Five-Minute Check (over Lesson 2-1)
Then/Now
New Vocabulary
Example 1: Graph Transformations of Monomial Functions
Key Concept: Leading Term Test for Polynomial End Behavior
Example 2: Apply the Leading Term Test
Key Concept: Zeros and Turning Points of Polynomial Functions
Example 3: Zeros of a Polynomial Function
Key Concept: Quadratic Form
Example 4: Zeros of a Polynomial Function in Quadratic Form
Example 5: Polynomial Function with Repeated Zeros
Key Concept: Repeated Zeros of Polynomial Functions
Example 6: Graph a Polynomial Function
Example 7: Real-World Example: Model Data Using Polynomial Functions
Over Lesson 2-1
A. Graph f(x) = 3x 3.
A.
C.
B.
D.
Over Lesson 2-1
B. Analyze f(x) = 3x 3.
A.
D = (–∞, ∞) R = [0, ∞), intercept: 0,
,
continuous for all real numbers, decreasing: (–∞, 0) ,
increasing: (0, ∞)
B.
D = (–∞, ∞) R = (–∞, ∞), intercept: 0,
,
continuous for all real numbers, decreasing: (–∞, 0),
increasing: (0, ∞)
C.
D = (–∞, ∞) R = (–∞, ∞), intercept: 0,
,
continuous for all real numbers, decreasing: (–∞, ∞)
D.
D = (–∞, ∞) R = (–∞, ∞), intercept: 0,
continuous for all real numbers, increasing: (–∞, ∞)
,
Over Lesson 2-1
Solve
A. 7
B. 3, 7
C.
D. no solution
.
Over Lesson 2-1
Solve
A. –14
B. –2
C. 88
D. no solution
.
Over Lesson 2-1
A. –3, 4
B. 3
C. 6, 1
D. 6
You analyzed graphs of functions. (Lesson 1-2)
• Graph polynomial functions.
• Model real-world data with polynomial functions.
• polynomial function
• quadratic form
• polynomial function of degree n • repeated zero
• leading coefficient
• leading-term test
• quartic function
• turning point
• multiplicity
Graph Transformations of Monomial
Functions
A. Graph f(x) = (x – 3)5.
This is an odd-degree function, so its graph is similar
to the graph of y = x 3. The graph of f(x) = (x – 3)5 is
the graph of y = x 5 translated 3 units to the right.
Answer:
Graph Transformations of Monomial
Functions
B. Graph f(x) = x 6 – 1.
This is an even-degree function, so its graph is similar
to the graph of y = x 2. The graph of f(x) = x 6 – 1 is the
graph of y = x 6 translated 1 unit down.
Answer:
Graph f(x) = 2 + x 3.
A.
C.
B.
D.
Apply the Leading Term Test
A. Describe the end behavior of the graph of
f(x) = 3x 4 – x 3 + x 2 + x – 1 using limits. Explain
your reasoning using the leading term test.
The degree is 4, and the leading coefficient is 3.
Because the degree is even and the leading
coefficient is positive,
.
Answer:
Apply the Leading Term Test
B. Describe the end behavior of the graph of
f(x) = –3x 2 + 2x 5 – x 3 using limits. Explain your
reasoning using the leading term test.
Write in standard form as f(x) = 2x 5 – x 3 – 3x 2. The
degree is 5, and the leading coefficient is 2. Because
the degree is odd and the leading coefficient is
positive,
.
Answer:
Apply the Leading Term Test
C. Describe the end behavior of the graph of
f(x) = –2x 5 – 1 using limits. Explain your reasoning
using the leading term test.
The degree is 5 and the leading coefficient is –2.
Because the degree is odd and the leading coefficient
is negative,
.
Answer:
Describe the end behavior of the graph of
g (x) = –3x 5 + 6x 3 – 2 using limits. Explain your
reasoning using the leading term test.
A. Because the degree is odd and the leading coefficient
negative,
.
B. Because the degree is odd and the leading coefficient
negative,
.
C. Because the degree is odd and the leading coefficient
negative,
.
D. Because the degree is odd and the leading coefficient
negative,
.
Zeros of a Polynomial Function
State the number of possible real zeros and
turning points of f(x) = x 3 + 5x 2 + 4x. Then
determine all of the real zeros by factoring.
The degree of the function is 3, so f has at most 3
distinct real zeros and at most 3 – 1 or 2 turning points.
To find the real zeros, solve the related equation f(x) = 0
by factoring.
x 3 + 5x 2 + 4x = 0
Set f(x) equal to 0.
x(x 2 + 5x + 4) = 0
Factor the greatest common
factor, x.
x(x + 4)(x + 1) = 0
Factor completely.
Zeros of a Polynomial Function
So, f has three distinct real zeros, x = 0, x = –4, and
x = –1. This is consistent with a cubic function having at
most 3 distinct real zeros.
CHECK You can use a
graphing calculator to
graph f(x) = x 3 + 5x 2 + 4x
and confirm these zeros.
Additionally, you can see
that the graph has
2 turning points, which is
consistent with cubic
functions having at most
2 turning points.
Zeros of a Polynomial Function
Answer: The degree is 3, so f has at most 3 distinct
real zeros and at most 2 turning points.
f(x) = x 3 + 5x 2 + 4x = x(x + 1)(x + 4), so
f has three zeros, x = 0, x = –1, and x = –4.
State the number of possible real zeros and
turning points of f(x) = x 4 – 13x 2 + 36. Then
determine all of the real zeros by factoring.
A. 4 possible real zeros, 3 turning points;
zeros 2, –2, 3, –3
B. 4 possible real zeros, 2 turning points;
zeros 4, 9
C. 3 possible real zeros, 2 turning points;
zeros 2, 3
D. 4 possible real zeros, 4 turning points;
zeros –2, –3
Zeros of a Polynomial Function in Quadratic
Form
State the number of possible real zeros and
turning points for h(x) = x 4 – 4x 2 + 3. Then
determine all of the real zeros by factoring.
The degree of the function is 4, so h has at most
4 distinct real zeros and at most 4 – 1 or 3 turning
points. This function is in quadratic form because
x 4 – 4x 2 + 3 = (x 2)2 – 4(x 2) + 3. Let u = x 2.
(x2)2 – 4(x2) + 3 = 0 Set h(x) equal to 0.
u2 – 4u + 3 = 0 Substitute u for x2.
(u – 3)(u – 1) = 0 Factor the quadratic
expression.
Zeros of a Polynomial Function in Quadratic
Form
(x 2 – 3)(x 2 – 1) = 0 Substitute x 2 for u.
Factor completely.
or x + 1 = 0 or x – 1 = 0
, x = –1, x = 1
Zero Product Property
Solve for x.
h has four distinct real zeros,
, –1, and 1. This
is consistent with a quartic function. The graph of
h(x) = x 4 – 4x 2 + 3 confirms this. Notice that there are
3 turning points, which is also consistent with a quartic
function.
Zeros of a Polynomial Function in Quadratic
Form
Answer: The degree is 4, so h has at most 4 distinct
real zeros and at most 3 turning points.
h(x) = x 4 – 4x 2 + 3 = (x 2 – 3)(x – 1)(x + 1),
so h has four distinct real zeros, x =
,
x = –1, and x = 1.
State the number of possible real zeros and
turning points of g(x) = x 5 – 5x 3 – 6x. Then
determine all of the real zeros by factoring.
A. 3 possible real zeros, 2 turning points;
real zeros 0, –1, 6
B.
5 possible real zeros, 4 turning points;
real zeros 0,
C.
3 possible real zeros, 3 turning points;
real zeros 0, –1, .
D.
5 possible real zeros, 4 turning points;
real zeros 0, 1, –1,
Polynomial Function with Repeated Zeros
State the number of possible real zeros and
turning points of h(x) = x 4 + 5x 3 + 6x 2. Then
determine all of the real zeros by factoring.
The degree of the function is 4, so it has at most 4
distinct real zeros and at most 4 – 1 or 3 turning
points. Find the real zeros.
x 4 + 5x 3 + 6x 2 = 0
Set h(x) equal to 0.
x 2(x 2 + 5x + 6) = 0
Factor the greatest
common factor, x 2.
x 2(x + 3)(x + 2) = 0
Factor completely.
Polynomial Function with Repeated Zeros
The expression above has 4 factors, but solving for
x yields only 3 distinct real zeros, x = 0, x = –3, and
x = –2. Of the zeros, x = 0 is repeated.
The graph of h(x) = x 4 + 5x 3 + 6x 2 shown here
confirms these zeros and shows that h has three
turning points. Notice that at x = –3 and x = –2, the
graph crosses the x-axis, but at x = 0, the graph is
tangent to the x-axis.
Polynomial Function with Repeated Zeros
Answer: The degree is 4, so h has at most 4 distinct
real zeros and at most 3 turning points.
h(x) = x 4 + 5x 3 + 6x 2 = x 2(x + 2)(x + 3), so
h has three zeros, x = 0, x = –2, and x = –3.
Of the zeros, x = 0 is repeated.
State the number of possible real zeros and
turning points of g(x) = x 4 – 4x 3 + 4x 2. Then
determine all of the real zeros by factoring.
A. 4 possible real zeros, 3 turning points;
real zeros 0, 2
B. 4 possible real zeros, 3 turning points;
real zeros 0, 2, –2
C. 2 possible real zeros, 1 turning point;
real zeros 2, –2
D. 4 possible real zeros, 3 turning points;
real zeros 0, –2
Graph a Polynomial Function
A. For f(x) = x(3x + 1)(x – 2) 2, apply the leadingterm test.
The product x(3x + 1)(x – 2)2 has a leading term of
x(3x)(x)2 or 3x4, so f has degree 4 and leading coefficient
3. Because the degree is even and the leading coefficient
is positive,
.
Answer:
Graph a Polynomial Function
B. For f(x) = x(3x + 1)(x – 2) 2, determine the zeros
and state the multiplicity of any repeated zeros.
The distinct real zeros are x = 0, x =
The zero at x = 2 has multiplicity 2.
Answer: 0,
, 2 (multiplicity: 2)
, and x = 2.
Graph a Polynomial Function
C. For f(x) = x(3x + 1)(x – 2) 2, find a few additional
points.
Choose x-values that fall in the intervals determined
by the zeros of the function.
Answer: (–1, 18), (–0.1, –0.3087), (1, 4), (3, 30)
Graph a Polynomial Function
D. For f(x) = x(3x + 1)(x – 2) 2, graph the function.
Plot the points you found. The end behavior of the
function tells you that the graph eventually rises to the
left and to the right. You also know that the graph
crosses the x-axis at nonrepeated zeros x =
and
x = 0, but does not cross the x-axis at repeated zero
x = 2, because its multiplicity is even. Draw a
continuous curve through the points as shown in the
figure.
Graph a Polynomial Function
Answer:
Determine the zeros and state the multiplicity of
any repeated zeros for f(x) = 3x(x + 2)2(2x – 1)3.
A.
0, –2 (multiplicity 2),
B.
2 (multiplicity 2), –
C.
4 (multiplicity 2),
D.
–2 (multiplicity 2),
(multiplicity 3)
(multiplicity 3)
(multiplicity 3)
(multiplicity 3)
Model Data Using Polynomial
Functions
A. POPULATION The table to
the right shows a town’s
population over an 8-year
period. Year 1 refers to the year
2001, year 2 refers to the year
2002, and so on. Create a
scatter plot of the data, and
determine the type of
polynomial function that could
be used to represent the data.
Model Data Using Polynomial
Functions
Enter the data using the list feature of a graphing
calculator. Let L1 be the number of the year. Then
create a scatter plot of the data. The curve of the
scatter plot resembles the graph of a cubic equation,
so we will use a cubic regression.
Answer: cubic function;
Model Data Using Polynomial
Functions
B. POPULATION The table
below shows a town’s
population over an 8-year
period. Year 1 refers to the
year 2001, year 2 refers to the
year 2002, and so on. Write a
polynomial function to model
the data set. Round each
coefficient to the nearest
thousandth, and state the
correlation coefficient.
Model Data Using Polynomial
Functions
Using the CubicReg tool on a graphing calculator and
rounding each coefficient to the nearest thousandth yields
f(x) = 10.020x 3 – 176.320x 2 + 807.469x + 4454.786.
The value of r 2 for the data is 0.89, which is close to 1, so
the model is a good fit. We can graph the complete
(unrounded) regression by sending it to the Y= menu. In
the Y= menu, pick up this regression equation by
entering VARS , Statistics, EQ. Graph this function and
the scatter plot in the same viewing window. The function
appears to fit the data reasonably well.
Model Data Using Polynomial
Functions
Answer: f (x) = 10.020x 3 – 176.320x 2 + 807.469x + 4454.786;
r 2 = 0.89
Model Data Using Polynomial
Functions
C. POPULATION The table
below shows a town’s
population over an 8-year
period. Year 1 refers to the
year 2001, year 2 refers to the
year 2002, and so on. Use the
model to estimate the
population of the town in the
year 2012.
Model Data Using Polynomial
Functions
Use 12 for the year 2012 and use the CALC feature
on a calculator to find f(12). The value of f(12) is
6069.1926, so the population in 2012 will be about
6069.
Answer: 6069
Model Data Using Polynomial
Functions
D. POPULATION The table
below shows a town’s
population over an 8-year
period. Year 1 refers to the
year 2001, year 2 refers to the
year 2002, and so on. Use the
model to determine the
approximate year in which the
population reaches 10,712.
Model Data Using Polynomial
Functions
Graph the line y = 10,712 for Y2. Then use
5: intersect on the CALC menu to find the point of
intersection of y = 10,712 with f(x). The intersection
occurs when x ≈ 15, so the approximate year in which
the population will be 10,712 is 2015.
Answer: 2015
BIOLOGY The number of fruit flies that hatched after day x is
given in the table. Write a polynomial function to model the
data set. Round each coefficient to the nearest thousandth, and
state the correlation coefficient. Use the model to estimate the
number of fruit flies hatched after 8 days.
A. y = 12.014x 2 – 72.940x + 5.3; r = 0.84; 190
B.
y = 20.833x 3 + 125.786x 2 + 251.238x + 195.714; r 2 = 0.99;
40,922
C.
y = 10x4 + 60.833x 3 + 141.5x 2 + 202.667x + 182; r 2 = 1;
82,966
D.
y = 70.893x – 20.672; r = 0.829; 346