Download Elementary Functions Definition of a polynomial Definition of a

Definition of a polynomial
A polynomial of degree n is a function of the form
f (x) = an xn + an−1 xn−1 + ...a2 x2 + a1 x + a0
Elementary Functions
where n is a nonnegative integer and the elements an , an−1 , ..., a2 , a1 , a0
are real numbers.
Part 2, Polynomials
Lecture 2.2a, Polynomials & Their Graphs
The integer n (the highest exponent on x) is the degree of the polynomial.
Dr. Ken W. Smith
The real numbers an , an−1 , ..., a2 , a1 , a0 are called coefficients of the
polynomial.
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The real number an is the leading coefficient of the polynomial.
The constant term is a0 ; it corresponds to the y-intercept of f (x).
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Definition of a polynomial
Polynomials are continuous and smooth
Example. The polynomial
Since f (x) does not involve square roots of the variable x, nor does it
have denominators involving the variable x, then the domain of a
polynomial is the entire real line: (−∞, ∞).
√
13
x + 23
7
has degree five, with leading coefficient 2 and constant term 23.
f (x) = 2x5 + πx3 +
3x2 −
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Graphs of polynomials are particularly nice.
They are continuous, without holes or gaps.
The simplest polynomials are the constant functions.
f (x) = a0
(whose graphs are straight lines) and the linear functions
f (x) = a1 x + a0 .
We have already looked at these, along with the functions of degree two,
the quadratics
f (x) = a2 x2 + a1 x + a0 .
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Polynomials are continuous and smooth
Polynomials are continuous and smooth
Continuous functions such as polynomials cover all y-values intermediate”
to f (a) and f (b).
Here is a picture (from Wikipedia) displaying this relationship.
Because the graph of a polynomial is continuous, it obeys the
Intermediate Value Theorem
This means that if the function takes on a particular y-value in one place
and a different y-value in another place, then the function takes on all
possible y-values between the two.
More explicitly, suppose a and b are two real numbers with f (a) < f (b).
Then given any real number u between f (a) and f (b), there is an x-value
c between a and b such that f (c) = u.
Continuous functions such as polynomials cover all y-values between f (a)
and f (b) (“intermediate” to f (a) and f (b).)
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Polynomials are continuous and smooth
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The following properties of a polynomial f (x) should be visible in the
The
green lines represent the y-values f (a) and f (b) and the
graphhorizontal
of y = f (x):
graph in blue covers all the y-values between f (a) and f (b), such as the
1 The domain is all real numbers, (−∞, ∞).
y-value u = f (c).
2 The function is continuous.
3
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2013
Summary
Graphs of polynomials are also “smooth”. They have no sharp corners or
cusps.
In the picture below, the graph on the left has a sharp corner at (1, 1).
The graph on the right has a cusp at the origin.
Neither of these graphs could be the graph of a polynomial.
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The function is “smooth.”
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End-behavior of the graphs of polynomials
End-behavior
Consider the simplest polynomials, the so-called power functions like
f (x) = x, f (x) = x2 , f (x) = x3 , f (x) = x4 , f (x) = x5 , ...
All of these function have a form like that of f (x) = x3 or f (x) = x4 .
If the degree of the polynomial is even but the leading coefficient is
negative then the end-behavior mimics that of a power function reflected
across the x-axis. The end behavior should look like the reflection, that is,
it will be
.&
If the exponent on a power function is even, then the y-values go to +∞
whether x is going to −∞ or ∞.
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End-behavior
On the other hand, if the graph is y = x3 or a similar graph where the
exponent on x is to an odd power, such as
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Turning points of a polynomial of degree n
For a general polynomial, the leading term will begin to dominate the
graph as x grows in absolute value (as x moves far away from the y-axis).
So, ultimately the graph of y = 2x5 + 23x4 − 77x3 + 2x2 − 100x + 40
5.
will
lookthe
likebehavior
the graph
of y = x(or
We say
at infinity
the “end behavior”) of the polynomial is
The end behavior of the fifth degree polynomial
-%
f (x) = 2x5 + 23x4 − 77x3 + 2x2 − 100x + 40 is the end behavior of x5 :
mimicking
the action of the graph away from the x-axis.
.
%.
A polynomial of even degree and positive leading coefficient (such as
f (x) = 3x6 + 2x − 7) has end behavior - % .
Since
it drops
from theisleft as we get close to the y-axis and then rises far
then the
end behavior
off to the right, it must turn around
.an
%odd number of times.
mimicking the action of the graph far away from the y-axis.
The local maximums and minimums, where the graph changes direction,
are called turning points and the number of turning points gives us some
clue to the degree of the polynomial.
But if the leading coefficient is negative then the end behavior of a
polynomial of odd degree is flipped over: - &.
For example, the graph of y = −2x5 + 23x4 − 77x3 + 2x2 − 100x + 40
rises off to the left of the y-axis and will drop off to the right of the y-axis.
In particular, the number of turning points is always less than the degree.
When x is large in absolute value, the polynomial
f (x) = −2x5 + 23x4 − 77x3 + 2x2 − 100x + 40 begins to look a lot like
−2x5 .
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Turning points of a polynomial of degree n
Turning points of a polynomial
The number of turning points is always less than the degree.
For example, the graph of the degree four polynomial
f (x) = x4 − 3x2 + x + 1 (below) crosses the x-axis four times (near
x = −1.7, x = −0.4, x = 1, x = 1.25) and has three obvious turning
points, around x = −1.3, x = 0.2, and x = 1.1.
Sometimes a pair of turning points can merge and disappear.
If we take the coefficient of x2 in the previous example
f (x) = x4 −3x2 + x + 1
and change −3 to −2
f (x) = x4 −2x2 + x + 1
or even change it to −1
f (x) = x4 −x2 + x + 1
a pair of turning points eventually disappear.
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Turning points of a polynomial
Turning points of a polynomial
Watch two turning points disappear.
Watch two turning points disappear.
The graph of
The graph of
f (x) =
x4 −3x2
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f (x) = x4 −2x2 + x + 1
+x+1
has three clear turning points (around x = −1.5, x = 0.25, x = 1.2)
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has three turning points (around x = −1, x = 0.25, x = 0.8)
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Turning points of a polynomial
Turning points of a polynomial
Watch two turning points disappear.
One more change, removing x2 all together...
The graph of
The graph of
f (x) = x4 −x2 + x + 1
f (x) = x4 +0x2 + x + 1
has one turning point (around x = −1)
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has one turning point (around x = −0.75)
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Turning points of a polynomial
Polynomials
Let’s see these again, three turning points of a degree 4 polynomial
softening into just one turning point.
Exercise. Consider the graph of a polynomial, below.
How many turning points does it have? 5
What do you think is the degree of this polynomial? 6
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f (x) =
x4 −3x2 + x + 1
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Polynomials
Elementary Functions
In the next lesson, we explore the Fundamental Theorem of Algebra and
also look at “sign diagrams.”
Part 2, Polynomials
Lecture 2.2b, The Fundamental Theorem of Algebra & Sign Diagrams
(END)
Dr. Ken W. Smith
Sam Houston State University
2013
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A first look at the Fundamental Theorem of Algebra
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Fundamental Theorem of Algebra
The x-intercepts of the graph of a polynomial f (x) are called the “zeroes”
(or “roots”) of the polynomial. They are the x-values for which f (x) = 0.
It is easy to create a polynomial with prescribed zeroes.
Suppose we wanted a polynomial with zeroes at x = −2, x = 1, x = 3 and
x = 4.
Just multiply x + 2 times x − 1 times x − 3 times x − 4.
f (x) = (x + 2)(x − 1)(x − 3)(x − 4)
If we evaluate this function at x = −2 then the first term is zero and so
f (−2) is zero. If we evaluate this function at x = 1 then the second term
is zero. And so on. Here is the graph of y = f (x).
One might observe that if we wanted four different zeroes (such as
x = −2, 1, 3 and 4 in this case) then the polynomial should have degree 4.
There is a vague sense in which the number of zeroes is the meaning of
degree.
We might hope that a polynomial of degree n has n zeroes.
This is almost true. We will elaborate on this more in a later lesson. Here
is a first draft of the Fundamental Theorem of Algebra.
Fundamental Theorem of Algebra (first version):
A polynomial of degree n has at most n zeroes.
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Fundamental Theorem of Algebra
Fundamental Theorem of Algebra
We can often find n zeroes if we are willing to count some zeroes more
than once.
Here are the graphs of
x4 − 4x2 + 3 = (x − 1)(x + 1)(x −
2 − 1)(x2 − 3) has four zeroes,
The polynomial x4 − 4x2 + 3 = (x√
√
occurring at x = −1, x = 1, x = − 3 and x = 3.
√
3)(x −
√
3)
&
x4 − 4x2 = (x − 0)(x − 0)(x − 2)(x + 2).
But if we alter the polynomial a little, dropping the constant term, we have
x4 − 4x2 = x2 (x2 − 4) = x2 (x − 2)(x + 2) = (x − 0)(x − 0)(x − 2)(x + 2).
This has zeroes at x = 2, −2 and 0. The zero at x = 0 occurs because of
the factor x2 ; we should count that zero twice.
We will say that f (x) = x4 − 4x2 = (x − 0)(x − 0)(x − 2)(x + 2) has
zeroes −2, 0, 0, 2.
If we count the zero x = 0 twice the number of zeroes is equal to the
degree.
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Two worked problems
1
2
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The sign diagram for a polynomial
Give a polynomial of degree 3 with roots (zeroes) x = 0, x = 1, x = 3.
Solution. All solutions will have the form a x(x − 1)(x − 3) where
a is some real number.
One answer is simply x(x − 1)(x − 3).
Give the polynomial of degree 3 with roots x = 0, x = 1, x = 3
passing through the point (−1, 16).
Solution Polynomials with roots x = 0, x = 1, x = 3 will have the
form ax(x − 1)(x − 3) where a is some real number. We need to find
a.
Substitute x = −1 into the expression f (x) = ax(x − 1)(x − 3) to
see that f (−1) = −8a.
A convenient aid to graphing a polynomial is to locate the zeroes of the
polynomial and then draw a “sign diagram.” A sign diagram keeps up with
the sign (+/-) of the polynomial in the regions between the zeroes.
Consider the polynomial
g(x) = −2(x − 1)2 (x + 3)(x − 4).
This polynomial has zeroes at x = 1, x = −3 and x = 4. In order, from
smallest to largest, these zeroes are −3, 1, and 4.
The IVT assures us that the only way the graph of the polynomial g(x)
crosses the x-axis is at a zero, so in each region between the zeroes,
(−∞, −3), (−3, 1), (1, 4), and (4, ∞),
The polynomial we are after has f (−1) = 16 so a = −2.
the polynomial has a particular sign; positive or negative.
Answer: −2x(x − 1)(x − 3)
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The sign diagram for a polynomial
The sign diagram for a polynomial
Let’s do the sign diagram again, a little faster.... Again, our function is
g(x) = −2(x − 1)2 (x + 3)(x − 4).
g(x) = −2(x − 1)2 (x + 3)(x − 4).
Visualize the zeroes x = −3, 1, 4 as fences, separating the regions
Visualize the zeroes x = −3, 1, 4 as fences, separating the regions
(−∞, −3), (−3, 1), (1, 4), (4, ∞).
(−∞, −3), (−3, 1), (1, 4), (4, ∞).
The sign diagram gives a sign, positive or negative, to each of these
regions.
(−)
|
−3
(+)
|
1
(+)
|
4
The sign diagram gives a sign, positive or negative, to each of these
regions.
(−)
(−)
|
−3
(+)
|
1
(+)
To the left of x = −3 we test x = −4 find the sign of g(−4).
g(−4) = −2(−5)2 (−1)(−8) is the product of five negative numbers.
Minus signs cancel in pairs to give a negative number.
If x is less than −3 then g(x) is negative.
Between x = −3 and x = 1, pick a nice number: x = 0 is the best!
Compute the sign of g(0). g(0) = −2(−1)2 (3)(−4) is positive.
If x is between 1 and 4 then g(x) is positive.
To the right of x = 4 pick a number, say x = 5 and find the sign of g(x).
The
sign diagram for a polynomial
It is negative in this case, due to the leading coefficient −2.
|
−3
(+)
|
1
(+)
|
4
(−)
From this diagram, we know that as x approaches −3 from the left, the
graph of g(x) rises to the x-axis and passes through the x-axis at x = −3,
then stays above the x-axis until x = 1 when it drops back to the axis,
kisses the x-axis and bounces back up, staying above the x-axis until
x = 4 when it passes through the x-axis and drops below it as x continues
to the right.
Below, is the true graph of y = g(x).
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(−)
If x is between −3 and 1 then g(x) is positive.
Between x = 1 and x = 4, pick a number like x = 2 and compute the sign
of g(2).
Here g(2) = −2(1)2 (5)(−2)
is positive.
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(−)
|
4
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If x is greater than 4 then g(x) is negative.
The sign
diagram is a nice aid Elementary
to graphing.
(Next slide.)
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A worked problem.
Let’s finish our analysis of the polynomial
g(x) = −2(x − 1)2 (x + 3)(x − 4), above. Here are some typical questions
one might be asked about g(x).
1
Describe the end behavior of the graph of y = g(x).
Solution. This is a fourth degree polynomial with leading coefficient
negative. So the end behavior is . & .
2
Find the real zeroes of g(x).
Solution. x = 1 (twice), x = −3, x = 4.
3
Find the y-intercepts of g(x).
Solution. (0, 24).
4
Determine the maximal number of turning points of the graph of
y = g(x).
Solution. Since the polynomial has degree four then it has at most
three turning points.
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A worked problem.
5
Draw the sign diagram of y = g(x) and then sketch the graph.
Solution. The sign diagram is
(−)
|
−3
(+)
|
1
(+)
|
4
(−)
The graph is
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In the next presentation, we explore the zeroes of polynomials.
(END)
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