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EVALUATING POLYNOMIAL FUNCTIONS
A polynomial function is a function of the form
f(x) = an x nn + an – 1 x nn – 11 +· · ·+ a 1 x + aa00
Where ann  00 and the exponents are all whole numbers.
For this polynomial function, aan is the leading coefficient,
coefficient
n
aa00 is the constant
constant term,
term and n is the degree.
degree
A polynomial function is in standard form if its terms are
descending order
order of
of exponents
exponents from
from left
left to
to right.
right.
written in descending
EVALUATING POLYNOMIAL FUNCTIONS
You are already familiar with some types of polynomial
functions. Here is a summary of common types of
polynomial functions.
Degree
Type
Standard Form
0
Constant
f (x) = a 0
1
Linear
f (x) = a1x + a 0
2
Quadratic
f (x) = a 2 x 2 + a 1 x + a 0
3
Cubic
f (x) = a 3 x 3 + a 2 x 2 + a 1 x + a 0
4
Quartic
f (x) = a4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0
Identifying Polynomial Functions
Decide whether the function is a polynomial function. If it is,
write the function in standard form and state its degree, type
and leading coefficient.
f (x) = 1 x 2 – 3x4 – 7
2
SOLUTION
The function is a polynomial function.
4
Its standard form is f (x) = – 3x +
1 2
x – 7.
2
It has degree 4, so it is a quartic function.
The leading coefficient is – 3.
Identifying Polynomial Functions
Decide whether the function is a polynomial function. If it is,
write the function in standard form and state its degree, type
and leading coefficient.
f (x) = x 3 + 3 x
SOLUTION
The function is not a polynomial function because the
x
term 3 does not have a variable base and an exponent
that is a whole number.
Identifying Polynomial Functions
Decide whether the function is a polynomial function. If it is,
write the function in standard form and state its degree, type
and leading coefficient.
–
f (x) = 6x 2 + 2x 1 + x
SOLUTION
The function is not a polynomial function because the term
2x –1 has an exponent that is not a whole number.
Identifying Polynomial Functions
Decide whether the function is a polynomial function. If it is,
write the function in standard form and state its degree, type
and leading coefficient.
f (x) = – 0.5x +  x 2 –
2
SOLUTION
The function is a polynomial function.
Its standard form is f (x) =  x – 0.5x –
2
It has degree 2, so it is a quadratic function.
The leading coefficient is .
2.
Identifying Polynomial Functions
Polynomial function?
f (x) = 12 x 2 – 3x 4 – 7
f (x) = x 3 + 3x
f (x) = 6x2 + 2x– 1 + x
f (x) = – 0.5x +  x2 –
2
Using Synthetic Substitution
One way to evaluate polynomial functions is to use
direct substitution. Another way to evaluate a polynomial
is to use synthetic substitution.
Use synthetic division to evaluate
f (x) = 2x 4 + -8x 2 + 5x - 7 when x = 3.
Using Synthetic Substitution
SOLUTION
2x 4 + 0x 3 + (–8x 2) + 5x + (–7)
Polynomial in
standard form
33•
2
0
–8
5
–7
Coefficients
6
18
30
105
6
10
35
98
x-value
2
The value of f (3) is the last number you write,
In the bottom right-hand corner.
GRAPHING POLYNOMIAL FUNCTIONS
The end behavior of a polynomial function’s graph
is the behavior of the graph as x approaches infinity
(+ ) or negative infinity (– ). The expression
x
+  is read as “x approaches positive infinity.”
GRAPHING POLYNOMIAL FUNCTIONS
END BEHAVIOR
GRAPHING POLYNOMIAL FUNCTIONS
CONCEPT
END BEHAVIOR FOR POLYNOMIAL FUNCTIONS
SUMMARY
x
+
+
f (x)
+
f (x)
–
f (x)
+
even
f (x)
–
f (x)
–
odd
f (x)
+
f (x)
–
an
n
x
>0
even
f (x)
>0
odd
<0
<0
–
Graphing Polynomial Functions
Graph f (x) = x 3 + x 2 – 4x – 1.
SOLUTION
To graph the function, make a table of
values and plot the corresponding points.
Connect the points with a smooth curve
and check the end behavior.
–2the leading
–1
0
1 is positive,
2
3
x
The degree
is–3odd and
coefficient
–3
23
+
+
(x)
as3x
so f (x)f(x) – –7 as x 3 – 3 and f–1
.
Graphing Polynomial Functions
Graph f (x) = –x 4 – 2x 3 + 2x 2 + 4x.
SOLUTION
To graph the function, make a table of
values and plot the corresponding points.
Connect the points with a smooth curve
and check the end behavior.
–3
0 coefficient
1
3
x
The degree
is even–2
and the–1leading
is2 negative,
f (x) ––21 as x0
––1 and0 f (x) 3 – –16
+
as x –105
so f (x)
.