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Some Lecture Notes and In-Class Examples for Pre-Calculus:
Section 2.7
Definition of a Quadratic Inequality
A quadratic inequality is any inequality that can be put in one of the forms
ax2 + bx + c < 0 ax2 + bx + c > 0
ax2 + bx + c < 0 ax2 + bx + c > 0
where a, b, and c are real numbers and a ≠ 0.
Procedure for Solving Quadratic Inequalities
•
Express the inequality in the standard form: ax2 + bx + c < 0
•
Solve the equation ax2 + bx + c = 0. The real solutions are the boundary points.
•
Locate these boundary points on a number line, thereby dividing the number line into test
intervals.
•
Choose one representative number within each test interval. If substituting that value into the
original inequality produces a true statement, then all real numbers in the test interval belong to
the solution set. If substituting that value into the original inequality produces a false statement,
then no real numbers in the test interval belong to the solution set.
•
Write the solution set; the interval(s) that produced a true statement.
or
ax2 + bx + c > 0.
Example 1
Solve and graph the solution set on a real number line: 2x2 – 3x > 2.
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
Step 2 Solve the related quadratic equation. Replace the inequality sign with an equal sign. Thus, we
will solve.
2x2 – 3x – 2
=
0
This is the related quadratic equation.
(2x + 1)(x – 2) =
0
Factor.
2x + 1 = 0
x–2
=
or
The boundary points are –1/2 and 2.
0
Set each factor equal to 0.
Step 3 Locate the boundary points on a number line. The number line with the boundary points is
shown as follows:
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 ( −∞, −1/ 2] , [ −1/ 2, 2] , & [ 2, ∞ ) .
Step 4 Take one representative number within each test interval and substitute that number into the
original inequality.
Step 5 The solution set are the intervals that produced a true statement. Our analysis shows that the
solution set is
( −∞, −1/ 2] ∪ [ 2, ∞ )
Example 2 Solve and graph the solution set on a real number line:
25 x 2 − 9 x < 0
Example 3 Solve and graph the solution set:
x +1
≤2
x+3
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.
Step 2 Find boundary points by setting the numerator and the denominator equal to zero.
Step 3 Locate boundary points on a number line.
Step 4 Take one representative number within each test interval and substitute that number into the
original equality.
Step 5 The solution set are the intervals that produced a true statement.
Example 4 Solve the inequality
(x
2
+ 1) ( x − 3)
x2 − 9
≥0
Example 5 Solve the inequality
( x + 5)
x 2 − 7 x + 12
≤0
Example 6 Solve the inequality
3
1
≥
5x + 1 x − 3
Section 3.5 Graphs of Functions
Definition of Even and Odd Functions
The function f is an even function if
f (-x) = f (x) for all x in the domain of f.
The right side of the equation of an even function does not change if x is replaced with -x.
The function f is an odd function if
f (-x) = -f (x) for all x in the domain of f.
Every term in the right side of the equation of an odd function changes sign if x is replaced by -x.
Example 7 Identify the following function as even, odd, or neither: f(x) = 3x2 - 2.
Solution:
We use the given function’s equation to find f(-x).
f(-x) = 3(-x) 2-2 = 3x2 - 2.
The right side of the equation of the given function did not change when we replaced x with -x. Because
f(-x) = f(x), f is an even function.
Even Functions and y-Axis Symmetry
--The graph of an even function in which f (-x) = f (x) is symmetric with respect to the y-axis.
Odd Functions and Origin Symmetry
-- The graph of an odd function in which f (-x) = - f (x) is symmetric with respect to the origin.
Example 8 Sketch the graph of the absolute value function, f ( x ) = x and:
a) Determine whether f is even or odd.
b) Sketch the graph.
c) Find the intervals on which the function is increasing or is decreasing.
Transformations of Functions
Vertical Shifts
Let f be a function and c a positive real number.
The graph of y = f (x) + c is the graph of y = f (x) shifted c units vertically upward.
The graph of y = f (x) – c is the graph of y = f (x) shifted c units vertically downward.
Example 9 Sketch the graphs of:
a.
f ( x ) = x2
b.
f ( x ) = x2 + 4
c.
f ( x ) = x2 − 4
Horizontal Shifts
Let f be a function and c a positive real number.
•
The graph of y = f (x + c) is the graph of y = f (x) shifted to the left c units.
•
The graph of y = f (x + c) is the graph of y = f (x) shifted to the right c units.
Example 10 Sketch the graphs of: f ( x ) = ( x − 4 )
2
and
f ( x ) = ( x + 2)
2
Reflection about the x-Axis
The graph of y = - f (x) is the graph of y = f (x) reflected about the x-axis.
Example 11 Graph f ( x ) = − x
Reflection about the y-Axis
The graph of y = f (-x) is the graph of y = f (x) reflected about the y-axis.
Example 12 Graph f ( x ) =
x and f ( x ) = − x
Stretching and Shrinking Graphs
Let f be a function and c a positive real number.
•
If c > 1, the graph of y = c f (x) is the graph of y = f (x) vertically stretched by multiplying each of
its y-coordinates by c.
•
If 0 < c < 1, the graph of y = c f (x) is the graph of y = f (x) vertically shrunk by multiplying each of
its y-coordinates by c.
Should say: g ( x ) = 2 x 2
Should say: g ( x ) =
Example 13 Graph f ( x ) = 4 x 2
Piecewise-defined functions
Example 14 Sketch the graph of:
1 2
x
4
2 x + 5

f ( x ) =  x2
 2

if x ≤ -1
if x < 1
if x ≥ 1
Example 14 Use the graph of f(x) = x3 to graph g(x) = (x+3)3 – 4
Section 3.6 Quadratic Functions
A function is quadratic if it has the form: f ( x ) = ax 2 + bx + c , where a, b, and c are real numbers.
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
To graph f (x) = a(x - h)2 + k:
1. Determine whether the parabola opens upward or downward. If a > 0, it opens upward. If a < 0,
it opens downward.
2. Determine the vertex of the parabola. The vertex is (h, k).
3. Find any x-intercepts by replacing f (x) with 0. Solve the resulting quadratic equation for x.
4. Find the y-intercept by replacing x with zero.
5. Plot the intercepts and vertex. Connect these points with a smooth curve that is shaped like a
cup.
Example 15 Graph the quadratic function f (x) = -2(x - 3)2 + 8.
Expressing a Quadratic Function as f (x) = a(x - h)2 + k
Example 16 Express f ( x ) = 3 x 2 + 24 x + 50 in f (x) = a(x - h)2 + k form
Example 17 Express f ( x ) = 2 x 2 − 6 x + 4 in f (x) = a(x - h)2 + k form
Example 18 Express f ( x ) = − x 2 − 2 x + 8 in f (x) = a(x - h)2 + k form
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
 2a ,

 −b  
f  
 2a  
Section 4.1 Polynomial Functions of Degree Greater Than Two
Definition of a Polynomial Function
Let n be a nonnegative integer and let an, an-1,…, 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 sharp
corners. By continuous, we mean that the graph has no breaks and can be drawn without lifting
your pencil from the rectangular coordinate system. These ideas are illustrated in the figure.
Note since all polynomial functions are continuous functions, which means that they can be
drawn without any breaks, we have an important theorem:
Intermediate Value Theorem for Polynomial Functions
If f is a polynomial function and f ( a ) ≠ f ( b ) for a < b , then f takes on every value between
f ( a ) and f ( b ) in the interval [ a, b] .
Example 19 Show that f ( x ) = x5 + 2 x 4 − 6 x3 + 2 x − 3 has a zero between 1 and 2.
The Leading Coefficient Test
Example 20 Use the Leading Coefficient Test to determine the end behavior of the graph of Graph the
quadratic function f (x) = x3 + 3x2 - x - 3.
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.
To graph polynomial functions by hand, we will need to know where the x-intercepts are located. These
points are referred to as zeros. To find zeros we simple set f ( x ) = 0 and solve.
Example 21 Find all zeros of f (x) = -x4 + 4x3 - 4x2
Once we are able to determine a function’s zeros, we can then sketch its graph with the help of the
properties we have talked about above. One helpful idea is to use the zeros to determine the intervals
upon which the function is negative and where it is positive.
Example 22 Sketch the graph of f ( x ) = x3 + x 2 − 4 x − 4
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 23 Find the x-intercepts and multiplicity of f(x) = 2(x+2)2(x-3)
Example 24 Sketch the graph of f ( x ) = x 4 − 4 x3 + 3 x 2
Graphing a Polynomial Function
f (x) = anxn + an-1xn-1 + an-2xn-2 + ¼ + a1x + a0 (an is not equal to 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).
Find the y-intercept by setting x equal to 0 and computing f (0).
3.
4. 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.
5.
Example 25 Graph: f (x) = x4 - 2x2 + 1.
Example 26 Use the graphing calculator to estimate the zeros of f ( x ) = x3 − 4.6 x 2 + 5.72 x − 0.656
Section 4.2 Properties of Division
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.
Example 27 Divide 4 – 5x – x2 + 6x3 by 3x – 2.
The Division Algorithm
Note: if d ( x ) = ( x − c ) and d is the remainder (which would have to be a constant since its degree
must be less than the divisor x – c) we can express f ( x ) in the following manner:
f ( x) = ( x − c) q ( x) + d
If we wanted to know what f ( c ) is, then we can use the above to derive the following:
f (c) = (c − c) q (c) + d = 0 + d = d
This leads to two theorems:
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.
Example 28 Use the remainder theorem to find f ( 2 ) for f ( x ) = x3 − 3 x 2 + x + 5 .
Example 29 Use the factor theorem to show x – 2 is a factor of f ( x ) = x3 − 4 x 2 + 3 x + 2 .
Example 30 Find a polynomial of degree three that has zeros 2, -1, and 3.
A faster method of polynomial division is possible. We call this method:
Synthetic Division
Example 31 Use synthetic division to divide 5x3 + 6x + 8 by x + 2.
Example 32 if f ( x ) = 3 x5 − 38 x3 + 5 x 2 − 1 , use synthetic division to find f ( 4 )
Example 33 Solve the equation 2x3 – 3x2 – 11x + 6 = 0 given that 3 is a zero of f (x) = 2x3 – 3x2 – 11x + 6.
Section 4.3 Zeros of Polynomials
The zeros of a polynomial f ( x ) are the solutions of the equation f ( x ) = 0 . Each zero is an xintercept of the graph of f.
Fundamental Theorem of Algebra: If a polynomial f ( x ) has positive degree and complex coefficients,
then f ( x ) has at least one complex zero.
***Remember that complex number are of the form: a + bi , so if b = 0 we actually have a real number.
Complete Factorization Theorem for Polynomials: if f ( x ) is a polynomial of degree n > 0, then there
exist n complex numbers c1 , c2 ,..., cn such that f ( x ) = a ( x − c1 )( x − c2 ) ⋅⋅⋅ ( x − cn ) where a is the
leading coefficient of f ( x ) .
Theorem on the Maximum Number of Zeros of a Polynomial: A polynomial of degree n > 0 has at most
n different complex zeros.
Example 34 Find a polynomial in factored form that has zeros -5, 2, and 4 and satisfies f ( 3) = −24 .
Multiplicity: If a factor ( x – c )appears m times in the factorization of the polynomial f ( x ) , c is a zero
of multiplicity m (or a root of multiplicity m).
Example 35 Find the zeros of f ( x ) =
1
3
2
( x − 2 )( x − 4 ) ( x − 1) and state their multiplicity and sketch
16
the graph.
Theorem on the Exact Number of Zeros of a Polynomial: If f ( x ) is a polynomial of degree n > 0 and if a
zero of multiplicity m is counted m times, then f ( x ) has precisely n zeros.
Example 36 Express f ( x ) = x5 − 4 x 4 + 13 x3 as a product of linear factors, and find the five zeros of
f ( x) .
Descartes’ 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.
Example 37 Determine the possible number of positive and negative real zeros of
f (x) = x3 + 2x2 + 5x + 4.
First Theorem on Bounds for Real Zeros of Polynomials: Suppose that f ( x ) is a polynomial with real
coefficients and a positive leading coefficient and that f ( x ) is divided synthetically by x – c.
1. If C > 0 and if all numbers in the third row of the division process are either positive or zero, then
c is an upper bound for the real zeros of f ( x ) .
2. If C < 0 and if all numbers in the third row of the division process are alternately positive and
negative (zero can be used as either), then c is a lower bound for the real zeros of f ( x ) .
Example 38 Find the upper and lower bounds for the real solutions of the equation f ( x ) = 0, where
f ( x ) = 2 x3 + 5x 2 − 8x − 7 .
Second Theorem on Bounds for Real Zeros of Polynomials: All of the real zeros of f ( x ) are in the
interval ( -M, M ) where
M=
max ( coefficients )
an
+1
Section 4.4 Complex and Rational Zeros of Polynomials
Theorem on Conjugate Pair Zeros of a Polynomial
If a polynomial f ( x ) of degree n > 1 has real coefficients and if z = a + bi with b ≠ 0 is a complex zero of
f ( x ) , then the conjugate z = a − bi is also a zero of f ( x ) .
Example 39 Find a polynomial f ( x ) with degree 4 that has real coefficients and zeros: 3 + 5i, -1 - i
A rule worth remembering:  x − ( a + bi )   x − ( a − bi )  = x 2 − 2ax + a 2 + b 2
Theorem on Expressing a Polynomial as a Product of Linear and Quadratic Factors
Every polynomial with real coefficients and positive degree n can be expressed as a product of linear and
quadratic polynomials with real coefficients such that the quadratic factors are irreducible over the set
of real numbers.
Example 40 Express x 5 − 4 x 3 + x 2 − 4 as a product of
a) linear and irreducible quadratic factors with real coefficients and
b) linear polynomials
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.
As an aid in listing the possible rational zeros, remember the following quotient:
Possible rational zeros =
± factors of a0
± factors of leading coefficient an
Example 41 Find all of the possible real, rational roots of f(x) = 2x3-3x2+5.
Example 42 Show that f ( x ) = 3 x3 − 4 x 2 + 7 x + 5 has no rational zeros.
Example 43 Find all rational solutions of f ( x ) = 6 x 4 + 5 x3 − 17 x 2 − 6 x
Example 44 Use the calculator to narrow the list of possible zeros and find all rational solutions of
f ( x ) = 3x 4 + 14 x3 + 14 x 2 − 8 x − 8
Section 4.5 Rational Functions
A function f is rational if f ( x ) =
g ( x)
h ( x)
, where g ( x ) and h ( x ) are polynomials. The domain of
rational functions is all real numbers except the zeros of the denominator h ( x ) .
Example 45 Find the Domain of f ( x) =
x+7
x−3
Arrow Notation
Symbol
Meaning
x→a+
x approaches a from the right.
x→a-
x approaches a from the left.
x→∞
x approaches infinity; that is, x increases without bound.
x→-∞
x approaches negative infinity; that is, x decreases without bound.
Question: What happens to the values of f ( x ) when x is close to a zero of the denominator?
Question: What can be said about the values of f ( x ) when x is very large positive or very large
negative?
Definition of a Vertical Asymptote
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
Locating Horizontal Asymptotes
Let f be the rational function given by
an x n + an −1 x n−1 + ... + 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, the graph of f has no horizontal asymptote (but it might have a slant asymptote if n = m
+1).
Example 46 Find the horizontal asymptotes for:
a.
f ( x) =
3x − 1
2
x − x−6
b. f ( x ) =
5x2 + 1
3x 2 − 4
c. f ( x ) =
2 x 4 − 3x 2 + 5
x2 + 1
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.
Example 47 Sketch the graph of f ( x ) =
2x − 3
5 x + 10
Example 48 Sketch the graph of f ( x ) =
3x + 4
2x − 5
Example 49 Sketch the graph of f ( x ) =
Example 50 Sketch the graph of f ( x ) =
(3 x + 4) ( x − 1)
(2 x − 5) ( x − 1)
x −1
x − x−6
2
Example 51 Sketch the graph and find the slant asymptote of
Section 5.2 Exponential Functions
f (x) =
x 2 − 4x − 5
x −3
EXPONENTIAL FUNCTION
A function f of the form:
f ( x) = ax,
a > 0 and a ≠ 1,
is called an exponential function with base a. The domain of the exponential function is ( −∞, ∞ ) .
Exponential Growth
If a > 1, then as x increases the function values increase exponentially which means the graph of the
function approaches positive infinity, and as x decreases the graph of the function approaches the x-axis
or y = 0 asymptotically .
Exponential Decay
If 0 < a < 1, then as x increases, the graph of the function approaches the x-axis asymptotically, and as x
decreases the graph approaches positive infinity.
Exponential functions are one-to-one:
This leads to two important ideas, namely:
1. If a x1 = a x2 , then x1 = x2
2. The exponential function has an inverse.
We can use the first property to help us solve exponential equations.
2
Example 52 Solve the equation: 9 x = 33 x+ 2
x
x
Example 53 Solve the equations: a. 25 = 125 b. 9 = 3
x +1
1
c.  
2
Graphing an Exponential Function with Base a > 1
Example 54 Graph the exponential function f ( x ) = 3x
Make a table of values and plot the points and draw a smooth curve.
6− x
=2
X
–3
–2
–1
0
1
2
3
y = 3x
1/27
1/9
1/3
1
3
9
27
1
Example 55 Graph the exponential function f ( x ) =  
2
x
Make a table of values and plot the points and draw a smooth curve.
X
–3
–2
–1
0
1
2
3
y = (1/2)x
8
4
2
1
1/2
1/4
1/8
Shifting the graph of an exponential equation
Example 56 Sketch the graph of f ( x ) = 3x − 2 and f ( x ) = 3x − 2
Example 57 Find an exponential function of the form f ( x ) = ba − x + c that has the horizontal
asymptote of y = 1, y intercept 4, and contains the point (1, 2.5).
Example 58 Sketch f ( x ) = 2 − x
2
Using the compound interest formula:
Interest A fee charged for borrowing a lender’s money is called the interest, denoted by I.
Principal The original amount of money borrowed is called the principal, or initial amount, denoted by P.
Time Suppose P dollars is borrowed. The borrower agrees to pay back the initial P dollars, plus the
interest, within a specified period. This period is called the time of the loan and is denoted by t.
Interest Rate The interest rate is the percent charged for the use of the principal for the given period.
The interest rate is expressed as a decimal and denoted by r. Unless stated otherwise, it is assumed to
be for one year; that is, r is an annual interest rate.
 r
The Compound Interest Formula: A = P 1 + 
 n
nt
A = amount after t years
P = principal
r = annual interest rate (expressed as a decimal)
n = number of times interest is compounded each year
t = number of years
Example 59 One thousand dollars is deposited in a bank that pays 5% interest compounded quarterly.
Find the future value A after three years. Would this amount be higher if we compounded it monthly
instead?
Bacterial Growth
Example 60 A technician for the French microbiologist Louis Pasteur noticed that a certain culture of
bacteria in milk doubles every hour. If the bacteria count B(t) is modeled by the
equation B ( t ) = 2000 × 2t , with t in hours, find
a. the initial number of bacteria,
b. the number of bacteria after 10 hours; and
c. the time when the number of bacteria will be 32,000.
Section 5.3 THE NATURAL EXPONENTIAL FUNCTION
Example 61 One hundred dollars is deposited in a bank that pays 5% annual interest. Find the future
value A after one year if the interest is compounded
(i)
Quarterly.
(ii)
Monthly.
(iii)
Daily.
(iv)
Hourly.
(v)
Minute by minute.
Notice the hourly compounding and the minute to minute compounding yields the same result to the
nearest penny. If we let n increase without bound in the formula for compound interest we are
compounding the interest continuously, which will yield a new equation.
Before we discuss this equation, let us first consider an important constant:
THE VALUE OF e
The number e, an irrational number, is sometimes called the Euler constant.
Mathematically speaking, e is the fixed number that the expression:
 1
 1 + 
h
h
Approaches as h gets larger and larger. The value of e to 15 places is e = 2.718281828459045.


To review: As n → ∞ 1 +
n
1
 → e ≈ 2.718281828459045
n
THE NATURAL EXPONENTIAL FUNCTION
The exponential function
f ( x ) = ex
with base e is so prevalent in the sciences that it is often referred to as the exponential function or the
natural exponential function.
CONTINUOUS COMPOUND-INTEREST FORMULA A = Pe rt
A = amount after t years
P = principal
r = annual interest rate (expressed as a decimal)
t = number of years
Example 62 Find the amount when a principal of 8300 dollars is invested at a 7.5% annual rate of
interest compounded continuously for eight years and three months.
Example 63 How much money did the government owe DeHaven’s descendants for 213 years on a
450,000-dollar loan at the interest rate of 6% compounded continuously?
A MODEL FOR EXPONENTIAL GROWTH OR DECAY A ( t ) = A0 ekt
A(t) = amount at time t
A0 = A(0), the initial amount
k = relative rate of growth (k > 0) or decay (k < 0)
t = time
Example 64 In the year 2000, the human population of the world was approximately 6 billion and the
annual rate of growth was about 2.1 percent. Using the model on the previous slide, estimate the
population of the world in the following years.
a. 2030
b. 1990
** The model predicts that the world had 4.86 billion people in 1990 (actual amount was 5.28 billion).
Example 65 Find the zeros of f ( x ) = − x 2 e − x + 2 xe − x
(e
Example 66 Simplify
x
2
− e− x ) − ( e x + e − x )
(e
x
+ e− x )
2
2
Section 5.4 Logarithmic Functions
DEFINITION OF THE LOGARITHMIC FUNCTION
For x > 0, a > 0, and a ≠ 1, y = log a x
if and only if
x = a y.
The function f (x) = loga x, is called the logarithmic function with base a. The logarithmic function is the
inverse function of the exponential function.
The two equations above are equivalent, so you may switch back and forth between the two forms
whenever it suits you. The form with the word log in it is called the logarithmic form, while the other
form is called the exponential form. To switch the form from log to exponential we drop the log and
switch the answer and the exponent (reverse the process to go the other way, exp to log). **Note the
base always remains the base!
Example 67 Write each exponential equation in logarithmic form.
4
3
a. 4 = 64
1
1
b.   =
 2  16
c. a −2 = 7
Example 68 Write each logarithmic equation in exponential form.
a. log 3 243 = 5
b. log 2 5 = x
What is the meaning of a log? It is an exponent. log a x is read as follows, “What power do you raise ‘a’
to in order to get x?”
Example 69 Find the value of each of the following logarithms.
a. log 5 25
b. log 2 16
c. log1 3 9
d. log 7 7
e. log 6 1
f. log 4
BASIC PROPERTIES OF LOGARITHMS
For any base a > 0, with a ≠ 1,
1. loga a = 1.
2. loga 1 = 0.
3. loga ax = x, for any real number x.
4. a log a x = x, for any x > 0.
Recall, that Exponential functions are one-to-one:
This leads to two important ideas, namely:
1
2
1. If a x1 = a x2 , then x1 = x2
2. The exponential function has an inverse.
The inverse of the exponential function is the logarithmic function, so …
Log functions are one-to-one:
This leads to two important ideas, namely:
1. If x1 = x2 , then log a x1 = log a x2
2. The exponential function is the inverse of the log function.
We will use the above idea to solve log equations. Please be aware that we can only solve equations
with at most one log on each side of the equation. If the equation is not of that form, we will need to
rewrite it so it does conform to that form.
Also, note that the domain for y = log a x is from ( 0, ∞ ) , so we must check that our answers are
elements of the set ( 0, ∞ ) .
Example 70 Solve log 6 ( 4 x − 5 ) = log 6 ( 2 x + 1)
Example 71 Solve log 4 ( 5 + x ) = 3
Sketching the Graph of Log Functions
DOMAIN OF LOGARITHMIC FUNCTION
Recall that the Domain of f (x) = ax is ( −∞, ∞ ) and the Range of f (x) = ax is ( 0, ∞ )
Since the logarithmic function is the inverse of the exponential function,
The Domain of f –1(x) = loga x is ( 0, ∞ )
Range of f –1(x) = loga x is ( −∞, ∞ )
Thus, the logarithms of 0 and negative numbers are not defined.
Example 72 Sketch the graph of y = log3 x.
Make a table of values.
X
y = log3 x
(x, y)
3–3 = 1/27
–3
(1/27, –3)
3–2 = 1/9
–2
(1/9, –2)
3–3 = 1/3
–1
(1/3, –1)
30 = 1
0
(1, 0)
31 = 3
1
(3, 1)
32 = 9
2
(9, 2)
Plot the ordered pairs and connect with a smooth curve to obtain the graph of y = log3 x.
The side by side comparison below shows the symmetry w.r.t. the line y = x that all functions and their
inverse share.
Note that the x-intercept is 1, and there is no y-intercept for the log graph. Also, the y-axis (x = 0) is the
vertical asymptote, and in general for y = log a x , the graph is Increasing if a > 1 and Decreasing if
0 < a < 1.
Example 73 Start with the graph of f (x) = log3 x and use transformations to sketch the graph of each
function. State the domain and range and the vertical asymptote for the graph of each function.
a. f ( x ) = log 3 x + 2
b. f ( x ) = log 3 ( x − 1)
c. f ( x ) = − log3 x
d. f ( x ) = log3 ( − x )
COMMON LOGARITHMS
The logarithm with base 10 is called the common logarithm and is denoted by omitting the base: log x =
log10 x. Thus, y = log x if and only if x = 10 y.
Applying the basic properties of logarithms
1. log 10 = 1.
2. log 1 = 0.
3. log 10x = x
4. 10log x = x
NATURAL LOGARITHMS
The logarithm with base e is called the natural logarithm and is denoted by ln x. That is, ln x = loge x.
Thus, y = ln x if and only if x = e y.
Applying the basic properties of logarithms
1. ln e = 1
2. ln 1 = 0
3. log ex = x
4. eln x = x
Example 74 Solve log x = 1.7959
NEWTON’S LAW OF COOLING
Newton’s Law of Cooling states that
T = Ts + (T0 − Ts ) e − kt , where T is the temperature of the object
at time t, Ts is the surrounding temperature, and T0 is the value of T at t = 0.
Example 75 The local McDonald’s franchise has discovered that when coffee is poured from a pot whose
contents are at 180ºF into a non-insulated pot in the store where the air temperature is 72ºF, after 1
minute the coffee has cooled to 165ºF. How long should the employees wait before pouring the coffee
from this non-insulated pot into cups to deliver it to customers at 125ºF?
GROWTH AND DECAY MODEL A = A0 e rt
A is the quantity after time t.
A0 is the initial (original) quantity (when t = 0).
r is the growth or decay rate per period.
t is the time elapsed from t = 0.
Example 76 In a large lake, one-fifth of the water is replaced by clean water each year. A chemical spill
deposits 60,000 cubic meters of soluble toxic waste into the lake.
a. How much of this toxin will be left in the lake after four years?
b. How long will it take for the toxic chemical in the lake to be reduced to 6000 cubic meters?
Solution
One-fifth (1/5) of the water in the lake is replaced by clean water every year, the decay rate for the toxin
is r = –1/5 and A0 = 60,000. So,
A = A0 ert
A = 60, 000e
 1
 − t
 5
where A is the amount of toxin (in cubic meters) after t years.
a. Substitute t = 4.
b. Substitute A = 6000 and solve for t.
Section 5.5 Properties of Logarithms
RULES OF LOGARITHMS
Let M, N, and a be positive real numbers with a ≠ 1, and let r be any real number.
1. Product Rule: log a ( MN ) = log a M + log a N
The logarithm of the product of two (or more) numbers is the sum of the logarithms of the
numbers.
M
2. Quotient Rule: log a 
N

 = log a M − log a N

The logarithm of the quotient of two (or more) numbers is the difference of the logarithms of
the numbers.
3. Power Rule:
log a M r = r log a M
The logarithm of a number to the power r is r times the logarithm of the number.
*Warning: log a ( M + N ) ≠ log a M + log a N Do not make this mistake!
Example 77 Write each expression in expanded form:
a. log 2
x 2 ( x − 1)
( 2 x + 1)
3
b. log c x 3 y 2 z 5
4
Example 78 Write each expression in condensed form:
a. log 3 x − log 4 y
c. 2 log 2 5 + log 2 9 − log 2 75
1
b. 2 ln x + ln ( x 2 + 1)
2
d.
1
ln x + ln ( x + 1) − ln ( x 2 + 1) 

3
Example 79 Solve* log 4 x + log 4 ( x + 1) = log 4 ( x − 1) + log 4 6
Example 80 Solve*:
a. log 2 ( x − 3) + log 2 ( x − 4 ) = 1
b. log 3 ( x − 1) − log 3 ( x − 3) = 1
*Please be aware that we can only solve equations with at most one log on each side of the equation. If
the equation is not of that form, we will need to rewrite it so it does conform to that form. Also, note
that the domain for y = log a x is from ( 0, ∞ ) , so we must check that our answers are elements of the
set ( 0, ∞ ) .
Example 81 Sketch the graph of f ( x ) = log 4 (16 x )
Section 5.6 Exponential and Logarithmic Equations
Example 82 Solve: a. 2 x = 15
b. 5 ⋅ 2 x − 2 = 17
PROCEDURE FOR SOLVING EXPONENTIAL EQUATIONS
Step 1. Isolate the exponential expression on one side of the equation.
Step 2. Take the common or natural logarithm of both sides of the equation in Step 1.
Step 3. Use the power rule log a M r = r log a M to “bring down the exponent.”
Step 4. Solve for the variable.
Example 83 Solve the equation 52x–3 = 3x+1 and approximate the answer to three decimal places.
Example 84 Solve the equation 3x – 8•3–x = 2.
CHANGE-OF-BASE FORMULA
Let a, b, and c be positive real numbers with a ≠ 1 and b ≠ 1. Then logb x can be converted to a
different base as follows:
log b x =
log a x
log a b
(base a )
log x
log b
=
(base 10)
=
ln x
ln b
(base e)
Example 85 Compute log513 by changing to (a) common logarithms and (b) natural logarithms:
Example 86 Solve log 3 x = log x for x.
Example 87 Solve y =
e x − e− x
for x in terms of y.
e x + e− x
Section 9.1 Systems of Equations
Definitions: A set of equations with common variables is called a system of equations.
If each equation is linear, then it is a system of linear equations or a linear system of equations.
If at least one equation is nonlinear, then it is called a nonlinear system of equations.
A system of equations is sometimes referred to as a set of simultaneous equations.
A solution of a system of equations in two variables x and y is an ordered pair of numbers (a, b) such
that when x is replaced by a and y is replaced by b, all resulting equations in the system are true.
The solution set of a system of equations is the set of all solutions of the system.
Example 88 Solve the following system:
 y = x2

 y = 2x + 3
Example 89 Graph the system to check your solution from above.
SUBSTITUTION METHOD
Step 1. Solve for one variable. Choose one of the equations, and express one of the variables in terms
of the other variable.
Step 2. Substitute. Substitute the expression obtained in Step 1 into the other equation to obtain an
equation in one variable.
Step 3. Solve the equation obtained in Step 2.
Step 4. Back-substitute. Substitute the value(s) you obtained in Step 3 back into the expression you
found in Step 1. This gives the solutions.
Step 5. Check. Check your answer(s) in the original equation.
Example 90 Solve the system of equations by the substitution method.
 4 x + y = −3
 2
− x + y = 1
Example 91 Check your solution from above graphically.
Example 92 Solve the system of equations by the substitution method.
 x 2 + y 2 = 25
 2
 x − y = 5
Solutions:
x + 2z = 1

Example 93 Solve the system 2 y − z = 4
 xyz = 0

Section 9.3 Systems of Inequalities
The statements x + y > 4, 2x + 3y < 7, y ≥ x, and x + y ≤ 9 are examples of linear inequalities in the
variables x and y.
A solution of an inequality in two variables x and y is an ordered pair (a, b) that results in a true
statement when x is replaced by a, and y is replaced by b in the inequality.
The set of all solutions of an inequality is called the solution set of the inequality. The graph of an
inequality in two variables is the graph of the solution set of the inequality.
PROCEDURE FOR GRAPHING A LINEAR INEQUALITY IN TWO VARIABLES
Step 1. Replace the inequality symbol by an equals (=) sign.
Step 2. Sketch the graph of the corresponding equation Step 1. Use a dashed line for the boundary if the
given inequality sign is < or >, and a solid line if the inequality symbol is ≤ or ≥.
Step 3. The graph in Step 2 will divide the plane into two regions. Select a test point in the plane. Be
sure that the test point does not lie on the graph of the equation in Step 1.
Step 4. (i) If the coordinates of the test point satisfy the given inequality, then so do all the points of the
region that contains the test point. Shade the region that contains the test point.
(ii) If the coordinates of the test point do not satisfy the given inequality, shade the region that does not
contain the test point.
The shaded region (including the boundary if it is solid) is the graph of the given inequality.
Example 94 Sketch the graph of each of the following inequalities.
a. 2x - y ≥ 2
b. y < 3
c. x + y < 4
SYSTEMS OF LINEAR INEQUALITIES IN TWO VARIABLES
An ordered pair (a, b) is a solution of a system of inequalities involving two variables x and y if and only
if, when x is replaced by a and y is replaced by b in each inequality of the system, all resulting
statements are true.
The solution set of a system of inequalities is the intersection of the solution sets of all the inequalities
in the system.
2 x + 3 y > 6
 y−x≤0
Example 95 Graph the solution set of the system of inequalities: 
3 x − 4 y ≥ 12
 x − 2y ≤ 2

Example 96 Graph the solution set of the system of inequalities: 
x≥9


y≤5
 x − 2 ≤ 5
 y − 4 > 2
Example 97 Graph the solution set of the system of inequalities: 
Section 9.5 Systems of Linear Equations in More Than Two Variables
A linear equation in the variables x1, x2, …, xn is an equation that can be written in the form:
a1 x1 + a2 x2 + ... + an xn = b
where b and the coefficients a1, a2, …, an, are real numbers. The subscript n may be any positive integer.
A system of linear equations (or a linear system) in three variables is a collection of two or more linear
equations involving the same variables. For example,
 x + 3y + z = 0

 2x − y + z = 5
3 x − 3 y + 2 z = 10

is a system of three linear equations in three variables x, y, and z.
An ordered triple (a, b, c) is a solution of a system of three equations in three variables x, y, and z if each
equation in the system is a true statement when a, b, and c are substituted for x, y, and z respectively.
Example 98 Determine whether the ordered triple (2, –1, 3) is a solution of the given linear system
5 x + 3 y − 2 z = 1

x− y+ z =6

 2 x + 2 y − z = −1

DEFINITION OF A MATRIX
A matrix is a rectangular array of numbers denoted by
 a11
a
A =  21
 
 am1
a12
a22
am 2
... a1n 
... a2 n 


... amn 
If a matrix A has m rows and n columns, then A is said to be of order m by n (written m 5 n).
The entry or element in the ith row and jth column is a real number and is denoted by the doublesubscript notation aij. The entry aij is sometimes referred as the (i, j)th entry or the entry in the (i, j)
position, and we often write A =  aij  . If m = n then A is called a square matrix of order n and is
denoted by An. The entries a11, a22, …, ann form the main diagonal of An. A 1 x n matrix is called a row
matrix, and an n x 1 matrix is called a column matrix.
We can display all the numerical information contained in a linear system in an augmented matrix of
the system.
 x − y − z =1

2 x − 3 y + z = 10
 x + y − 2z = 0

 1 −1 −1 1 


 2 −3 1 10 
1 1 −2 0 
OPERATIONS THAT PRODUCE EQUIVALENT SYSTEMS
1. Interchange the position of any two equations.
2. Multiply any equation by a nonzero constant.
3. Add a nonzero multiple of one equation to another.
ELEMENTARY ROW OPERATIONS
Two matrices are row equivalent if one can be obtained from the other by a sequence of elementary
row operations.
Row Operation
In Symbols
Description
Interchange two rows
Ri ↔ Rj
Interchange the ith and jth rows
Multiply a row by a nonzero
constant
cRj
Multiply the jth row by c.
Add a multiple of one row to
another row
cRi + Rj → Rj
Replace the jth row by adding c times jth
row to it.
We would like to get the system into a form that will allow us to identify the solutions quite easily. That
form is called triangular form. A procedure called the Gaussian elimination method is used to convert
systems into triangular form.
GAUSSIAN ELIMINATION METHOD
Step 1. Rearrange the equations, if necessary, to obtain an x-term with a nonzero coefficient in the first
equation. Then multiply the first equation by the reciprocal of the coefficient of the x-term to get 1 as a
leading coefficient.
Step 2. By adding appropriate multiples of the first equation, eliminate any x-terms from the second
and third equations. Multiply the resulting second equation by the reciprocal of the coefficient of the yterm to get 1 as the leading coefficient.
Step 3. If necessary by adding appropriate multiple of the second equation from Step 2, eliminate any yterm from the third equation. Solve the resulting equation for z.
Step 4. Back-substitute the values of z from Steps 3 into one of the equations in Step 3 that contain only
y and z, and solve for y.
Step 5. Back-substitute the values of y and z from Steps 3 and 4 in any equation containing x, y, and z,
and solve for x.
Step 6. Write the solution set and check your work.
Example 99 Solve the system of equations.
 x − 2 y + 3z = 4

 2x + y − 4z = 3
 −3 x + 4 y − z = −2

Example 100 Solve the system of equations.
2x + y + z = 6


 −3 x − 4 y + 2 z = 4

x + y − z = −2

INCONSISTENT SYSTEMS
If, in the process of converting a linear system to triangular form, an equation of the form 0 = a occurs,
where a ≠ 0, then the system has no solu[on and is inconsistent.
Example 101 Solve the system of equations.
 x − y + 2z = 5

 2x + y + z = 7
3 x − 2 y + 5 z = 20

DEPENDENT EQUATIONS
If, in the process of converting a linear system to triangular form,
(i)
an equation of the form 0 = a (a ≠ 0) does not occur, but
(ii)
an equation of the form 0 = 0 does occur, then the system of equations has infinitely many
solutions and the equations are dependent.
Example 102 Solve the system of equations.
 x −y +z=7

 3 x + 2 y − 12 z = 11
4 x + y − 11z = 18

GEOMETRIC INTERPRETATION
The graph of a linear equation in three variables, such as ax + by + cz = d (where a, b, and c are not all
zero), is a plane in three-dimensional space. Following are the possible situations for a system of three
linear equations in three variables.
a.
Three planes intersect in a single point. The system has only one solution.
b.
Three planes intersect in one line. The system has infinitely many solutions.
c. Three planes coincide with each other. The system has only one solution.
d. There are three parallel planes. The system has no solution.
e. Two parallel planes are intersected by a third plane. The system has no solution.
f.
Three planes have no point in common. The system has no solution.
Section 9.6 The Algebra of Matrices
EQUALITY OF MATRICES
Two matrices A = [aij] and B = [bij] are said to be equal, written A = B, if
1. A and B have the same order m x n (that is, A and B have the same number m of rows and the
same number n of columns.)
2. aij = bij for all i and j. (The (i, j)th entry of A is equal to the corresponding (i, j)th entry of B.)
MATRIX ADDITION
If A =[aij] and B = [bij] are two m x n matrices, their sum A + B is the matrix defined by the m x n matrix
defined by
A + B =  aij + bij  , for all i and j.
Example 103 Add the following matrices:
 2 −3   1 2 
 0 1  +  7 −4 

 

SCALAR MULTIPLICATION
If A = [aij] be an m x n matrices, and let c be a real number. Then the scalar product of A and c is denoted
by cA and is defined by cA = caij  .
Example 104 Perform the scalar multiplication:
1 2 
3

 7 −4 
MATRIX SUBTRACTION
If A and B are two m x n matrices, their difference is defined by
A − B = A + ( −1) B
Example 105 Perform the indicated operation:
 2 − 3  1 2 
 0 1  −  7 −4 

 

MATRIX ADDITION AND SCALAR MULTIPLICATION PROPERTIES
Let A, B, and C be m x n matrices and c and d be scalars.
1.
A+B=B+A
2.
A + (B + C) = (A + B) + C
3.
A+0=0+A=A
4.
A + (–A) = (–A) + A = 0
5.
(cd)A = c(dA)
6.
1A = A
7.
c(A + B) = cA + cB
8.
(c + d)A = cA + dA
Example 106 Solve the matrix equation 3A + 2X = 4B for X, where
 2 0
1 3 
A=
and B = 

.
 4 6
5 2 
RULE FOR DEFINING THE PRODUCT AB
In order to define the product AB of two matrices A and B, the number of columns of A must be equal to
the number of rows of B. If A is an m x p matrix and B is a p x n matrix, then the product AB is an m x n
matrix
PRODUCT OF 1 x n AND n x 1 MATRICES
 b1 
b 
2
Suppose A = [ a1 a2 a3 an ] is a 1 x n row matrix, and B =   is a n x 1 column matrix. We

 
bn 
define the product AB = a1b1 + a2b2 + ... + anbn .
MATRIX MULTIPLICATION
Let A = [aij] be an m x p matrix and B = [bij] be a p x n matrix. Then the product AB is the m x n matrix C =
[cij], where the entry cij of C is obtained by multiplying the ith row (matrix) of A by the jth column
(matrix) of B.
The definition of the product AB says that cij = ai1b1 j + ai 2b2 j + ... + aip bpj .
 1 2
 3 −2 1 
and B = 


 −1 3 
1 2 3
Example 107 Find the products AB and BA, where A = 
PROPERTIES OF MATRIX MULTIPLICATION
Let A, B, and C be matrices and let c be a scalar. Assume that each product and sum is defined. Then
1. (AB)C = A(BC)
2. (i) A(B + C) = AB + AC
(ii) (A + B)C = AC + BC
3.
c(AB) = (cA)B = A(cB)
1 2 3
1 0 2




Example 108 Find the products AB and BA, where A = −2 −3 −1 and B = 2 3 0 .




 4 5 6 
 0 3 1 
Section 9.7 The Inverse of a Matrix
INVERSE OF A MATRIX
If A be an n x n matrix and let In be the n x n identity matrix that has 1’s on the main diagonal and 0s
elsewhere.
If there is an n x n matrix B such that AB = I n
and
BA = I n , then B is called the inverse of A and
–1
we write B = A (read “A inverse”).
PROCEDURE FOR FINDING THE INVERSE OF A MATRIX
Let A be an n x n matrix.
1. Form the n x 2n augmented matrix [A|I], where I is the n x n identity matrix.
2. If there is a sequence of row operations that transforms A into I, then this same sequence of
row operations will transform [A|I] into [I|B], where B = A–1.
3. Check your work by showing that AA–1 = I.
4. If it is not possible to transform A into I by row operations, then A does not have an inverse.
(This occurs if, at any step in the process, you obtain a matrix [C|D] in which C has a row of
zeros.)
1 1 0


Example 109 Find the inverse (if it exists) of the matrix A = 0 3 1 .


 2 3 3 
Solution:
1 1 0 1 0 0 


Start with  A I  =  0 3 1 0 1 0  …
 2 3 3 0 0 1 
1 3 

2 4
Example 110 Find the inverse (if it exists) of the matrix A = 
A RULE FOR FINDING THE INVERSE OF 2 x 2 MATRIX
a b 
 is invertible if and only if ad – bc ≠ 0. Moreover, if ad – bc ≠ 0, then the
c d 
1  d −b 
inverse is given by A−1 =
. If ad – bc = 0, the matrix does not have an inverse.
ad − bc  −c a 
The matrix A = 
SOLVING SYSTEMS OF LINEAR EQUATIONS BY USING MATRIX INVERSES
Matrix multiplication can be used to write a system of linear equations in matrix form.
3 x − 2 y = 4  3 −2   x   4 
→

  =  
 4 x − 3 y = 5  4 −3   y   5 
Solving a system of linear equations amounts to solving the matrix equation of the form AX = B.
The solution to this equation is X = A−1 B.
Example 111 Use a matrix to solve the linear system.
=4
 x +y

3y + z = 7

2 x + 3 y + 3 z = 21

Section 9.8 Determinants
DETERMINANT OF A 2 x 2 MATRIX
a b
a b 
is denoted by det (A), or

c d
c d 
The determinant of the matrix A = 
det ( A ) = A =
a
b
c
d
and is defined by
= ad − bc.
 −3 2 
.
 1 5
Example 112 Find the determinant of A = 
MINORS AND COFACTORS IN AN n x n MATRIX
Let A be an n x n square matrix. The minor Mij of the element aij is the determinant of the (n – 1) x (n
– 1) matrix obtained by deleting the ith row and the jth column of A.
The cofactor of Aij of the entry aij is given by: Aij = ( −1)
i+ j
M ij
1 2 3


For example, consider A = 4 5 6 the Minor of the 1st row and the 3rd column is M 13 . It is


7 8 9 
1 2 3


found by crossing out the 1 row and the 3 column of A: A = 4 5 6 , then find the


7 8 9 
st
rd
1+ 3
determinant (4x8 – 5x7) = -3. The Cofactor A13 = ( −1)
M 13 is then = 1*-3 = -3.
DETERMINANT OF A SQUARE MATRIX
Let A be a square matrix of order n ≥ 3. The determinant of A is the sum of the entries in any row of
A (or column of A), multiplied by their respective cofactors.
 1 −3 3 

Example 113 Find the determinant of A = 4
2 0  .

 −2 −7 5 
2 5 1
 −4 0 −3
Example 114 Find the determinant of A = 
 3 −2 1

 −1 4 2
0
0 
.
6

0
Theorem on Row of Zeros: if every element in a row or column of a square matrix is zero, then
A = 0.
Theorem on Matrix Invertibility: If A is a square matrix, then A is invertible if and only if A ≠ 0 .
Section 9.9 Properties of Determinants
You can see from example 114 that this method of finding determinants is tedious. It would be even
worse for a matrix of higher order, so we need more efficient methods.
Theorem on Row or Column Transformations of a Determinant:
If a matrix B is obtained from interchanging two rows or columns of A, then B = − A .
If a matrix B is obtained from multiplying A by multiplying one row or column by a scalar k, then
B =k A.
If a matrix B is obtained from A by adding k times any row or column of A to another row or column
of A, then B = A .
Theorem on Identical Rows: if two rows or columns of a square matrix are the same, then A = 0 .
Example 115 Remove the common factors from the rows, then find the determinant for
 14 −6 4 
A =  4 −5 12 
 −21 9 −6 
CRAMER’S RULE FOR SOLVING TWO EQUATIONS IN TWO VARIABLES
a1 x + b1 y = c1
of two equations of two variables has a unique solution (x, y) given by
a2 x + b2 y = c2
The system 
x=
Dx
D
and
where D =
a1
a2
y=
Dy
D
provided that D ≠ 0,
b1
c
, Dx = 1
b2
c2
b1
a
, and Dy = 1
b2
a2
c1
.
c2
5 x + 4 y = 1
2 x + 3 y = 6
Example 116 Use Cramer’s rule to solve the system: 
CRAMER’S RULE FOR SOLVING THREE EQUATIONS IN THREE VARIABLES
a1 x + b1 y + c1 z = k1

The system a2 x + b2 y + c2 z = k2 of three equations in three variables has a unique solution (x, y, z)
a x + b y + c z = k
3
3
3
 3
given by
x=
Dy
Dx
, y=
D
D
a1 b1 c1
D = a2
a3
b2
b3
Dz
provided that D ≠ 0, where,
D
b1 c1
a1 k1 c1
and z =
k1
c2 , Dx = k2
c3
k3
b2
b3
c2 , Dy = a2
c3
a3
k2
k3
a1
b1
k1
c2 , and Dz = a2
c3
a3
b2
b3
k2 .
k3
7 x + y + z − 1 = 0

Example 117 Use Cramer’s rule to solve the system of equations:  5 x + 3 z − 2 y = 4
 4x − z + 3y = 0

Section 9.10 Partial Fractions
Consider the fraction:
We can verify that
7x − 2
x2 − 4
3
4
+
is equal to the fraction above. We are going to learn a method for
x−2 x+2
‘decomposing’ a given rational expression into the sum of simpler rational expressions.
Definitions:
5x + 7
2
3
=
+
( x − 1)( x + 3) x + 3 x − 1
Each of the two fractions on the right is called a partial fraction. Rewriting a single fraction as the
sum of two (or more) fractions is called the partial-fraction decomposition of the rational
expression.
**In order to apply this method of fraction decomposition, we must have a rational expression
where the degree of the numerator is less than the degree of the denominator. If this is not the
case we can use long division to rewrite the given expression so that it conforms to our
requirement.
The Methods
CASE 1: THE DENOMINATOR IS THE PRODUCT OF DISTINCT (NONREPEATED) LINEAR FACTORS
Suppose Q(x) can be factored as Q ( x ) = ( x − a1 )( x − a2 ) ... ( x − an ) , with no factor repeated. The
P ( x)
partial-fraction decomposition of
Q ( x)
P ( x)
Q ( x)
is of the form:
=
An
A1
A2
+
+ ... +
,
x − a1 x − a2
x − an
where A1, A2, …, An, are constants to be determined.
PROCEDURE FOR PARTIAL-FRACTION DECOMPOSITION
Step 1 Write the form of the partial-fraction decomposition with the unknown constants A, B, C, … in
the numerators of the decomposition.
Step 2 Multiply both sides of the equation in Step 1 by the original denominator. Use the distributive
property and eliminate common factors. Simplify.
Step 3 Write both sides of the equation in Step 2 in descending powers of x, and equate the
coefficients of like powers of x.
Step 4 Step 3 will result in a system of linear equations in A, B, C,… . Solve this linear system for the
constants A, B, C,… .
Step 5 Substitute the values for A, B, C,… obtained in Step 4 into the equation in Step 1, and write the
partial-fraction decomposition.
Example 118 Find the partial-fraction decomposition of the expression:
10 x − 4
x3 − 4 x
Solution
(
)
Factor the denominator: x3 − 4 x = x x 2 − 4 = x ( x − 2 )( x + 2 ) .
Step 1 Write the partial-fraction.
10 x − 4
A
B
C
= +
+
x ( x − 2 )( x + 2 ) x x − 2 x + 2
Step 2 Multiply by original denominator.


10 x − 4
B
C 
A
=
x ( x − 2 )( x + 2 ) 
+
 = x ( x − 2 )( x + 2 )  +
 x x − 2 x + 2 
 x ( x − 2 )( x + 2 ) 
10 x − 4 = A ( x − 2 )( x + 2 ) + Bx ( x + 2 ) + Cx ( x − 2 )
10 x − 4 = A ( x − 2 )( x + 2 ) + Bx ( x + 2 ) + Cx ( x − 2 )
= A ( x2 − 4) + B ( x2 + 2 x ) + C ( x2 − 2 x )
= Ax 2 − 4 A + Bx 2 + 2 Bx + Cx 2 − 2Cx
10 x − 4 = ( A + B + C ) x 2 + ( 2 B − 2C ) x − 4 A
Step 3 Now use the fact that two polynomials are equal if and only if the coefficients of the like powers
are equal.
0 x 2 + 10 x − 4 = ( A + B + C ) x 2 + ( 2 B − 2C ) x − 4 A
Equating corresponding coefficients leads to the system of equations.
A+ B + C = 0

2 B − 2C = 10

 −4 A
= −4

Step 4 Solve the system of equations in Step 3 to obtain A = 1, B = 2, and C = –3.
Step 5 Since A = 1, B = 2, and C = –3, the partial-fraction decomposition is
10 x − 4 1
2
−3
1
2
3
= +
+
= +
−
3
x − 4x x x − 2 x + 2 x x − 2 x + 2
An alternative (and sometimes quicker) method of finding the constants is to substitute well-chosen
values for x in the equation (identity) obtained in Step 2.
Alternative Solution
Start with equation (1) from Step 2.
10 x + 4 = A ( x − 2 )( x + 2 ) + Bx ( x + 2 ) + Cx ( x − 2 )
Substitute x = 2 (because it causes the terms with A and C to be 0) in equation (1) to obtain
10 ( 2 ) + 4 =
A ( 2 − 2 )( 2 + 2 ) + B ( 2 )( 2 + 2 ) + C ( 2 )( 2 − 2 )
16 = 8 B
2=B
Substitute x = –2 (because it causes the terms with A and B to be 0) in equation (1) to obtain
10 ( −2 ) + 4 =
A ( −2 − 2 )( −2 + 2 ) + B ( −2 )( −2 + 2 ) + C ( −2 )( −2 − 2 )
−24 = 8C
−3 = C
Substitute x = 0 (because it causes the terms with B and C to be 0) in equation (1) to obtain
10 ( 0 ) + 4 =
A ( 0 − 2 )( 0 + 2 ) + B ( 0 )( 0 + 2 ) + C ( 0 )( 0 − 2 )
−4 = −4 A
1= A
Thus, A = 1, B = 2 and C = –3 and the partial-fraction decomposition is given by
10 x − 4 1
2
−3
1
2
3
= +
+
= +
−
3
x − 4x x x − 2 x + 2 x x − 2 x + 2
CASE 2: THE DENOMINATOR HAS A REPEATED LINEAR FACTOR
Let (x – a)m be the linear factor (x – a) that is repeated m times in Q(x). Then the portion of the partialfraction decomposition of
P ( x)
Q ( x)
that corresponds to the factor (x – a)m is
An
A1
A2
+
+ ... +
, where A1, A2, …, An, are constants to be determined.
m
2
x − a ( x − a)
( x − a)
Example 119 Find the partial-fraction decomposition of the expression:
x+4
( x + 3)( x − 1)
2
Solution:
Step 1 (x – 1) is repeated twice, (x + 3) is nonrepeating, the partial-fraction decomposition has the form
x+4
( x + 3)( x − 1)
2
=
A
B
C
+
+
x + 3 x − 1 ( x − 1) 2
A=
1
1
5
x+4
1
1
5
, B = − , and C = →
=
−
+
2
2
16
16
4
( x + 3)( x − 1) 16 ( x + 3) 16 ( x − 1) 4 ( x − 1)
CASE 3: THE DENOMINATOR HAS A DISTINCT(NONREPEATED) IRREDUCIBLE QUADRATIC FACTOR
Suppose ax2 + b + c is an irreducible quadratic factor of Q(x). Then the portion of the partial-fraction
decomposition of
P ( x)
Q ( x)
that corresponds to ax2 + bx + c has the form
Ax + B
, where A and B are
ax + bx + c
2
constants to be determined.
Example 120 Find the partial-fraction decomposition of
3x 2 − 8 x + 1
.
( x − 4 ) ( x 2 + 1)
Solution:
Step 1 (x – 4) is linear, (x2 + 1) is irreducible; thus, the partial-fraction decomposition has the form
3x 2 − 8 x + 4
A
Bx + C
=
+ 2
. ...
2
( x − 4 ) ( x − 1) x − 4 x − 1
CASE 4: THE DENOMINATOR HAS A REPEATED IRREDUCIBLE QUADRATIC FACTOR
Suppose the denominator Q(x) has a factor (ax2 + bx + c) m where m ≥ 2 is an integer and ax2 + bx + c is
irreducible. Then the portion of the partial-fraction decomposition of
factor ax2 + bx + c has the form
P ( x)
Q ( x)
that corresponds to the
Am x + Bm
A1 x + B1
A2 x + B2
+
+ ... +
, where A1, B1, A2,
2
m
2
ax + bx + c ( ax 2 + bx + c )
( ax2 + bx + c )
B2, …, Am, Bm are constants to be determined.
Example 121 Find the partial-fraction decomposition of
2 x 4 − x3 + 13 x 2 − 2 x + 13
( x − 1) ( x 2 + 4 )
2
.
Solution:
Step 1 (x – 1) is a nonrepeating linear factor, (x2 + 4) is an irreducible quadratic factor repeated twice,
the partial-fraction decomposition has the form:
2 x 4 − x3 + 13 x 2 − 2 x + 13
( x − 1) ( x 2 + 4 )
Section 10.1
2
=
A
Bx + C
Dx + E
+
+ 2
...
( x − 1) x + 4 ( x 2 + 4 )2
Sequences and Series
DEFINITION OF A SEQUENCE
An infinite sequence is a function whose domain is the set of positive integers. The function values,
written as
a1, a2, a3, a4, … , an, …
are called the terms of the sequence. The nth term, an, is called the general term of the sequence.
Example 122 Write the first six terms of the sequence defined by: bn = ( −1)
n +1
1
 .
n
Solution:
Replace n with each integer from 1 to 6. …
DEFINITION OF FACTORIAL
For any positive integer n, n factorial (written n!) is defined as n ! = n ⋅ ( n − 1) ⋅⋅⋅ 4 ⋅ 3 ⋅ 2 ⋅1.
As a special case, zero factorial (written 0!) is defined as 0! = 1.
Example 123 Write the first five terms of the sequence whose general term is: an
( −1)
=
n +1
n!
.
SUMMATION NOTATION
The sum of the first n term of a sequence a1, a2, a3, …, an, … is denoted by
n
∑a
i
= a1 + a2 + a3 + + an .
i =1
The letter i in the summation notation is called the index of summation, n is called the upper limit, and
1 is called the lower limit, of the summation.
9
Example 124 Find each sum. a.
7
∑i
∑(2 j
b.
i =1
2
2k
c. ∑
k =0 k !
4
− 1)
j =4
SUMMATION PROPERTIES
Let ak and bk, represent the general terms of two sequences, and let c represent any real number. Then
n
1.
n
∑c = c⋅n
2.
k =1
n
4.
n
∑ cak = c∑ ak
k =1
n
k =1
n
∑ ( ak − bk ) = ∑ ak −∑ bk
k =1
k =1
3.
5.
n
n
n
k =1
k =1
k =1
∑ ( ak + bk ) = ∑ ak +∑ bk
n
j
n
k =1
k =1
k = j +1
∑ ak = ∑ ak + ∑ ak , for 1 ≤ j < n
k =1
DEFINITION OF A SERIES
Let a1, a2, a3, … , ak, … be an infinite sequence. Then
1. The sum of the first n terms of the sequence is called the nth partial sum of the sequence and is
n
denoted by a1 + a2 + a3 + + an =
∑a .
i
i =1
2. The sum of all terms of the infinite sequence is called an infinite series and is denoted by
∞
a1 + a2 + a3 + + an + = ∑ ai .
i =1
Example 125 Write each sum in summation notation.
a. 3 + 5 + 7 + + 21
b.
1 1
1
+ + +
4 9
49
Section 10.2 Arithmetic Sequences
DEFINITION OF AN ARITHMETIC SEQUENCE
The sequence a1, a2, a3, a4, … , an, … is an arithmetic sequence, or an arithmetic progression if there is a
number d such that each term in the sequence except the first is obtained from the preceding term by
adding d to it. The number d is called the common difference of the arithmetic sequence. We have
d = an + 1 – an, n ≥ 1.
For example, 1, 4, 7, 10, 13, …
RECURSIVE DEFINITION OF AN ARITHMETIC SEQUENCE
An arithmetic sequence
a1, a2, a3, a4, … , an, …
can be defined recursively. The recursive formula
an + 1 = an + d for n ≥ 1
defines an arithmetic sequence with first term a1 and common difference d.
For Example, an +1 = an + 3 for n ≥ 1
nTH TERM OF AN ARITHMETIC SEQUENCE
If a sequence a1, a2, a3, … is an arithmetic sequence, then its nth term, an, is given by
an = a1 + (n – 1)d,
where a1 is the first term and d is the common difference.
Example 126 Find the common difference d and the nth term an of an arithmetic sequence whose 5th
term is 15 and whose 20th term is 45.
SUM OF n TERMS OF AN ARITHMETIC SEQUENCE
Let a1, a2, a3, … an be the first n terms of an arithmetic sequence with common difference d. The sum Sn
 a1 + an 
 where an = a1 + (n – 1)d.
 2 
of these n terms is given by S n = n 
Example 127 Find the sum of the arithmetic sequence of numbers:
1 + 4 + 7 + … + 25
10
Example 128 Find the sum
1

∑  4 k + 3 
k =1
Example 129 Express the sum in terms of summation notation: 3 + 8 + 13 + 18 + . . . + 463
Section 10.3 Geometric Sequences
DEFINITION OF A GEOMETRIC SEQUENCE
The sequence a1, a2, a3, a4, … , an, … is a geometric sequence, or a geometric progression, if there is a
number r such that each term except the first in the sequence is obtained by multiplying the previous
term by r. The number r is called the common ratio of the geometric sequence. We
have
an +1
= r , n ≥ 1.
an
RECURSIVE DEFINITION OF A GEOMETRIC SEQUENCE
A geometric sequence
a1, a2, a3, a4, … , an, …
can be defined recursively. The recursive formula
an + 1 = ran, n ≥ 1
defines a geometric sequence with first term a1 and common ratio r.
THE GENERAL TERM OF A GEOMETRIC SEQUENCE
Every geometric sequence can be written in the form
a1, a1r, a1r2, a1r3, … , a1rn+1, …
where r is the common ratio. Since a1 = a1(1) = a1r0, the nth term of the geometric sequence is
an = a1rn–1, for n ≥ 1.
Example 130 For the geometric sequence 1, 3, 9, 27, …, find each of the following:
a. a1
b. r
c. an
SUM OF THE TERMS OF A FINITE GEOMETRIC SEQUENCE
Let a1, a2, a3, … an be the first n terms of a geometric sequence with first term a1 and common ration r.
The sum Sn of these terms is
n
S n = ∑ a1r
i =1
i −1
=
a1 (1 − r n )
1− r
, r ≠ 1.
15
Example 131 Find each sum: a.
∑ 5 ( 0.7 )
i =1
i −1
15
b.
∑ 5 ( 0.7 )
i =1
i
VALUE OF AN ANNUITY
Let P represent the payment in dollars made at the end of each of n compounding periods per year, and
let i be the annual interest rate. Then the value of A of the annuity after t years is:
nt
 i
1 +  − 1
n
A= P
i
n
Example 132 Finding the Value of an Annuity
An individual retirement account (IRA) is a common way to save money to provide funds after
retirement. Suppose you make payments of 1200 dollars into an IRA at the end of each year at an
annual interest rate of 4.5% per year, compounded annually. What is the value of this annuity after 35
years?
n
The nth partial sum of a geometric sequence is given by: S n =
∑a r
1
i =1
SUM OF THE TERMS OF AN INFINITE GEOMETRIC SEQUENCE
If |r| < 1, the infinite sum
a1 + a1r + a1r2 + a1r3 + … + a1rn–1 + …
is given by
∞
S = ∑ a1r i −1 =
i =1
a1
.
1− r
Example 133 Finding the Sum of an Infinite Geometric Series
Find the sum 2 +
3 9 27
+ +
+.
2 8 32
Section 10.4 Mathematical Induction
i −1
1− rn
.
= a1
1− r
Our goal is to learn to prove statements by mathematical induction.
THE PRINCIPLE OF MATHEMATICAL INDUCTION
Let Pn be a statement that involves the natural number n with the following properties:
1.
P1 is true (the statement is true for the natural number 1), and
2.
If Pk is true statement, the Pk+1 is a true statement.
Then the statement Pn is true for every natural number n.
DETERMINING THE STATEMENT Pk+1 FROM THE STATEMENT Pk
Suppose the given statement is Pk : k ≥ 1 then we havePk+1: k + 1 ≥ 1.
That is, Pk+1 asserts the same property for k + 1 that Pk asserts for k.
Example 134 Use mathematical induction to prove that, for all natural numbers n,
2 + 4 + 6 + + 2n = n ( n + 1) .
Solution:
First verify that the statement is true for n = 1.
2 (1) = 1(1 + 1)
2=2
The second condition requires two steps.
Step 1 Assume the formula is true for k.
Pk : 2 + 4 + 6 + + 2k = k ( k + 1) .
Step 2 On the basis of the assumption that Pk is true, show that Pk+1 is true.
Pk +1 : 2 + 4 + 6 + + 2 ( k + 1) = ( k + 1) ( k + 1) + 1
Begin by using Pk , the statement assumed to be true and add 2(k + 1) to both sides
2 + 4 + 6 + + 2k = k ( k + 1)
2 + 4 + 6 + + 2k + 2 ( k + 1) = k ( k + 1) + 2 ( k + 1)
2 + 4 + 6 + + 2k + 2 ( k + 1) = ( k + 1)( k + 2 )
2 + 4 + 6 + + 2k + 2 ( k + 1) = ( k + 1)( k + 2 )
2 + 4 + 6 + + 2k + 2 ( k + 1) = ( k + 1) ( k + 1) + 1
This last statement says that Pk+1 is true if Pk is assumed to be true. Therefore, by the principle of
mathematical induction, the statement 2 + 4 + 6 + + 2n = n ( n + 1) is true for every natural number n.
Example 135 Use mathematical induction to prove that 2n > n for every natural number n.
Solution:
First verify that the statement is true for n = 1.
21 > 1
Step 1 Assume the formula is true for k.
Pk : 2 k > k
Step 2 Use Pk to prove that the following is true.
Pk +1 : 2k +1 > k + 1
By the product rule of exponents, 2k+1 = 2k•21, so multiply both sides by 2.
2k > k
2k ⋅ 2 > 2 ⋅ k
2k +1 = 2k ⋅ 2 > 2k = k + k ≥ k + 1
Thus, 2k+1 > k + 1 is true.
By mathematical induction, the statement 2n > n is true for every natural number n.
Section 10.5 The Binomial Theorem
PASCAL’S TRIANGLE
When expanding (x + y)n the coefficients of each term can be determined using Pascal’s Triangle. The top
of the triangle, that is, the first row, which contains only the number 1, represents the coefficients of (x
+ y)0 and is referred to as Row 0. Row 2 represents the coefficients of (x + y)1. Each row begins and ends
with 1. Each entry of Pascal’s Triangle if found by adding the two neighboring entries in the previous
row.
PASCAL’S TRIANGLE
Example 136 Using Pascal’s Triangle to Expand a Binomial Power
Expand (4y – 2x)5
n
r 
DEFINITION OF  
n
r 
If r and n are integers with 0 ≤ r ≤ n, the we define   =
n
0
Note:   = 1
and
n
  =1
n
n!
r !( n − r ) !
THE BINOMIAL THEOREM
If n is a natural number, then the binomial expansion of (x + y)n is given by
n
 n  n  n  n −1  n  n − 2 2
n n
 n  n−r r
n
x
+
y
=
x
+
x
y
+
x
y
+
+
y
=
(
)    
∑
 
 
 x y .
i =0  r 
0
1 
2
n
The coefficient of xn–ryr is
n!
.
r !( n − r ) !
PARTICULAR TERM IN A BINOMIAL EXPRESSION
 n  r n−r
x y .
n−r
The term containing the factor xr in the expansion of (x + y)n is 
Finding a Particular Term in a Binomial Expansion
Example 137 Find the term containing x10 in the expansion of (x + 2a)15 .
Solution:
Use the formula for the term containing xr.
 n  r n − r  15  10
15 
15 −10
5
=   x10 ( 2a )

x y =
 x ( 2a )
n−r
15 − 10 
5
Section 11.1 Parabolas
PARABOLA
Let l be a line and F a point in the plane not on the line l. The set of all points P in the plane that are the
same distance from F as they are from the line l is called a parabola. Thus, a parabola is the set of all
points P for which d(F, P) = d(P, l), where d(P, l) denotes the distance between P and l.
Line l is the directrix.
Point F is the focus.
The line through the focus, perpendicular to
the directrix is the axis or axis of symmetry.
The point where the axis intersects
the parabola is the vertex.
EQUATION OF A PARABOLA
y 2 = 4ax
The equation y2 = 4ax is called the standard equation of a parabola with vertex (0, 0) and focus (a, 0).
Similarly, if the focus of a parabola is placed on the negative x-axis, we obtain the equation y2 = – 4ax as
the standard equation of a parabola with vertex (0, 0) and focus (– a, 0).
x 2 = 4ay
By interchanging the roles of x and y, we find that the equation x2 = 4ay is the standard equation of a
parabola with vertex (0, 0) and focus (0, a). Similarly, if the focus of a parabola is placed on the negative
x-axis, we obtain the equation x2 = – 4ay as the standard equation of a parabola with vertex (0, 0) and
focus (0, – a).
LATUS RECTUM
The line segment that passes through the focus of a parabola, is perpendicular to the axis of the
parabola, and has endpoints on the parabola is called the latus rectum of the parabola. Following are
figures that show that the length of the latus rectum for the graphs of y2 = ±4ax and x2 = ±4ay for a > 0 is
4a.
Example 138 Find the standard equation of a parabola with vertex (0, 0) and satisfying the given
description.
a. The focus is (–3, 0).
b. The axis of the parabola is the y-axis, and the graph passes through the point (–4, 2).
Solution
a.
Vertex (0, 0) and focus (–3, 0) are both on the x-axis, so parabola opens left and the equation
has the form y2 = – 4ax with a = 3.
y 2 = − 4 ax
y 2 = −4 ( 3 ) x
The equation is y 2 = −12 x.
b.
Vertex is (0, 0), axis is the y-axis, and the point (–4, 2) is above the x-axis, so parabola opens up
and the equation has the form x2 = – 4ay and x = –4 and y = 2 can be substituted in to obtain
x 2 = 4 ay
( −4 )
2
= 4a ( 2 )
x 2 = 4ay
The equation is x 2 = 4 ( 2 ) y
16 = 8a
a=2
x 2 = 8 y.
Main facts about a parabola with vertex (h, k) and a > 0
Standard Equation
(y – k)2 = 4a(x – h)
Equation of axis
y=k
Description
Opens right
Vertex
(h, k)
Focus
(h + a, k)
Directrix
x=h–a
Main facts about a parabola with vertex (h, k) and a > 0
Standard Equation
(y – k)2 = –4a(x – h)
Equation of axis
y=k
Description
Opens left
Vertex
(h, k)
Focus
(h – a, k)
Directrix
x=h+a
Main facts about a parabola with vertex (h, k) and a > 0
Standard Equation
(x – h)2 = 4a(y – k)
Equation of axis
x=h
Description
Opens up
Vertex
(h, k)
Focus
(h, k + a)
Directrix
y=k–a
Main facts about a parabola with vertex (h, k) and a > 0
Standard Equation
(x – h)2 = – 4a(y – k)
Equation of axis
x=h
Description
Opens down
Vertex
(h, k)
Focus
(h, k – a)
Directrix
y=k+a
Example 139 Find the vertex, focus, and the directrix of the parabola 2y2 – 8y – x + 7 = 0. Sketch the
graph of the parabola.
Solution
Complete the square on y.
2 ( y2 − 4 y ) = x − 7
2 ( y 2 − 4 y + 4) = x − 7 + 8
2
2 ( y − 2) = x + 1
( y − 2)
2
=
1
( x + 1)
2
2
Compare ( y − 2 ) =
1
2
( x + 1) with the standard form ( y − k ) = 4a ( x − h ) . We have h = –1, k = 2, and
2
1
1
4a = , or a = . Vertex: (h, k) = (–1, 2)
2
8


1
8
  7
  8


Focus: (h + a, k) =  −1 + , 2  =  − , 2 
Directrix: x = h – a = −1 −
1
9
=−
8
8
REFELCTING PROPERTY OF PARABOLAS
A property of parabolas that is useful in applications is their reflecting property. The reflecting property
says that if a reflecting surface has parabolic cross sections with a common focus, then all light rays
entering the surface parallel to the axis will be reflected through the focus. This property is used in
reflecting telescopes and satellite antennas, since the light rays or radio waves bouncing off a parabolic
surface are reflected to the focus, where they are collected and amplified.
Conversely, if a light source is located at the focus of a parabolic reflector, the reflected rays will form a
beam parallel to the axis. This principle is used in flashlights, searchlights, and other such devices.
Section 11.2 The Ellipse
ELLIPSE
An ellipse is the set of all points in the plane, the sum of whose distances from two fixed points is a
constant. The fixed points are called the foci (the plural of focus) of the ellipse.
EQUATION OF AN ELLIPSE
x2 y2
+
=1
a2 b2
is called the standard form of the equation of an ellipse with center (0, 0) and foci (–c, 0) and (c, 0),
where b2 = a2 – c2.
( x − h)
2
a2
( y −k)
+
2
b2
= 1 is the standard form of the equation of an ellipse with center (h, k) and its
major axis is parallel to a coordinate axis.
Main facts about horizontal ellipses with center (h, k)
Standard Equation
( x − h)
Center
(h, k)
Eq’n of major axis
y=k
Length of major axis
2a
Eq’n of minor axis
x=h
Length of minor axis
2b
2
a2
( y −k)
+
b2
Vertices
(h + a, k), (h – a, k)
Endpts. minor axis
(h, k – b), (h, k + b)
Foci
(h + c, k), (h – c, k)
Eq’n a, b, and c
c2 = a2 – b2
Symmetry
about x = h and y = k
2
=1
Main facts about vertical ellipses with center (h, k)
Standard Equation
( x − h)
b
2
2
( y −k)
+
a
2
2
=1
Vertices
(h, k + a), (h, k – a)
Center
(h, k)
Endpts minor axis
(h – b, k), (h + b, k)
Eq’n of major axis
x=h
Foci
(h, k + c), (h, k – c)
Length of major axis
2a
Eq’n with a, b, c
c2 = a2 – b2
Eq’n of minor axis
y=k
Symmetry
about x = h and y = k
Length of minor axis
2b
Example 140 Find an equation of the ellipse that has foci (–3, 2) and (5, 2), and has a major axis of
length 10.
Solution:
Foci lie on the line y = 2, so it is a horizontal ellipse.
 −3 + 5 2 + 2 
,
 = (1, 2 )
2 
 2
The center is the midpoint of foci 
The length of the major axis =10, the vertices are at a distance of a = 5 units from the center.
The Foci are at a distance of c = 4 units from the center.
Use b2 = a2 – c2 to obtain b2.
b2 = (5)2 – (4)2 = 25 – 16 = 9 to obtain b2.
Major axis is horizontal so standard form is
( x − 1)
25
2
( y − 2)
+
9
2
= 1 Center: (1, 2)
a = 5, b = 3, c = 4
Example 141 Find the center, vertices, and foci of the ellipse with equation 3x2 + 4y2 +12x – 8y – 32
= 0.
Solution:
Complete the squares on x and y.
( 3x + 12 x ) + ( 4 y − 8 y ) = 32
3 ( x + 4 x ) + 4 ( y − 2 y ) = 32
3 ( x + 4 x + 4 ) + 4 ( y − 2 y + 1) = 32 + 12 + 4
2
2
2
2
2
2
2
2
3 ( x + 2 ) + 4 ( y − 1) = 48
2
2
3 ( x + 2 ) + 4 ( y − 1) = 48
( x + 2)
16
2
( y − 1)
+
12
2
=1
This is standard form. Center: (–2, 1), a2 = 16, b2 = 12, and c2 = a2 – b2 = 16 – 12 = 4. Thus, a = 4,
and c = 2.
Length of major axis is 2a = 8.
Length of minor axis is 2b = 4 3.
Center: (h, k) = (–2, 1)
Foci: (h ± c, k) = (–2 ± 2, 1) = (–4, 1) and (0, 1)
Vertices: (h ± a, k) = (–2 ± 4, 1)
Endpoints of minor axis:
( h, k ± b ) = ( −2,1 ± 2
(
= −2,1 + 2
)
3 ) and ( −2,1 − 2 3 )
3
= ( −2, 4.46 ) and ( −2, −2.46 )
APPLICATIONS OF ELLIPSES
1. The orbits of the planets are ellipses with the sun at one focus.
2. Newton reasoned that comets move in elliptical orbits about the sun.
3. An electron in an atom moves in an elliptical orbit with the nucleus at one focus.
4. The reflecting property for an ellipse says that a ray of light originating at one focus will be
reflected to the other focus.
REFELCTING PROPERTY OF ELLIPSES
Sound waves also follow such paths. This property is used in the construction of “whispering
galleries,” such as the gallery at St. Paul’s Cathedral in London. Such rooms have ceilings whose
cross sections are elliptical with common foci. As a result, sounds emanating from one focus are
reflected by the ceiling to the other focus. Thus, a whisper at one focus may not be audible at all at a
nearby place, but may nevertheless be clearly heard far off at the other focus.
Section 11.3 The Hyperbola
HYPERBOLA
A hyperbola is the set of all points in the plane, the difference of whose distances from two fixed
points is constant. The fixed points are called the foci of the hyperbola.
Here is a hyperbola in standard position, with foci F1(–c, 0) and F2(c, 0) on the x-axis at equal
distances from the origin. The two parts of the hyperbola are called branches.
EQUATION OF A HYPERBOLA
x2 y2
−
=1
a2 b2
is called the standard form of the equation of a hyperbola with center (0, 0). The x-intercepts are (–
a, 0) and (a, 0), which correspond to the vertices. The line segment joining the two vertices is the
transverse axis. The center is the midpoint of the transverse axis. The line segment joining (0, –b)
and (0, b) is the conjugate axis.
x2 y2
−
=1
a2 b2
y 2 x2
− =1
a 2 b2
EQUATION OF A HYPERBOLA
( x − h)
2
a2
( y −k)
−
2
=1
b2
is the standard equation of a hyperbola with center (h, k), opening to the left and right.
( y − h)
a2
2
(x − k)
−
2
b2
=1
is the standard equation of a hyperbola with center (h, k), opening up and down.
Main properties of hyperbolas, opening to the left and right, centered at (h, k)
Standard Equation
( x − h)
Eq’n transverse axis
y=k
a2
2
( y −k)
−
b2
2
=1
Length transverse axis
2a
Eq’n conjugate axis
x=h
Length conjugate axis
2b
Center
(h, k)
Vertices
(h – a, k), (h + a, k)
Endpts. conjugate axis
(h, k – b), (h, k + b)
Foci
(h – c, k), (h + c, k)
Eq’n for a, b, and c
c2 = a2 + b2
Asymptotes
y−k = ±
b
( x − h)
a
Main properties of hyperbolas, opening up and down, centered at (h, k)
Standard Equation
(y −k)
Eq’n transverse axis
x=h
Length transverse axis
2a
Eq’n conjugate axis
y=k
Length conjugate axis
2b
a2
2
( x − h)
−
b2
2
=1
Center
(h, k)
Vertices
(h, k – a), (h, k + a)
Endpts. conjugate axis
(h – b, k), (h + b, k)
Foci
(h, k – c), (h, k + c)
Eq’n for a, b, and c
c2 = a2 + b2
Asymptotes
y−k = ±
a
( x − h)
b
PROCEDURE FOR SKETCHING THE GRAPH OF A HYPERBOLA CETNERED AT (h, k)
Step 1 Plot the center (h, k), and draw horizontal and vertical dashed lines through the center.
Step 2 Locate the vertices and the endpoints of the conjugate axis. Lightly sketch the fundamental
rectangle, with sides parallel to the coordinate axes, through these points.
Step 3 Sketch dashed lines through opposite vertices of the fundamental rectangle. These are the
asymptotes.
Step 4 Draw both branches of the hyperbola, through the vertices and approaching the asymptotes.
Step 5 The foci are located on the transverse axis, c units from the center, where
c2 = a2 + b2.
Example 142 Show that 9x2 – 16y2 + 18x + 64y –199 = 0 is an equation of a hyperbola, and then graph
the hyperbola.
Solution:
Complete the squares on x and y.
(9x
2
+ 18 x ) + ( −16 y 2 + 64 y ) = 199
9 ( x 2 + 2 x ) − 16 ( y 2 − 4 y ) = 199
9 ( x 2 + 2 x + 1) − 16 ( y 2 − 4 y + 4 ) = 199 + 9 − 64
2
2
9 ( x + 1) − 16 ( y − 2 ) = 144
2
2
9 ( x + 1) − 16 ( y − 2 ) = 144
( x + 1)
16
Steps 1-2
2
( y − 2)
−
9
2
=1
Locate the vertices.
Center (–1, 2); a2 = 16, a = 4; b2 = 9, b = 3
(h – a, k) = (–1– 4, 2) = (–5, 2)
(h + a, k) = (–1+ 4, 2) = (3, 2)
Draw the fundamental rectangle.
Vertices: (3, –1), (3, 5), (–5, 5), (–5, –1)
Step 3 Sketch the asymptotes. Extend the diagonals of the rectangle obtained in Step 2 to sketch the
asymptotes:
y−2 =
3
3
( x + 1) and y − 2 = − ( x + 1)
4
4
Step 4 Sketch the graph. Draw two branches opening left and right, starting from the vertices (–5, 2)
and (3, 2) and approaching the asymptotes (without crossing them).
APPLICATIONS OF HYPERBOLAS
1. Comets that do not move in elliptical orbits around the sun almost always move in hyperbolic
orbits.
2. Boyle’s Law states that if a perfect gas is kept at a constant temperature, then its pressure P and
volume V are related by the equation PV = c, where c is constant. The graph of this
equation is a hyperbola.
3. The definition of a hyperbola forms the basis of several important navigational systems, such as
LORAN (Long Range Navigation).
4. The hyperbola has the reflecting property that a ray of light from a source at one focus of a
hyperbolic mirror (a mirror with hyperbolic cross sections) is reflected along the line through
the other focus.
5. The reflecting properties of the parabola and hyperbola are combined into one design for a
reflecting telescope. The parallel rays from a star are finally focused at the eyepiece at F2.
(see below)