College Algebra
Exam 4 Material
Quadratic Function
• A function that can be written in the form:
f(x) = ax2 + bx + c
(standard form)
where “a”, “b”, and “c” are real numbers with
a≠0
• A second degree polynomial function
• Examples:
f(x) = x2 – 8x + 15
f(x) = -4x2 – 16x + 33
Graphs of Quadratic Functions
• Every quadratic function has a graph that is a
“U-shaped” curve called a “parabola” that opens
upward when “a” is positive and downward when
“a” is negative
• The highest point on a parabola that opens
downward and the lowest point on a parabola
that opens upward is called the “vertex”
• The vertical line through the vertex is called the
“axis”
• The parabola is always “symmetric” with
respect to its axis (if the graph were folded along
the axis it would lay on top of the other half
“Vertex Form”
of a Quadratic Function
• Every quadratic function can be written in “vertex form”:
f(x) = a(x – h)2 + k
where “a”, “h”, and “k” are real numbers with a ≠ 0, and
where (h, k) is the vertex of the parabola
• As before, parabola opens upward when “a” is positive
and downward when “a” is negative
• Examples:
f(x) = 2(x – 3)2 – 4
parabola that opens upward with vertex (3, -4)
f(x) = -3(x + 5)2 + 1
parabola that opens downward with vertex (-5, 1)
Converting a Quadratic Function
to Vertex Form
• Given f(x) = ax2 + bx + c, factor “a” from variable
terms leaving variables in parentheses
• Complete the square on the expression inside
the parentheses and balance on the same side
of the equation
• Factor the expression inside parentheses and
combine like terms outside parentheses to get
vertex form:
f x   ax  h   k
2
Example One:
Converting to Vertex Form
f(x) = x2 – 8x + 15
Factor “a” from variable terms:
f(x) = 1(x2 – 8x
) + 15
Complete the square on the expression inside
parentheses and balance on the same side of
the equation:
f(x) = 1(x2 – 8x + 16) + 15 – 16
Factor the expression inside parentheses and
combine like terms on outside:
f(x) = 1(x – 4)2 – 1
Note: parabola opens upward with vertex (4, -1)
Example Two:
Converting to Vertex Form
f(x) = -4x2 – 16x + 33
f(x) = -4(x2 + 4x
) + 33
f(x) = -4(x2 + 4x + 4) + 33 + 16
f(x) = -4(x + 2)2 + 49
Note: parabola opens downward with
vertex
(-2, 49)
Graphing a Quadratic Function
• Convert to “vertex form” to find vertex, (h, k)
• Note the “x” value of vertex, h, and pick two
other x values that are bigger than h, or two
values that are smaller than h, and calculate
corresponding y values
• Plot these three points on a graph and use
symmetric properties of parabola to plot two
more points on the other side of the axis
• Connect points with smooth curve with arrows at
both ends
Example of
Graphing Quadratic Function
f(x) = x2 – 8x + 15
f(x) = 1(x – 4)2 – 1
Parabola opens upward with vertex (4, -1)
Note: h = 4 (x coordinate of vertex)
If we pick x = 3, we get y = 32 – 8(3) + 15
(3, 0) is a point on parabola
If we pick x = 2, we get y = 22 – 8(2) + 15
(2, 3) is a point on parabola
Graph f(x) = 1(x – 4)2 – 1
Symmetric Property
Points : (3,0) and (2,3)
Vertex : (4,-1)
Example of
Graphing Quadratic Function
f(x) = -2x2 + 4x + 3
f(x) = -2(x - 1)2 + 5
Parabola opens downward with vertex:
(1, 5)
If x = 0, then y = -2(0)2 + 4(0) + 3 = 3
(0, 3) is a point on the graph
If x = -1, then y = -2(-1)2 + 4(-1) + 3 = -3
(-1, -3) is a point on the graph
Graph f(x) = -2(x - 1)2 + 5
Vertex : (1,5)
Points : (0,3) and (-1,-3)
Symmetric Property
Equation of Axis, Domain and
Range of Quadratic Functions
• Given vertex (h, k), the equation of the axis is:
x=h
• Unless otherwise specified, the domain of a quadratic
function is “all real numbers”
• The range depends on whether the parabola opens
upward or downward:
– If upward the range is from “k” to positive infinity
– If downward, the range is from negative infinity to “k”
• Example:
If vertex is (-2, 3) and “a” is positive:
Equation of axis is: x  2
Domain is:  , 
Range is: [3,  )
Homework Problems
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Section:
Page:
Problems:
3.1
303
All: 1 – 4, Odd: 13 – 25
MyMathLab Assignment 3.1 for practice
MyMathLab Quiz 3.1 is due for a grade on
the date of our next class meeting
Polynomial Functions
• A polynomial function in “x” is defined as f(x) = a
finite sum of terms, each of which has the form
anxn where “n” represents a whole number and
an is the coefficient of the variable xn
• The largest exponent in the finite sum of terms
is called the degree of the function (the only
exception being where the polynomial function is
of the form: f(x) = 0. This is called the “zero
polynomial” and has no degree.)
Examples of Polynomial
Functions
f x   3
Degree : 0
Special Name : Constant Function
f x   2 x  5
Degree : 1
Special Name : Linear Function
f x   4 x 2  5 x  1
Degree : 2
Special Name : Quadratic Function
f x   x 3  x 2  4 x  4
Degree : 3
Special Name : Cubic Function
Dividing Polynomial Function by
“x – k” Using Long Division
• This method has been discussed in
previous classes:
– First write each polynomial in descending
powers
– If a term of some power is missing, write
that term with a zero coefficient
– Complete the problem exactly like a long
division problem in basic math
Example
 2 x  3x 150 x  4
3x  2x  0x 150 x  4
2
3
3
2
10
3x 2  10x  40 
x4
x  4 3x 3  2 x 2  0 x  150
 ( 3x 3  12 x 2 )
10x 2  0x
 ( 10 x 2  40 x)
40x  150
 ( 40x 160)
10
Dividing Polynomial Function by
“x – k” Using Synthetic Division
• First write each polynomial in descending powers
• If a term of some power is missing, write that term with a zero
coefficient
• Set the problem up as a long division problem, but write only
the coefficients of the dividend inside division symbol and “k”
outside
• Put a blank line under dividend coefficients with an underline
• Drop the first coefficient below this line
• Multiply this coefficient by “k” and write the answer above the
line below the second coefficient
• Add the two numbers and put answer beneath the line
• Repeat this process until you reach the end
• The numbers beneath the line represent coefficients of the
answer and the remainder where the first term of the answer is
of degree one less than the dividend and the last number is the
remainder
Last Example by Synthetic
Division
 2x  3x 150 x  4
3x  2x  0x 150 x  4
2
3
3
2
3  2  0 150
12 40 160
10
3 10 40
4
From last row answer is :
10
3x  10 x  40 
x4
2
Same answer as from long division,
but found much easier! !!
Homework Problems
• Section:
• Page:
• Problems:
3.2
319
Odd: 1 – 17
• MyMathLab Assignment 3.2(a) for practice
Remainder Theorem
• When a polynomial, f x  ,is divided by
x  k  the remainder is f k 
• Consider the last example where
f x   2 x 2  3x 3  150 was divided by x  4
using synthetic division: 4 3  2  0 150
12 40 160
• The Theorem says that
10
3 10 40
f 4  10
• And it is:
f 4  24  34  150
2
3
f 4  32  192 150  10
Evaluating a Polynomial
• It is usually easiest to evaluate a
polynomial for a specific value of “x” by
using synthetic division and the remainder
theorem
• Example: Given f x   x5  2 x 4  x3  5
5 1 2 1 0
0
5
• Find: f 5
1
f 5  1995
5 15 80 400 2000
3 16 80 400 1995
Homework Problems
• Section:
• Page:
• Problems:
3.2
319
Odd: 27 – 37
• MyMathLab Assignment 3.2(b) for practice
Zeros of Polynomial Functions
• Zeros of polynomial functions are the
values of “x” that make “y = 0”
• “k” is a zero of f(x) if f(k) = 0
• If a zero, k, of a polynomial function is a
real number, then k is also an x-intercept
of the graph of f(x)
• Polynomial functions may also have zeros
that are non-real complex numbers (these
are not x-intercepts of the graph)
Relationship Between Zeros and
Factors
• If k is a zero of f(x) and f(x) has degree “n”,
then x – k is one factor of f(x), and the
other factor is the polynomial of degree
“n – 1” whose coefficients are shown in the
bottom row of synthetic division
3
2


• Example: Given: f x  x  4 x  x  6
• Find: f 3
Example Continued
Given : f x   x  4 x  x  6, find f 3
3
2
Divide by x  3, using Synthetic Division t o find f 3
3 1 4
1
6
3 3 6
1 1  2 0
• Synthetic Division shows that f 3 is a
zero of f x   x 3  4 x 2  x  6 , so x  3 is
one factor, and the other factor is:
x2  x  2
Homework Problems
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Section:
3.2
Page:
320
Problems: Odd: 39 – 55
MyMathLab Assignment 3.2(c) for practice
MyMathLab Quiz 3.2 is due for a grade on the date of
our next class meeting
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Section:
3.3
Page:
329
Problems: Odd: 5 – 21
MyMathLab Assignment 3.3(a) for practice
Rational Zeros Theorem
• Given a polynomial function f x , with
integer coefficients, where a n is the
coefficient of the highest degree term and
a0 is the coefficient of the constant term,
rational zeros must be the ratio of some
factor of a0 to some factor of a n
• Example: Find all possible rational zeros:
2
f x   6 x  10 x  4
Example Continued
Find all possible rational zeros : f x   6 x 2  10 x  4
 1,  2,  4
 1,  2,  3,  6
All factors of 6 :
a0
All possible ratios of
:
an
1 1 1 1 2 2 2 2 4 4 4 4
 , , , , , , , , , , ,
1 2 3 6 1 2 3 6 1 2 3 6
1 1 1
2 1
4 2
 1,  ,  ,  ,  2,  1,  ,  ,  4,  2,  , 
2 3 6
3 3
3 3
1 1 1 2
4
Simplified Possibilit ies :  ,  ,  ,  ,  1,  ,  2,  4
6 3 2 3
3
All factors of - 4 :
Using Rational Zeros Theorem
on Polynomials with
Non-Integer Coefficients
• “k” is a zero of f(x), if and only if “k” is also
a zero of the polynomial formed by
multiplying a constant by f(x)
• To find zeros of f(x) when f(x) does not
have integer coefficients, find zeros of
af(x) where “a” is a factor that forms a
polynomial with integer coefficients
Relationship Between Degree of a
Polynomial Function and the
Number of Distinct Zeros
• A polynomial function of degree “n” has at
most “n” distinct zeros
• Examples: A polynomial of:
degree 2, has at most ___
2 distinct zeros
degree 5, has at most ___
5 distinct zeros
Other Examples
• Given f x   x 2  3x  4
• We can find zeros by solving f x  0
x 2  3x  4  0
x  4x  1  0
x  4, x  1 (Two distinct zeros)
• Given f x   x 2  2 x  1
• We can find zeros by solving f x  0
x2  2x 1  0
x  1x  1  0
x  1, x  1 (One distinct zero)
Still meets criteria of at most two distinct zeros
Non-Real Complex Zeros of
Polynomial Functions
• If a non-real complex number is a zero of
f(x), then its conjugate is also a zero of f(x)
• Examples:
3  2i is also a zero
If 3 - 2i is a zero, then _____
1 5i is also a zero
If 1+ 5i is a zero, then _____
Exact Number of Zeros in a
Polynomial of Degree “n”
• A polynomial function, f(x), of degree “n”
has exactly “n” zeros, but they may not all
be distinct
• A number “k” is said to be a zero of
multiplicity “m” if “m” is greater than 1
and is the largest exponent for which
(x – k)m is a factor of f(x)
• In this case “k” counts as “m” of the “n”
zeros of f(x)
Example
•
•
•
•
•
•
Given: f  x   x  5 x  2   x  1
Degree: 6
Exact Number of Zeros: 6
Number of Distinct Zeros: 3
Distinct Zeros:  5, 2,  1
What is the multiplicity of each zero?
3
2
2 is a zero of multiplici ty 3
- 1 is a zero of multiplici ty 2
Finding Zeros of a Polynomial
Function
• Use Rational Zeros Theorem to find all possible rational
zeros
• Use Synthetic Division and Remainder Theorem to try to
find one rational zero (remainder will be zero)
• If “n” is a rational zero, factor original polynomial as
(x – n)q(x)
• Test remaining possible rational zeros in q(x). If one is
found, factor again as in previous step
• Continue in this way until all rational zeros have been
found
• See if additional irrational or non-real complex zeros can
be found by solving a quadratic equation
Example
• Find all zeros of: f x   x  2 x  2 x  3x  2
• Find all solutions to: x 5  2 x 3  2 x 2  3x  2  0
• Rational Zeros Theorem says the only
possible rational zeros are:  1 and  2
• See if -1 is a zero:  1 1 0  2  2  3  2
5
3
2
1 1
1
1 1 1 1
• Conclusion:
1
2
- 1 is a zero, (x  1) is a factor and another factor is :
x
4
3
2
-x -x -x-2

2
0
Example Continued
• This new factor x 4 - x 3 - x 2 - x - 2 has the
same possible rational zeros:  1 and  2
• Check to see if -1 is also a zero of this:
1
• Conclusion:
1 1 1 1  2
1 2 1
2
1 2 1 2
0
- 1 is a zero, (x  1) is a factor and another factor is :
x
3
- 2x  x - 2
2

Example Continued
x
 has as
• This new factor
possible rational zeros:  1 and  2
• Check to see if -1 is also a zero of this:
3
- 2x 2  x - 2
1
• Conclusion:
1 2
1
1 3
1 2
3 4
4 6
- 1 is NOT a zero, so try another possible zero : 1
Example Continued
• Check to see if 1 is a zero:
1
1 2 1 2
1 1
0
1 1 0  2
• Conclusion:
- 1 is NOT a zero, so try another possible zero : 2
Example Continued
• Check to see if 2 is a zero:
2
1 2
2
1
0
1 2
0
2
1
0
• Conclusion:
2 is a zero, (x  2) is a factor and another factor is :
x
2

1
Example Continued
• Summary of work done:
f x   x 5  2 x 3  2 x 2  3x  2


f x   x  1 x  2 x  1
2
2
- 1 is a zero of multiplici ty two, 2 is a zero, and the
other two zeros can be found by solving : x 2  1  0
x 1  0
2
x 2  1
x  i
Distinct z eros: - 1, 2, i, -i
Homework Problems
• Section:
• Page:
• Problems:
3.3
330
Odd: 29 – 47
• MyMathLab Assignment 3.3(b) for practice
• MyMathLab Quiz 3.3 (Shortened Version)
is due for a grade on the date of our next
class meeting
Further Hints on Finding All Zeros
of a Polynomial Function
• In trying to find all zeros of a polynomial function, it
would be useful to know the number of positive and
negative real zeros
• Descartes’ Rule of Signs - If f(x) is a polynomial
function with real coefficients, written in descending
powers, with a non-zero constant term
– The number of positive real zeros is equal to the number of sign
changes in the terms of f(x), or is less than the number of sign
changes by an even positive integer
– The number of negative real zeros is equal to the number of sign
changes in the terms of f(-x), or is less than the number of sign
changes by an even positive integer
– In both considerations, missing terms do not count as a sign
change
Application of
Descartes’ Rule of Signs
• Find the number of positive and negative
real zeros of: f x   3x5  4 x3  x 2  5x  2
• How many sign changes? 3
• Number of positive real zeros? 3 or 1
f  x   3x5  4 x3  x 2  5x  2
• Find f(-x):
• How many sign changes? 2
• Number of negative real zeros? 2 or 0
Further Consideration
of Previous Example
• Consider all possible zeros:
• Total Zeros:
Positive
3
3
1
1
5
f  x   3x 5  4 x 3  x 2  5 x  2
Negative
Non-Real*
2
0
2
0
0
2
2
4
* Since non-real zeros occur only as conjugate pairs there
must always be an even number
Homework Problems
• Section:
• Page:
• Problems:
3.3
331
All: 73 – 78
• MyMathLab Assignment 3.3(c) for practice
Example of Finding Polynomial
Functions with Specific Zeros
• Given the fact that k is a zero of f(x) if and
only if (x – k ) is a factor:
• Find a polynomial with zeros: 3, 1, and -2
f x  x  3x 1x  2
f x   x 3  2 x 2  5 x  6
• Note: Other polynomial functions with the
same zeros will be non-zero multiples of
this one
f x   mx 3  2 x 2  5 x  6, m  0
• Other examples: f x   2 x3  4 x 2  10 x  12
f x   3x3  6 x 2  15x  18
Additional Considerations Relative
to Last Example
• For the previous example, if we are also
told that f(2) = 8, find the polynomial
function with the specified zeros


f x   m x 3  2 x 2  5 x  6 , m  0


f 2  m 23  22  52  6  8
2
m8  8 10  6  8
m 4  8
m  2
f x   2x3  2 x 2  5x  6, m  0
f x   2 x3  4 x 2  10 x  12
Homework Problems
• Section:
• Page:
• Problems:
3.3
330
Odd: 49 – 71
• MyMathLab Assignment 3.3(d) for practice
• MyMathLab Quiz 3.3 is due for a grade on
the date of our next class meeting
Turning Points
of Polynomial Functions
• The point at which the graph of a
polynomial function changes from
increasing to decreasing or vice versa is
called a turning point of the function
How many turni ng points?
3
Turning Points
of Polynomial Functions
• A polynomial function of degree n has
at most n – 1 turning points with at
least one turning point between
successive zeros
• Given that the following function has zeros
at 1 and -1, what do you know about its
turning points? f x   x5  x 3  2 x 2  2
• It has at most how many? 4
• Where is at least one of those located?
Between (-1, 0) and (1, 0)
Graphs of Polynomial Functions
• The domain of every polynomial function is:
  , 
• Polynomial functions are continuous over their
domain (entire graph can be drawn without lifting
pencil)
• Even polynomial functions,  f  x  f x , are
symmetric with respect to y-axis
• Odd polynomial functions,  f  x   f x , are
symmetric with respect to origin
• Note: Many polynomial functions are neither
even nor odd
Graphs of Polynomial Functions
• End behavior of graphs of polynomial functions is
determined by characteristics of highest degree term:
– If degree of polynomial is odd and coefficient of highest degree
is positive, ends look like:
– If degree of polynomial is odd and coefficient of highest degree
is negative, ends look like:
– If degree of polynomial is even and coefficient of highest
degree is positive, ends look like:
– If degree of polynomial is even and coefficient of highest
degree is negative, ends look like:
Note : Difference between even/odd function and even/odd degree function!
Graphs of Polynomial Functions
• x-intercepts of polynomial functions are
real zeros of the function
• The y-intercept of a polynomial function,
f(x), is f(0)
Sketching Graphs of
Polynomial Functions
• Find the real zeros of the function and plot
them as x-intercepts
• Find and plot the y-intercept
• Use synthetic division to find values of the
polynomial function between zeros
• Show end behavior of graph based on
characteristics of highest degree term
Sketch Graph:
f x   2 x 3  3x 2  11x  6
• Descartes’ Rule of Signs:
2 positive real zeros or none; 1 negative real zero
1 3
• Possible rational roots:  1,  2,  3,  6,  , 
2 2
• Test for rational roots:  2 2  3  11
4
2 7


f  x    x  2 2 x  7 x  3  0
2
6
14  6
3 0
Example Continued
• Solving equation by zero factor:
 x  2 2 x 2  7 x  3  0
x  22x 1x  3  0
1
x  2, x  , x  3 (x - intercepts of graph)
2
3
2


f
x

2
x

3
x
 11x  6
• Find y-intercept:
f 0  6


• Find values of function between zeros:
For example, find f 2
Example Continued
• Find f(2) by synthetic division:
2
f 2  12
2  3  11
6
4
2  18
2 1  9  12
Sketch of Graph
f x   2 x  3x  11x  6
3
2
Note : Scale factor on y - axis is 2
y - intercept : 6
1
Zeros : - 2, , 3
2
Point between zeros : 2,-12
End behavior based on odd degree with positive coefficien t
Note : Graph has maximum number of turning points!
Homework Problems
• Section:
• Page:
• Problems:
3.4
343
Odd: 21 – 27, All: 29, 30,
37 – 40
• MyMathLab Assignment 3.4(a) for practice
Intermediate Value Theorem
• If “a” and “b” are in the domain of a
polynomial function, f(x), and f(a) and
f(b) have opposite signs, then f(x) has a
least one real zero between “a” and “b”
• We can use this theorem to approximate
the value of irrational zeros to any desired
degree of accuracy as shown by the
following example
Find a Zero to the Nearest Tenth:
f  x   x  2 x  3x  6
3
• Find f(1)
1
1
1
• Find f(2)
2
2
2 3 6
1 3
0
3
0 6
2 3 6
2
8 10
4
5
4
1
1
f 1  6
f 2  4
• There must be a zero between 1 and 2
f 1.5  2.265
• Find f(1.5) 1.5 1 2  3  6
1
1.5
3.5
5.25 3.375
2.25  2.265
• Now we know there is a zero between:
1.5 and 2
Find a Zero to the Nearest Tenth:
f  x   x  2 x  3x  6
3
• Find f(1.8)
1. 8
1
1
2
2
3 6
1.8
6.84 6.912
3.8
3.84 0.912
f 1.8  0.912
• There must be a zero between: 1.5 and 1.8
• Find f(1.7) 1.7 1 21.7  63.29  65.593 f 1.7  0.407
1
3.7
3.29  0.407
• There must be a zero between: 1.7 and 1.8
6
• Find f(1.75) 1.75 1 2  3
1
1.75
3.75
6.5625
3.5625
6.234375
0.234375
f 1.75  0.234375
Find a Zero to the Nearest Tenth:
f  x   x  2 x  3x  6
3
2
• Since f 1.7  0.407 and f 1.75  0.234375
• We have now found that, to the nearest
tenth, a zero is: 1.7
Homework Problems
• Section:
3.4
• Page:
343
• Problems: Odd: 43 – 51
Also, without using a graphing calculator,
approximate the zero discussed in
problems 45 and 47 to the nearest tenth
• MyMathLab Assignment 3.4(b) for practice
Boundedness Theorem
• Given a polynomial function, f(x), with real
coefficients, of degree 1 or more, and a real
number “c” in the domain of f(x), and finding
f(c) by synthetic division:
– If c > 0 and all numbers on bottom row of
synthetic division are non-negative, then f(x) has
no zero greater than c
– If c < 0 and all numbers on bottom row of
synthetic division alternate in sign (with 0
considered positive or negative as needed), then
f(x) has no zero less than c
Example of Boundedness Theorem
• In a previous example:
f  x   x  2 x  3x  6
3
2
we found f(2) by synthetic division:
2
1
1
2 3 6
2
8 10
4
5
4
f 2  4
• Notice bottom row. What does this tell us
about zeros?
f x has no zero bigger tha n 2
Example of Boundedness Theorem
• Using the same function:
f  x   x  2 x  3x  6
3
2
find f(-3) by synthetic division:
3
2 3 6
3
3
0
1 1
0 6
1
f  3  6
• Notice bottom row. What does this tell us
about zeros?
f x has no zero less than  3
Homework Problems
• Section:
• Page:
• Problems:
3.4
343
Odd: 53 – 59
• MyMathLab Assignment 3.4(c) for practice
• MyMathLab Quiz 3.4 is due for a grade on
the date of our next class meeting
Rational Functions
• A “rational function” is a function that can be
written as:
p x 
f x  
, where px  and qx  are polynomial functions and qx   0
q x 
• Any “real “ zeros of qx  will make the rational
function undefined and will establish “vertical
asymptotes” for the graph of the function
• When the zeros of qx  are non-real complex
numbers, there will be no vertical asymptotes
Vertical Asymptote
• A “vertical asymptote” is a vertical line located at
each real zero of the denominator polynomial, q x
• On either side a vertical asymptote the value of
the rational function, f x ,approaches   as
as x gets closer to the real zero of q x
• The possible behaviors of the graph of a rational
function near a vertical asymptote are illustrated
on the next slide



Behavior of Rational Function
Graphs Near Vertical Asymptotes
Finding Vertical Asymptotes of
Rational Functions
• Given a rational function:
px 
f x  
qx 
• Find zeros of qx 
• For each number, k , that is a real zero of
qx  , x  k is a vertical asymptote
Example of Finding
Vertical Asymptotes
• Find the vertical asymptotes:
x 3
f x   2
x  2x  8
• Find zeros of denominator:
x2  2x  8  0
x  4x  2  0
x  4  0 or x  2  0
x  4 or x  2 Equations of Asymptotes
Approx. Graph:
x 3
f x   2
x  2x  8
Homework Problems
• Section:
• Page:
• Problems:
3.5
364
37 – 45 (Vertical Asymptote)
• No MyMathLab Assignment
Linear Systems of Equations
• When two or more linear equations are considered
simultaneously they are referred to as a system of
equations
• Example of System of Linear Equations in Two
Variables:
2x – y = 7
3x + 7y = 2
• The solution to this system is the set of all points, (x, y)
pairs, that make both equations true at the same time
• If these represent different non-parallel lines, the solution
is a single point. If they represent parallel lines, there is
no solution. If they are different forms of the same line,
then any point on the line is a solution.
Solving a System of Linear
Equations in Two Variables by
Substitution
• Solve either equation for one variable (choose
easiest)
• Substitute this value into the second equation
• Solve the second equation
• Substitute the solution to the second equation
back into the first equation and solve it
• The solution to the system is shown as an (x, y)
pair
Example
2x – y = 7
3x + 7y = 2
It is easiest to solve first equation for y:
2x – 7 = y
Substitute into the second equation:
3x + 7(2x – 7) = 2
3x + 14x – 49 = 2
17x = 51
x=3
Substitute into first equation:
2(3) – y = 7
-y=1
y = -1
Solution: (3, -1)
Homework Problems
• Section:
• Page:
• Problems:
5.1
484
Odd: 7 – 17
• MyMathLab Assignment 5.1(a) for practice
Solving a System of Linear
Equations in Two Variables by
Elimination
• Multiply one or both equations by a constant so
that when the two equations are added, the
result is a single equation in one variable
• Solve the resulting equation for that variable
• Substitute that value back into either original
equation to find value of other variable
• Show solution as an ordered pair
Example of Solving a System of
Linear Equations by Elimination
2x – y = 7
3x + 7y = 2
Multiply the first equation by 7 and add the result to the
second equation to eliminate “y”:
14x – 7y = 49
3x + 7y = 2
17x
= 51
x=3
Substitute into either equation (first):
2(3) – y = 7
y = -1
Solution: (3, -1) (Same as before)
Homework Problems
• Section:
• Page:
• Problems:
5.1
484
Odd: 19 – 29
• MyMathLab Assignment 5.1(b) for practice
• MyMathLab Quiz 5.1(Shortened Version) is due
for a grade on the date of our next class meeting
• This is the last assignment for a grade!!!!
Linear Equations in Three
Variables
• Equations of the form:
Ax + By + Cz = D
• Solutions to these equations are “ordered
triples”
• Example:
4x + 3y – z = -5
There are an infinite number of ordered triples
that are solutions.
One solution is (3, -5, 2)
Solving a System of Three Linear
Equations by Elimination
• Multiply one or two equations by a constant and
add to eliminate one variable
• Use the same process on two different
equations to eliminate the same variable
• Now find the solution to this system of two
equations in two variables as previously learned
• Find the number solution for the third variable by
substituting into any of the three original
equations
Example
3x + 9y + 6z = 3
2x + y – z = 2
x+ y+ z=2
Multiply second by 6 and add to first to eliminate
“z”:
3x + 9y + 6z = 3
12x + 6y – 6z = 12
15x + 15y
= 15
Also add second and third to eliminate “z”:
3x + 2y
= 4
Example Continued
Solve the system:
15x + 15y
= 15
3x + 2y
= 4
Multiply second by -5 and add to first:
15x + 15y
= 15
-15x – 10y
= -20
5y
= -5
y
= -1
Substitute into second equation at top to get:
3x + 2(-1)
= 4
3x = 6
x=2
Example Continued
Substitute both x = 2 and y = -1 into any of
the original equations (last):
x + y + z=2
(2) + (-1) + z = 2
1+z=2
z=1
Solution is:
(2, -1, 1)
Homework Problems
• Section:
• Page:
• Problems:
5.1
485
Odd: 47 – 57
• MyMathLab Assignment 5.1(c) for practice
• MyMathLab Quiz 5.1 is due for a grade on
the date of our next class meeting
Determinant Solution of Linear
Systems
• A “matrix” is a rectangular arrangement of
numbers
• A matrix is designated as being an “m x n matrix”
with “m” representing the number of rows, and
“n” representing the number of columns
• Examples:
A 2 x 2 matrix has two rows and two columns
A 3 x 4 matrix has three rows and four columns
Determinant of n x n Matrix
• Every matrix, A, that has an equal number
of rows and columns has a real number
associated with it called its “determinant”
• The determinant of a matrix, A, is indicated
by the symbol: |A|
• The way in which the determinant is found
depends on the number of rows and
columns
Determinant of a 2 x 2 Matrix
• The determinant of a 2 x 2 matrix:
a b
c d
is defined as: ad – bc
• Example:
If matrix B is:
4 -2
3 5 then |B| is:
(4)(5) – (3)(-2) = 20 + 6 = 26
Solving a System of Two Linear
Equations in Two Variables using
Cramer’s Rule
• Write each equation in standard form
• Make a 2 x 2 “coefficient” matrix by taking the coefficients of “x” as
the first column and the coefficients of “y” as the second column
• Call the determinant of this “coefficient matrix” by the name “D”
• Make an “X” 2 x 2 matrix by substituting in the “coefficient” matrix
the constants on the right side of the equal sign for the “x”
coefficients and find the determinant of this matrix, Dx
• Make a “Y” 2 x 2 matrix by substituting in the “coefficient” matrix the
constants on the right side of the equal sign for the “y” coefficients
and find the determinant of this matrix, Dy
• The solution to the equation will be:
x = Dx/D and y = Dy/D
except in the case where D = 0 when this method fails
Example of Solving a System by
Cramer’s Rule
2x – y = 7
3x + 7y = 2
Coefficient Matrix:
2
-1
3
7
D = (2)(7) – (3)(-1) = 17
“X” Matrix:
7
-1
2
7
Dx = (7)(7) – (2)(-1) =49 + 2 = 51
“Y” Matrix:
2
7
3
2
Dy = (2)(2) – (3)(7) = 4 – 21 = -17
Example Continued
From previous page:
D = 17, Dx = 51, and Dy = -17
x = Dx / D
x = 51 / 17 = 3
y = Dy / D
y = -17 / 17 = -1
Solution for System: (3, -1)
Cramer’s Rule Applied
to “Larger Systems”
• Cramer’s Rule can be applied to systems
containing “n” equations and “n” variables,
but will not be discussed at this time.
Homework Problems
• Section:
• Page:
• Problems:
5.1
484
Odd: 7 – 17
(Solve by Cramer’s Rule)
• No MyMathLab Assignment
Solving Systems of Equations
by Gauss-Jordan Method
• Must have same number of equations as
variables
• Put each equation in standard form
(variable terms in alphabetical order on left
side of = sign and constants on right side)
• Make an “augmented matrix” consisting of
a rectangular arrangement of variable
coefficients, a vertical line, and constants
as appear on the right side of the = sign
Gauss-Jordan Method Continued
• With a goal of achieving an arrangement
of “ones” down the diagonal and “zeros” in
all other positions on the left side of the
vertical line in the augmented matrix:
– Interchange any two rows
– Multiply numbers in any row by any nonzero
number
– Replace any row by adding to its numbers the
multiples of the numbers of another row
Gauss-Jordan Method Continued
• When “ones” have been obtained down
the diagonal from upper left to lower right,
and “zeros” are in all other positions, the
solutions to the system will be seen in
alphabetical order on the right side of the
vertical bar arranged from top to bottom
Example of Gauss-Jordan Method
Applied to System with 2 Variables
2x + 3y = -1
5x – 2y = 26
• Augmented Matrix:
2
3 1
5
2
26
1
5
3
2
2
1

2
26
R1  2
- 5R1  R2
Example Continued
1
0
1
0
3
2
19

2
1

2
57
2
3
2
1
1

2
3
2
- R2
19
3
- R2  R1
2
Example Continued
1
0
0
1
Solution :
4
3
4, - 3
Example of Gauss-Jordan Method
Applied to System with 3 Variables
x + y– z= 0
2x + 3y + z = 5
-3x – y +2z = -4
Augmented Matrix:
1
2
-3
1
3
-1
-1
1
2
0
5
-4
Example Continued
1
2
-3
1
3
-1
-1
1
2
0
5
-4
1
0
0
1
1
2
-1
3
-1
0
5
-4
- 2R1  R2
3R1 R3
- 1R2  R1
- 2R2  R3
Example Continued
1
0
0
0
1
0
-4
3
-7
-5
5
-14
1
0
0
0
1
0
-4
3
1
-5
5
2
R3  (-7)
4R3  R1
- 3R3  R2
Example Continued
1
0
0
0
1
0
0
0
1
3
-1
2
Solution to system:
x = 3, y = -1, z = 2
Correctly written as an ordered triple:
(3, -1, 2)
Homework Problems
• Section:
• Page:
• Problems:
5.1
484
Odd: 19 – 25, 47 – 53
(Solve by Gauss-Jordan)
• No MyMathLab Assignment
Direct Variation
• We say that y varies directly as x (or y is directly
proportional to x) if there is a constant, k, such
that:
y = kx
• To solve a problem involving direct variation, the
key is to find the value of k, from knowing one
pair of (x, y) values
• Once the k value is determined, it establishes a
formula, that can then find a value of y for any
value of x
Direct Variation Example
• Assuming that the pressure on an object
underground varies directly as its depth,
and the pressure is 5 psi when the depth
is 10 feet, what would the pressure be at a
depth of 30 feet?
p  kd
5  k10
1
k
2
1
p d
2
1
p  30  15 psi
2
Inverse Variation
• We say that y varies inversely as x (or y is
inversely proportional to x) if there is a constant,
k, such that:
k
y
x
• To solve a problem involving inverse variation,
the key is to find the value of k, from knowing
one pair of (x, y) values
• Once the k value is determined, it establishes a
formula, that can then find a value of y for any
value of x
Inverse Variation Example
• Assuming that the illumination of light on
an object varies inversely as the square of
its distance from the object, and the
illumination is 50 candela at 5 meters,
what would the illumination be at 20
meters?
1250
1250
k
I 2
D
k
50  2
5
2550  k
1250  k
I
D
2
1250
I
20 2
I
400
I  3.125
Joint Variation
• Joint variation occurs when y varies with multiple
variables, directly with some and inversely with others
• In such cases, we write y as being equal to a constant, k,
times the product of the other variables or inverses of the
other variables, as appropriate
• To solve a problem involving joint variation, the key is to
find the value of k, from knowing the value of one set of
variables
• Once the k value is determined, it establishes a formula,
that can then find a value of y given the values of the
other variables
Joint Variation Example
• Assuming that the gravitational attraction of an
object varies directly as its mass and inversely
as the square of the distance from the center of
the mass, and the gravitational attraction is 33
newtons at 21 meters from the center when the
mass is 2079 kilograms. What would the
gravitational attraction be for the same mass at a
distance of 3 meters from the center?
kM
G 2
D
k 2079
33 
212
3321
2
2079
k
7k
7M
G 2
D
.
72079
G
 1617 N
2
3
Homework Problems
• Section:
• Page:
• Problems:
3.6
373
Odd: 11 – 41
• No MyMathLab Assignment