Transcript Polynomial Functions ppt.

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
most it can have
f x  x  3x 15x  19x  30
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?
xx 119xx 30
0f xx x 23xx 315
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)
x 23xx 315
xx 119xx 30
0f x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.
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
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au