Transcript a - x

4.1 Objective:
Students will look at polynomial functions of
degree greater than 2, approximate the
zeros, and interpret graphs.
Remember integers are … –2, -1, 0, 1, 2 … (no decimals
or fractions) so positive integers would be 0, 1, 2 …
A polynomial function is a function of the form:
n must be a positive integer
f  x   an x  an 1 x
n
n 1
   a1 x  ao
All of these coefficients are real numbers
The degree of the polynomial is the largest
power on any x term in the polynomial.
Determine which of the following are polynomial
functions. If the function is a polynomial, state its
degree.
f x   2 x  x
4
g x   2 x 0
h x   2 x  1
3
2
F x    x
x
A polynomial of degree 4.
We can write in an x0 since this = 1.
A polynomial of degree 0.
Not a polynomial because of the
square root since the power is NOT
1
an integer
x  x2
Not a polynomial because of the x in
the denominator since the power is
1
1
negative
x
x
Graphs of polynomials are smooth and continuous.
No sharp corners or cusps No gaps or holes, can be drawn
without lifting pencil from paper
This IS the graph
of a polynomial
This IS NOT the graph
of a polynomial
Let’s look at the graph of
even integer.
g x   x 4
f x   x
2
Notice each graph
looks similar to x2
but is wider and
flatter near the
origin between –1
and 1
f x   x
n
where n is an
hx   x 6
and grows
steeper on either
side
The higher the
power, the flatter
and steeper
Let’s look at the graph of
integer.
Notice each graph
looks similar to x3
but is wider and
flatter near the
origin between –1
and 1
f x   x
g x   x 5
n
where n is an odd
and grows
steeper on
either side
hx   x 7
f x   x 3
The higher the
power, the flatter
and steeper
Let’s graph
f x    x  2
Reflects about
the x-axis
4
Looks like x2
but wider near
origin and
steeper after 1
and -1
So as long as the function
is a transformation of xn,
we can graph it, but what if
it’s not? We’ll learn some
techniques to help us
determine what the graph
looks like in the next slides.
Translates up 2
LEFT
and
RIGHT
HAND BEHAVIOUR OF A GRAPH
The degree of the polynomial along with the sign of the
coefficient of the term with the highest power will tell us
about the left and right hand behaviour of a graph.
Even degree polynomials rise on both the left and
right hand sides of the graph (like x2) if the coefficient
is positive. The additional terms may cause the
graph to have some turns near the center but will
always have the same left and right hand behaviour
determined by the highest powered term.
left hand
behaviour: rises
right hand
behaviour: rises
Even degree polynomials fall on both the left and
right hand sides of the graph (like - x2) if the
coefficient is negative.
turning points
in the middle
left hand
behaviour: falls
right hand
behaviour: falls
Odd degree polynomials fall on the left and rise on
the right hand sides of the graph (like x3) if the
coefficient is positive.
turning Points
in the middle
left hand
behaviour: falls
right hand
behaviour: rises
Odd degree polynomials rise on the left and fall on
the right hand sides of the graph (like x3) if the
coefficient is negative.
turning points
in the middle
left hand
behaviour: rises
right hand
behaviour: falls
A polynomial of degree n can have at most n-1 turning
points (so whatever the degree is, subtract 1 to get
the most times the graph could turn).
Let’s determine left and right hand behaviour for the
graph of the function:
doesn’t mean it has that many
4
3 turning2points but that’s the
f x  x  3x most
15 xit can
 19
x

30
have

degree is 4 which is even and the coefficient is positive so the
graph will look like x2 looks off to the left and off to the right.
The graph can
have at most 3
turning points
How do we
determine
what it looks
like near the
middle?
x 23xx 315
xx 119xx 30
0f x x 
5
4
3
2
x and y intercepts would be useful and we know how
to find those. To find the y intercept we put 0 in for x.
f 0  0  30  150  190  30  30
4
3
2
To find the x intercept we put 0 in for y.
Finally we need a smooth
curve through the
intercepts that has the
correct left and right hand
behavior. To pass through
these points, it will have 3
turns (one less than the degree
so that’s okay)
(0,30)


f
x

x

3
x

15
x

19
x

30
0  x  2 x  3 x 1 x  5

4

3

2


We found the x intercept by putting 0 in for f(x) or y (they
are the same thing remember). So we call the x intercepts
the zeros of the polynomial since it is where it = 0. These
are also called the roots of the polynomial.
4.2 Dividing
Polynomials
© 2002 by Shawna Haider
4.2 Properties of Division
Objective: We will learn how to use synthetic
division which will help us in finding
imaginary roots of polynomial functions.
Dividing by a Monomial
If the divisor only has one term, split the polynomial up into a
fraction for each term.
18 x 4  24 x 3  6 x 2  12 x 18 x 4 24 x 3 6 x 2 12 x




6x
6x
6x
6x
6x
divisor
Now reduce each fraction.
3x3
4x2
4
x
3
2
2
18 x
24 x 6 x 12 x




6x
6x
6x
6x
1
1
1
1
 3x  4 x  x  2
3
2
Long Division
If the divisor has more than one term, perform long division. You do the
same steps with polynomial division as with integers. Let's do two
problems, one with integers you know how to do and one with polynomials
and copy the steps.
28
Subtract
(which

Now multiply by
x + 11
21
x3
changes
the
sign
Bring
down
the
theMultiply
divisor and
put
and
2 + 8x - 5
32 698
x
3
x
of
each
term
in
next
or
the
answer
below.
putnumber
below
Remainder
2 – 3x
64
x
the
polynomial)
term
subtract
added here
58 3 into 6
First divide
or
x into x2 11x - 5
over divisor
32
Now divide
is the x into 11x 11x - 33
26 3 into 5 This or
remainder
28
So we found the answer to the problem x2 + 8x – 5  x – 3
or the problem written another way: x 2  8 x  5
x 3
Let's Try Another One
If any powers of terms are missing you should write them in with
zeros in front to keep all of your columns straight.
12
Subtract
(which

y-2
Write out with
y2
changes
the
sign
Multiply
and
Bring
Multiply
down
and
the 2 y + 2 y2 + 0y + 8
long
division
2
Divide
Divide
yyinto
intoiny-2y
of
each
term
put
below
next
put
term
below
including 0y for
Remainder
2 + 2y
y
the
polynomial)
subtract
missing term
added here
y 8
y2
This is the
remainder
-2y + 8 over divisor
- 2y - 4
12
Synthetic Division
There is a shortcut for long division as long as the divisor is x – k
where k is some number. (Can't have any powers on x).
Set divisor = 0 and
3
2
1 x  6 x  8x  2
solve. Put answer
here.
x3
x + 3 = 0 so x = - 3
-3
1
6
8
-2
up3these
Bring
number
down
below
Addupthese
line up
- 3 firstAdd
- 9theseAdd
Multiply
Multiply
these
these and
and
2 +3 x - 1
This is the remainder
1
x
1
put
answer
put answer
above
line
above line
Put
variables
back
in (one
was of
divided
outthe
in
Sonext
the
Listanswer
all
coefficients
is:
(numbers
in xfront
x's) and
in
in next process so first number is one less power than
2 top. If a term is missing, put in a 0.
constant along the
column
column original problem).
1
x  3x  1 
x3
Let's try another Synthetic Division
Set divisor = 0 and
solve. Put answer
here.
4
1
0 x3
0x
1 x  4x  6
4
2
x4
x - 4 = 0 so x = 4
0
-4
0
6
up48
Bring
number
down
these
below
Add
upthese
line
Add
up these up
4 firstAdd
16theseAdd
192
Multiply
Multiply
Multiply
these
these and
and
3 + 4 x2 + 12 x + 48 198
This is the
these
and
1
x
put
answer
put answer
remainder
put
answer
above
line
above line Now put variables back in (remember one x was
above
lineanswer
Sonext
the
List
all coefficients
is:
(numbers in front of x's) and the
in
in next divided out 3in process2so first number is one less
in next
constant along the top. Don't forget the 0's for missing
column
column power than original problem so x3).
column
terms.
198
x  4 x  12 x  48 
x4
Let's try a problem where we factor the polynomial
completely given one of its factors.
4 x 3  8 x 2  25 x  50
-2
4
factor : x  2
You want to divide
the factor into the
polynomial so set
divisor = 0 and solve
for first number.
8 -25 -50
up
Bring
number
down
below
Addupthese
line up
- 8 firstAdd
0theseAdd
50these
Multiply
Multiply
these
No remainder so x + 2
these and
and
2
4 x + 0 x - 25
0
put
IS a factor because it
put answer
answer
above
line
divided in evenly
above line
Put
variables
back
in
(one
x
was
divided
outthe
in
Sonext
the
Listanswer
all coefficients
is the divisor
(numbers
times in
thefront
quotient:
of x's) and
in
in next
process
sothe
first
number
is one
less power
You could
check
this
byIf a term
constant
along
top.
is missing,
putthan
in a 0.
column
2
column
original
problem).
multiplying
them out
and getting
original polynomial
x  24 x
 25

4.3 Zeros of Polynomials
Objective:
Students will find the zeros of a polynomial
and be able to discuss how many zeros are
real & imaginary, positive & negative,
rational & irrational and large or small in
value.
4.3 Zeros of Polynomials
Can you find the zeros
of the polynomial?
g ( x)  x  1 x  2 x  3
3
2
There are repeated factors. (x-1) is to the 3rd power so it
is repeated 3 times. If we set this equal to zero and solve
we get 1. We then say that 1 is a zero of multiplicity 3
(since it showed up as a factor 3 times).
What are the other
zeros and their
multiplicities?
-2 is a zero of multiplicity 2
3 is a zero of multiplicity 1
So knowing the zeros of a polynomial we can plot them on
the graph. If we know the multiplicity of the zero, it tells us
whether the graph crosses the x axis at this point (odd
multiplicities CROSS) or whether it just touches the axis
and turns and heads back the other way (even multiplicities
TOUCH). Let’s try to graph:
f x   x  1x  2
2
What would the left and
right hand behavior be?
You don’t need to multiply this out but figure out what the
highest power on an x would be if multiplied out. In this
case it would be an x3. Notice the negative out in front.
What would the y
intercept be?
Find the zeros and
their multiplicity
(0, 4)
1 of mult. 1
(so crosses axis
at 1)
-2 of mult. 2
(so touches at 2)
Steps for Graphing a Polynomial
•Determine left and right hand behaviour by looking at
the highest power on x and the sign of that term.
•Determine maximum number of turning points in graph by
subtracting 1 from the degree.
•Find and plot y intercept by putting 0 in for x
•Find the zeros (x intercepts) by setting polynomial = 0 and
solving.
•Determine multiplicity of zeros.
•Join the points together in a smooth curve touching or
crossing zeros depending on multiplicity and using left and
right hand behavior as a guide.
Let’s graph:
f x   x x  3x  4
2
•Determine
left
and right
hand
behavior
by
looking
atand
•Find
•Determine
and
plot
y
maximum
intercept
number
by
putting
of
turns
0
in
for
in
x
graph
by
•Find
•Join
the
the
points
zeros
(x
together
intercepts)
in
a
smooth
by
setting
curve
polynomial
touching
=
or
0
•Determine
multiplicity
of xzeros.
0sign
multiplicity
2 (touches)
the
highest
power
on
and
the
of
that
term.
subtracting
the degree.
solving.
crossing
zeros1depending
on multiplicity
and using
left and
2from
2
3
multiplicity
1
(crosses)
0behaviour
 x out,
x as
3
x

4
Zeros
are:
0,
4 3, -4
right Multiplying
hand
a
guide.
highest
power
would
be
x
-4 multiplicity
Degree is 4 so maximum number
of turns1is(crosses)
3
f 0  0 0  30  4  0
Here is the actual graph. We did pretty good. If we’d wanted to be
more accurate on how low to go before turning we could have
plugged in an x value somewhere between the zeros and found the y
value. We are not going to be picky about this though since there is a
great method in calculus for finding these maxima and minima.
What is we thought backwards? Given the
zeros and the degree can you come up with a
polynomial? Find a polynomial of degree 3
that has zeros –1, 2 and 3.
What would the function look like in factored form to have
the zeros given above?
f x  x  1x  2x  3
Multiply this out to get the polynomial. FOIL two of them
and then multiply by the third one.
f x   x  4 x  x  6
3
2
4.4 Complex and Rational Zeros of
Polynomials
Objective:
In this lesson, we will:
• Learn that imaginary roots always come in
pairs and those pairs are conjugates
• Find the possible and actual rational zeros
of a polynomial function
4.4 Complex and
Rational Zeros of
Polynomials
Using the Rational Root Theorem
to Predict the Rational Roots of a
Polynomial
Find the Roots of a Polynomial
For higher degree polynomials, finding the complex
roots (real and imaginary) is easier if we know one
of the roots.
The table below can help get you started. Complete
the table below:
Polynomial
y  x 4  2x2  3
3
2
y  x  7x 17x 15
4
3
2
y  3x  x  3x  x 1
# + Real # - Real # Imag.
Roots
Roots
Roots
The Rational Root Theorem
The Rational Root Theorem gives us a tool to predict
the Values of Rational Roots:
If P(x)  a0 x n  a1 x n1  ...  an 1 x  an , where the
coeffiecients are all integers,
p
& a rational zero of P(x) in reduced form is
, then
q
 p must be a factor of an (the constant term) &
 q must be a factor of a0 (the leading coefficient).
List the Possible Rational Roots
For the polynomial:
All possible values of:
3
2
f (x)  x  3x  5x  15
p  1,3,5,15
q  1
All possible Rational Roots of the form p/q:
p
 1,3,5,15
q
Narrow the List of Possible Roots
For the polynomial:
Graph the polynomial:
3
2
f (x)  x  3x  5x  15
# + real zeros: 1
# - real zeros: 0
# imag. zeros: 2
All possible Rational Roots of the form p/q:
p
 1,3,5,15
q
Find a Root That Works
For the polynomial:
3
2
f (x)  x  3x  5x  15
Substitute each of our possible rational roots into f(x).
If a value, a, is a root, then f(a) = 0. (Roots are
solutions to an equation set equal to zero!)
f (1)  1 3  5  15  12
f (3)  27  27  15  15  0
f (5)  125  75  25  15  60
*
Find the Other Roots
Now that we know one root is x = 3, do the other two
roots have to be imaginary? What other category have
we left out?
To find the other roots, divide the factor that we know
into the original polynomial:

3
2
x  3 x  3x  5x  15
Find the Other Roots (con’t)
x2  5
3
2
x  3 x  3x  5x  15

The resulting polynomial is a quadratic, but it doesn’t
have real factors. Solve the quadratic set equal to zero
by either using the quadratic formula, or by isolating the
x and taking the square root of both sides.
Find the Other Roots (con’t)
The solutions to the
quadratic equation:
x  i 5, i 5
3
2
For the polynomial:
f (x)  x  3x  5x  15
The three complex
roots of the
polynomial are:
x  3, i 5,  i 5
4.5 Rational Functions
Objective:
In this lesson, we will:
• Sketch graphs of rational functions
• Look more in depth at the vertical &
horizontal asymptotes
• Find equations of rational functions
4.5 Rational Functions
and Their Graphs
Example
Find the Domain of this Function.
x7
f ( x) 
x3
Solution:
The domain of this function is the set of all real
numbers not equal to 3.
Arrow Notation
Symbol
Meaning
xa
x approaches a from the right.
xa
x
x approaches a from the left.
x approaches infinity; that is, x
increases without bound.
x
is, x
x approaches negative infinity; that
decreases without bound.
Definition of a Vertical Asymptote
The line x  a is a vertical asymptote of the graph of a function f if f(x)
increases or decreases without bound as x approaches a.
f (x)   as x  a 
f (x)   as x  a 
y
y
f
f
x
a
x=a
a
x
x=a
Thus, f (x)   or f(x)    as x approaches a from either the left or the
right.
Definition of a Vertical Asymptote
The line x  a is a vertical asymptote of the graph of a function f if f(x)
increases or decreases without bound as x approaches a.
f (x)  as x  a 
f (x)   as x  a 
y
y
x=a
x=a
x
a
f
a
x
f
Thus, f(x)   or f(x)    as x approaches a from either the left or the
right.
Locating Vertical Asymptotes
If
f ( x) 
p ( x)
q ( x)
is a rational function in which
p(x) and q(x) have no common factors and a is a
zero of q(x), the denominator, then x = a is a
vertical asymptote of the graph of f.
Definition of a Horizontal
Asymptote
The line y = b is a horizontal asymptote of the graph of a function f if f(x)
approaches b as x increases or decreases without bound.
y
y
y
y=b
y=b
f
f
x
f
y=b
x
x
f(x)  b as x  
f(x)  b as x  
f (x)  b as x 
Locating Horizontal Asymptotes
Let f be 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,t he graph of f has no horizontal
asymptote.
Strategy for Graphing a Rational Function
p ( x)
Suppose that f ( x)  q( x) where p(x) and q(x) are
polynomial functions with no common factors.
1. Find any vertical asymptote(s) by solving the equation
2.
3.
q (x)  0.
Find the horizontal asymptote (if there is one) using the
rule for determining the horizontal asymptote of a rational
function.
Use the information obtained from the calculators graph
and sketch the graph labeling the asymptopes.
Sketch the graph of
2x  3
f ( x) 
5 x  10
2x  3
f ( x) 
5 x  10
• The vertical asymptote is x = -2
• The horizontal asymptote is y = 2/5
2x  3
f ( x) 
5 x  10
10
8
6
4
2
-10 -8 -6 -4 -2
2
-2
-4
-6
-8
-10
4
6
8 10
Section 4.5
Rational Functions
• Find an equation of a rational function f that
satisfies
•
•
•
•
x-intercept : 4
Vertical asymptote: x=-2
Horizontal asymptote: y= -3/5
hole at x=1
Section 4.5
Rational Functions
X-intercept -- x-4 must be factor in numerator
(x-4)
Vertical asymptote-- x+2 is factor in denominator
x-4
x+2
Horizontal asymptote--Mult. Numerator by -3 and
denominator by 5
-3(x-4)
5(x+2)
Hole--x-1 must be factor in both num. & den.
f(x)= -3(x-4)(x-1)
5(x+2)(x-1)
Ch. 4 Review Answers