
2
.
f
(0) 0
2
6 • 0 2 The
y
-intercept is –2. The parabola passes through (0, 2). 10
Step 5 Graph the parabola.
8 6 4 2 -10 -8 -6 -4 -2 -2 -4 -6 -8 -10 2 4 6 8 10
• 1.
2.
Minimum and Maximum: Quadratic Functions
Consider f(x) = ax 2 + bx +c.
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)).
1.
2.
3.
4.
5.
Strategy for Solving Problems Involving Maximizing or Minimizing Quadratic Functions 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) = ax 2 + 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
1 be a nonnegative integer and let
a n
,
a n
,…,
a
2 ,
a
1 ,
a
0 , be real numbers with
a n
0. The function defined by
f
(
x
)
a n x n
a n-1 x n
-1 …
a
2
x
2
a
1
x
a
0 is called a
polynomial function of x of degree n .
The number
a n
, 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 sharp corners. By
continuous,
we mean that the graph has no Smooth rounded corner
y y
Smooth rounded corner breaks and can be drawn without lifting your pencil from the rectangular coordinate system. These ideas are illustrated in the figure.
Smooth rounded corner
x
Smooth rounded corner
x
The Leading Coefficient Test
As
x
increases or decreases without bound, the graph of the polynomial function
f
(
x
)
a n x n
a n
-1
x n
-1
a n
-2
x n
-2 …
a
1
x
a
0 (
a n
0) eventually rises or falls. In particular,
1.
For
n
odd:
a n
> 0
a n
< 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
)
a n x n
a n
-1
x n
-1
a n
-2
x n
-2 …
a
1
x
a
0 (
a n
0) eventually rises or falls. In particular,
1.
For
n
even:
a n
> 0
a n
< 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
)
x
3
3
x
2
x
3.
Rises right
y
(
Solution
n
Because the degree is odd 3) and the leading coefficient, 1, is positive, the graph falls to the left and rises to the right, as shown in the figure.
x
Falls left
Text Example
Find all zeros of
f
(
x
)
x
4 4
x
3 4
x
2 .
Solution
x
2 0
x
0 We find the zeros of
f
by setting
f
(
x
) equal to 0.
x
4 4
x
3 4
x
2 0
We now have a polynomial equation.
x
4 4
x
3 4
x
2 0
Multiply both sides by
1. (optional step)
x
2 (
x
2 4
x
4) 0
Factor out x 2 .
x
2 (
x
2) 2 0 or (
x
2) 2 0
x
2
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
1.
2.
Graphing a Polynomial Function
f
(
x
)
a n x n
a n
-1
x n
-1
a n
-2
x n
-2
a
1
x
a
0 (
a n
Use the Leading Coefficient Test to determine the 0) graph's end behavior. 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
a.
(
x
), then: If
k
is even, the graph touches the
x
-axis at
r
around. and turns
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
computing
f
(0). equal to 0 and
4.
5.
Graphing a Polynomial Function
f
(
x
)
a n x n
a n
-1
x n
-1
a n
-2
x n
-2
a
1
x
a
0 (
a n
0) Use symmetry, if applicable, to help draw the graph: a. y -axis symmetry:
f
(
x
)
b.
Origin symmetry:
f
(
x
)
f f
( (
x 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
)
x
4
2
x
2
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
)
x
4
2
x
2
1 .
(
Solution Step 2 Find the x-intercepts (zeros of the function) by setting f (x)
0.
x
4 2
x
2 1 0 (
x
2 1) (
x
2 1) 0
Factor.
(
x
1)(
x
1)(
x
1)(
x
1) 0
Factor completely.
x
1)
x
2 0 1 (
x
1) 2 (
x
1) 2 0
Express the factoring in more compact notation.
or (
x
1) 2 0
x
1
Set each factor equal to zero.
Solve for x.
Text Example cont.
Graph:
f
(
x
)
x
4 2
x
2 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 left
1 1
Rises right
x
Text Example cont.
Graph:
f
(
x
)
x
4
2
x
2
1 .
Solution Step 3 Find the y-intercept.
Replace
x
with 0 in
f
(
x
)
x
4
x
1
.
f
(0) 0 4 2 • 0 2 1 1 There is a
y
-intercept at 1, so the graph passes through (0, 1).
y
Rises left
1
Rises right
x
1 1
Text Example cont.
Graph:
f
(
x
)
x
4
2
x
2
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
)
x
4 2
x
2 1
Replace x with
x.
f
(
x
) (
x
) 4 2(
x
) 2 1
x
4 2
x
2 1 Because
f
(
x
)
f
(
x
)
,
the graph of following figure shows the graph.
f is
symmetric with respect to the
y
-axis. The
Text Example cont.
Graph:
f
(
x
)
x
4
2
x
2
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 ne
x
t term in the original dividend and write it ne
x
t to the remainder to form a new dividend.
• Use this new e
x
pression 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 e
x
ponent on a variable in the remainder) is less than the degree of the divisor.
Divide 4 – 5
x
–
Text Example
x
2 + 6
x
3 by 3
x
– 2.
Solution
We begin by writing the divisor and dividend in descending powers of
x
. Next, we consider how many times 3x divides into 6x 3 .
Multiply.
3
x
– 2 6
x
3 6
x
3 2
x
2 –
x
2 – 4
x
2 3
x
2 – 5
x
– 5
x
+ 4 Divide: 6
x
3 /3
x
= 2
x
2 .
Multiply: 2
x
2 (3
x
– 2) = 6
x
3 Subtract 6
x
3 – 4
x
2 from 6
x
3 – 4
x
2 .
–
x
2 and bring down –5
x
.
Now we divide 3
x
2 by 3
x
to obtain
x
, multiply
x
and the divisor, and subtract.
Multiply.
3
x
– 2 6
x
3 6
x
3 2
x
2 –
x
2 – 4
x
2 +
x
– 5
x
3
x
2 – 5
x
3
x
2 – 2
x
-3
x
+ 4 + 4 Divide: 3
x
2 /3
x
=
x
.
Multiply:
x
(3
x
Subtract 3
x
2 – 2) = 3
x
2 – 2
x
from 3
x
2 – 2
x
.
– 5
x
and bring down 4.
Text Example cont.
Divide 4 – 5
x
–
x
2 + 6
x
3 by 3
x
– 2.
Solution
Now we divide –3 divisor, and subtract.
x
by 3
x
to obtain –1, multiply –1 and the Multiply.
3
x
– 2 6
x
3 6
x
3 2
x
2 –
x
2 – 4
x
2 +
x
– 1 – 5
x
+ 4 3
x
2 – 5
x
3
x
2 – 2
x
-3
x
+ 4 -3
x
+ 2 2 Divide: -3
x
/3
x
= -1.
Multiply: -1(3
x
– 2) = -3
x
+ 2.
Subtract -3
x
+ 2 from -3
x
+ 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: 3
x
3
x
2
x
2 2 4
x
3
x
3 3
Example cont.
x
2 3
x
3 3
x
3 2
x
2 4
x
3
Example cont.
x
2 3
x
3
3x
3
x
3 2
x
2 4
x
3
Example cont.
x
2 3
x
3
3x
3
x
3 2
x
2 4
x
3
3x
3
+ 9x
2
+ 9x
Example cont.
x
2 3
x
3
3x
3
x
3 2
x
2
3x
3
+ 9x
2
+ 9x
4
x
3
-11x
2
- 5x - 3
Example cont.
x
2
3x -11
3
x
3 3
x
3 2
x
2 4
-(3x
3
+ 9x
2
+ 9x)
x
3
-11x
2
- 5x - 3
Example cont.
3x -11
x
2 3
x
3 3
x
3 2
x
2
3x
3
+ 9x
2
+ 9x
4
x
3
-11x
2
- 5x - 3 -11x
2
- 33x - 33
Example cont.
3x -11
x
2 3
x
3 3
x
3 2
x
2 4
3x
3
+ 9x
2
+ 9x
x
3
-11x
2
- 5x - 3 -11x
2
- 33x - 33 28x+30
Example cont.
3
x
3
x
2
x
2 2 4
x
3
x
3 3 ( 3
x
11 )
x
28
x
30 2 3
x
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 x 3 + 4x 2 – 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 Multiply by 3.
1
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.
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. 3 1 1 4 3 7 -5 Add.
5 3 3 1 1 1 1 4 3 7 4 3 7 -5 21 16 5 Add.
Multiply by 3.
-5 21 16 5 48 53 Add.
Multiply by 3.
1
x
2 + 7
x
+ 16 +
x
53 – 3
Text Example
Use synthetic division to divide 5
x
3 + 6
x
+ 8 by
x
+ 2.
Solution
The divisor must be in the form
x
– (-2). This means that
c
– c. Thus, we write
x
+ 2 as = -2. Writing a 0 coefficient for the missing
x
2 -
x
term in the dividend, we can express the division as follows:
x
– (-2) 5
x
3 + 0
x
2 + 6
x
+ 8 .
Now we are ready to set up the problem so that we can use synthetic division. This is c in x-(-2).
Use the coefficients of the dividend in descending powers of x.
-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.
1. Bring down 5.
-2 5 0 6 8 5
4. Multiply:
-2(-10) = 20.
-2 5 5 0 -10 6 -10 20 8
6. Multiply:
-2(26) = -52.
-2 5 5 0 6 -10 20 -10 26 8 -52
2. Multiply:
-2(5) = -10.
-2 5 0 -10 6 5 8
3. Add:
0 + (-10) = -10.
-2 5 5 0 -10 -10 6 Add.
8
5. Add:
6 + 20 = 26.
-2 5 5 0 -10 6 20 -10 26 8 Add.
7. Add:
8 + (-52) = -44.
-2 5 5 0 -10 6 20 -10 26 8 -52 -44 Add.
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 5
x
2 . -2 5 5 0 6 -10 20 -10 26 8 -52 -44 Thus, 5
x
2 – 10
x
+ 26 –
x
+ 2 5
x
3 + 0
x
2 + 6
x
+ 8 44
x
+ 2
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
• If
x
– c is a factor of – c is a factor of
f
(
x
), then
f f
(
x
).
( c) = 0.
Text Example
Solve the equation 2 3 is a zero of
f
x
3 (
x
) = 2
x
3 – 3
x
2 – 3
x
2 – 11
x
– 11
x
+ 6 = 0 given that + 6.
Solution
We are given that
f
(3) = 0. The Factor Theorem tells us that
x
is a factor of
f
(
x
). We’ll use synthetic division to divide
f
(
x
) by
x
– 3. – 3 3 2 2 -3 6 3 -11 9 -2 6 -6 0
x
– 3 2
x
2 + 3
x
– 2 2
x
3 – 3
x
2 – 11
x
+ 6 Equivalently, 2
x
3 – 3
x
2 – 11
x
+ 6 = (
x
– 3)(2
x
2 + 3
x
– 2)
Text Example cont.
Solution
Now we can solve the polynomial equation. 2
x
3 – 3
x
2 – 11
x
+ 6 = 0 (
x
– 3)(2
x
2 + 3
x
– 2) = 0 (
x
– 3)(2
x
– 1)(
x
+ 2) = 0
x
– 3 = 0 or 2
x
– 1 = 0 or
x
+ 2 = 0 This is the given equation.
Factor using the result from the synthetic division.
Factor the trinomial.
Set each factor equal to 0.
x
= 3
x
= 1/2
x
= -2 Solve for x.
The solution set is {-2, 1/2 , 3}.
Zeros of Polynomial Functions
The Rational Zero Theorem
• If
f
(
x
)
a n x n
a n-1 x n
-1 …
a
1
x
a
0 has
integer
coefficients and
p/q
(where
p/q
is reduced) is a rational zero, then
p
is a factor of the constant term
a
0 and
q
is a factor of the leading coefficient
a n
.
Example
Find all of the possible real, rational roots of f(x) = 2x 3 -3x 2 +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) = x 3 +12x 2 +21x+10 p/q = 1, 2, 5, 10 f(1) = 44 f(-1) = 0 Divide out -1 to get x 2 +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
Solve:
Text Example
x
4
6
x
2
8
x
+ 24 0.
Solution
The graph of
f
(
x
)
x
4
6
x
2
8
x
+ 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 • 0 6 8 24 2 4 4 1 2 2 12 24 0
The zero remainder indicates that 2 is a root of x 4
6x 2
8x + 24
0.
x -intercept: 2
Solve:
Text Example cont.
x
4
6
x
2
8
x
+ 24 0.
Solution
Now we can rewrite the given equation in factored form.
x
4 6
x
2 8
x
+ 24 0
This is the given equation.
(
x
– 2)(
x
3 2
x
2 2
x
12) 0
This is the result obtained from the synthetic division.
x
– 2 0 or
x
3 2
x
2 2
x
12 0
Set each factor equal to zero.
Solve:
Text Example cont.
x
4
6
x
2
8
x
+ 24 0.
Solution
We can use the same approach to look for rational roots of the polynomial equation
x
3 2
x
2 2
x
12 = 0, listing all possible rational roots. However, take a second look at the figure of the graph of
x
4
6
x
2
8
x
+ 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
x
3 2
x
2 2
x
12 = 0, confirmed by the following synthetic division.
These are the coefficients of x 3
2x 2
2x
12 = 0.
• 1 2 2 12 2 8 12 1 4 6 0
x -intercept: 2 The zero remainder indicates that 2 is a root of x 3
2x 2
2x
12 = 0.
Solve:
Text Example cont.
x
4
6
x
2
8
x
+ 24 0.
Solution
Now we can solve the original equation as follows.
x
4 6
x
2 8
x
+ 24 0
This is the given equation.
(
x
– 2)(
x
3 2
x
2 2
x
12) 0
This was obtained from the first synthetic division.
(
x
– 2)(
x
– 2)(
x
2 4
x
6) 0
This was obtained from the second synthetic division.
x
– 2 0 or
x
– 2 0 or
x
2 4
x
6 0
x
2
x
2
x
2 4
x
6 0
Set each factor equal to zero.
Solve.
Solve:
Text Example cont.
x
4
6
x
2
8
x
+ 24 0.
Solution
We can use the quadratic formula to solve
x
2 4
x
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
)
a
n
x
n
a
n
1
x
n
1
…
a
2
x
2
polynomial with real coefficients.
1.
a
1
x
The number of positive real zeros of
f a
0
be a 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
)
x
3 2
x
2 5
x
+ 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
)
x
3 2
x
2 5
x
+ 4
This is the given polynomial function.
Replace x with
x.
f
(
x
) (
x
) 3 2(
x
) 2 5(
x
) 4
x
3 2
x
2 5
x
+ 4
•Text Example cont.
Determine the possible number of positive and negative real zeros of
f
(
x
)
x
3 2
x
2 5
x
+ 4.
Solution
Now count the sign changes.
f
(
x
)
x
3 2
x
2 5
x
+ 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.
f
(
x
)
x
7
x
3
Symbol
x
a
x
a
x
x
Arrow Notation
Meaning
x
approaches
a
from the right.
x
approaches
a
from the left.
x
approaches infinity; that is,
x
increases without bound.
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
f (x)
as x
a
approaches
a
.
f (x)
as x
a
y y
f f
a
x = a
x a x
x = a
Thus,
f
(
x
) right. or
f
(
x
) as
x
approaches
a
from either the left or the
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)
y
as x
a
y
f (x)
as x
a
x = a x = a
x x a a
f f Thus,
f
(
x
) right. or
f
(
x
) as
x
approaches
a
from either the left or the
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 approaches
b
as
x
increases or decreases without bound.
f
if
f
(
x
)
y
y = b
y y
y = b
f
x
f
y = b
f
x x
f (x)
b as x
f (x)
b as x
f (x)
b as x
f
Locating Horizontal Asymptotes
Let f be the rational function given by (
x
)
a n x n
...
a
1
x
a
0
b m x m
a n
1
x n
1
b m
1
x m
1 ...
b
1
x
b
0 ,
a n
0,
b m
0 The degree of the numerator is n. The degree of the denominator is m.
1.
If n 2. 3. If n=m, the line y = a n /b m is the horizontal asymptote of the graph of f. If n>m,t he graph of f has no horizontal asymptote. • 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 f ( x ) f ( x ): y -axis symmetry has symmetry. f ( x ) f ( x ): origin symmetry 2. 3. 4. 5. Find the y -intercept (if there is one) by evaluating f (0) . Find the x -intercepts (if there are any) by solving the equation p ( x ) Find any vertical asymptote(s) by solving the equation q ( x ) 0. 0. Find the horizontal asymptote (if there is one) using the rule for 6. 7. determining the horizontal asymptote of a rational function. Plot at least one point between and beyond each x -intercept and vertical asymptote. Use the information obtained previously to graph the function between and beyond the vertical asymptotes. f ( x ) 5 2 x x 3 10 f ( x ) 5 2 x x 3 10 • the graph has no symmetry f ( x ) 5 2 x x 3 10 • the graph has no symmetry • The y-intercept is (0,-3/10) f ( x ) 5 2 x x 3 10 • the graph has no symmetry • The y-intercept is (0,-3/10) • The x-intercept is (3/2, 0) f ( x ) 5 2 x x 3 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 f ( x ) 5 2 x x 3 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 f ( x ) 5 2 x x 3 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) f ( x ) 5 2 x x 3 10 -10 -8 -6 -4 -2 -2 -4 -6 -8 -10 10 4 2 8 6 2 4 6 8 10 Find the slant asymptote of Solution f ( x ) x 2 x 4 x 3 5 Because the degree of the numerator, 2, is exactly one more than the degree of the denominator, 1, the graph of into x 2 4 x 5. f has a slant asymptote. To find the equation of the slant asymptote, divide x 3 The equation of the slant asymptote is y x 1. Using our strategy for -2 -1 -1 -2 -3 2 1 6 5 4 3 Slant asymptote: y = x - 1 1 2 3 4 5 6 7 8 Vertical asymptote: x = 3 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 where f is a polynomial function. )> 0 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 negative, then f(x)<0 for all numbers, x, in the interval. f is • Write the solution set; selecting the interval(s) that satisfy the given inequality. Solve and graph the solution set on a real number line: 2 x 2 – 3 x > 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. 2 x 2 – 3 x 2 x 2 – 3 x – 2 > 2 – 2 – 2 > 0 Step 2 Solve the related quadratic equation . Replace the inequality sign with an equal sign. Thus, we will solve. 2 x x (2 x = -1/2 2 x 2 – 3 x + 1)( x – 2 = 0 – 2) = 0 + 1 = 0 or or x – 2 = 0 x = 2 This is the related quadratic equation. Factor. Set each factor equal to 0. Solve for x . The boundary points are –1/2 and 2. Solve and graph the solution set on a real number line: 2 x 2 – 3 x > 2. Solution Step 3 Locate the boundary points on a number line . The number line with the boundary points is shown as follows: -1/2 2 -5 -4 -3 -2 -1 0 1 2 3 4 5 x 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 ). Solve and graph the solution set on a real number line: 2 x 2 – 3 x > 2. Solution Step 4 Take one representative number within each test interval and substitute that number into the original inequality . Test Interval (-ºº, -1/2] [-1/2, 2] Representative Number -1 0 Substitute into 2 x∆ – 3 x > 2 2(-1) 2 – 3(-1) > 2 5 > 2, True 2(0) 2 – 3(0) > 2 0 > 2, False [2, ºº) 3 2(3) 2 – 3(3) > 2 9 > 2, True Conclusion ( -ºº, -1/2] belongs to the solution set. [-1/2, 2] does not belong to the solution set. [2, ºº ) belongs to the solution set. Solve and graph the solution set on a real number line: 2 x 2 – 3 x > 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 ) -5 -4 -3 -2 -1 0 1 2 2 ( 3 4 5 x Solve and graph the solution set: x 1 x 3 2 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 x 3 2 This is the giv en inequality. x 1 x 3 2 0 Subt ract 2 from both sides, obtaining 0 on the right. x 1 x 3 2( x x 3) 3 0 The least common d enomin ator is x + 3. Express 2 in terms of this denomin ator. x 1 2( x x 3 3) 0 Subt ract rational expressions. x 1 2 x 6 x 3 0 x 5 x 3 0 Apply the distributive property. Simpli fy. Solve and graph the solution set: x x 1 3 2 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. Step 3 Solve and graph the solution set: x 1 x 3 2 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). Solve and graph the solution set: x 1 x 3 2 Step 4 Take one representative number within each test interval and substitute that number into the original equality . Test Interval Representative Number Substitute into ( x ) x x 5 3 Conclusion (-ºº, -5) (-5,-3) (-3, ºº) -6 -4 0 f ( 6) ( 6) 5 6 3 1/ 3 f ( 4) ( 4) 4 3 5 1 f (0) 5/ 3 0 0 5 3 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) Solve and graph the solution set: x 1 x 3 2 Step 5 The solution set are the intervals that produced a true statement . Our analysis shows that the solution set is [-5, -3). • 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 + v 0 t + s 0 • where s is the height in feet, v 0 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 s 0 is the original height (initial height) of the object in feet. 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 – -16 t 2 + 80 t – 64 > 0 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. 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 t -16 t 2 = 1 + 80 t t > 64 = 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. 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 Test Interval (-ºº, 1) Representative Number 0 (1, 4) (4, ºº) 2 5 Substitute into ( x – 1)( x – 4) < 0 ( 0 – 1)( 0 – 4) < 0 4 < 0, False ( 2 – 1)( 2 – 4) < 0 -2 < 0, True ( 5 – 1)( 5 – 4) < 0 4 < 0, False Conclusion ( -ºº, 1) does not belong to the solution set. (1, 4) belongs to the solution set. (4, ºº ) does not belong to the solution set. The object will be above 64 feet between 1 and 4 seconds. • • 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. 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 in step 1. into the equation 4. Use the equation from step 3 to answer the problem's question. 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. We know that y expressed as y kx. varies directly as x is 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 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 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 20 P Replace k, the constant of variation, with 20. 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 G 20 P and solve for G . in G = 20 P 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. • • 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. • • 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. • Describe in words the variation shown by the given equation: H kT Q Solution: H varies directly as T and inversely as Q • 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 ." 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 kilogramsStrategy for Graphing a Rational Function
Sketch the graph of
Text Example
Polynomial and Rational Inequalities
Definition of a Polynomial Inequality
Example
Example cont.
Example cont.
Example cont.
Text Example
Text Example cont.
Text Example cont.
Text Example cont.
Text Example cont.
The Position Formula for a Free-Falling Object Near Earth’s Surface
Example
Example cont.
Example cont.
Modeling Using Variation
Direct Variation
Solving Variation Problems
Text Example
Text Example cont.
Text Example cont.
Text Example cont.
Direct Variation with Powers
Inverse Variation
Example
Joint Variation
Example