Transcript Algebra 2 Interactive Chalkboard Copyright © by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION Glencoe/McGraw-Hill 8787 Orion Place Columbus, Ohio 43240

Algebra 2 Interactive Chalkboard
Copyright © by The McGraw-Hill Companies, Inc.
Send all inquiries to:
GLENCOE DIVISION
Glencoe/McGraw-Hill
8787 Orion Place
Columbus, Ohio 43240
Lesson 7-1 Polynomial Functions
Lesson 7-2 Graphing Polynomial Functions
Lesson 7-3 Solving Equations Using
Quadratic Techniques
Lesson 7-4 The Remainder and Factor Theorems
Lesson 7-5 Roots and Zeros
Lesson 7-6 Rational Zero Theorem
Lesson 7-7 Operations on Functions
Lesson 7-8 Inverse Functions and Relations
Lesson 7-9 Square Root Functions and Inequalities
Example 1 Find Degrees and Leading Coefficients
Example 2 Evaluate a Polynomial Function
Example 3 Functional Values of Variables
Example 4 Graphs of Polynomial Functions
State the degree and leading coefficient of
in one variable. If it is not a polynomial
in one variable, explain why.
Answer: This is a polynomial in one variable. The
degree is 3 and the leading coefficient is 7.
State the degree and leading coefficient of
in one variable. If it is not a
polynomial in one variable, explain why.
Answer: This is not a polynomial in one variable.
It contains two variables, a and b.
State the degree and leading coefficient of
in one variable. If it is not a
polynomial in one variable, explain why.
Answer: This is not a polynomial in one variable. The
term 2c–1 is not of the form ancn, where n is
a nonnegative integer.
State the degree and leading coefficient of
in one variable. If it is not a polynomial
in one variable, explain why.
Rewrite the expression so the powers of y are in
decreasing order.
Answer: This is a polynomial in one variable with
degree of 4 and leading coefficient 1.
State the degree and leading coefficient of each
polynomial in one variable. If it is not a polynomial
in one variable, explain why.
a.
Answer: degree 3, leading coefficient 3
b.
Answer: This is not a polynomial in one variable.
It contains two variables, x and y.
c.
Answer: This is not a polynomial in one variable.
The term 3a–1 is not of the form ancn, where
n is nonnegative.
d.
Answer: degree 3, leading coefficient 1
Nature Refer to Example 2 on page 347 of your
textbook. A sketch of the arrangement of hexagons
shows a fourth ring of 18 hexagons, a fifth ring of 24
hexagons, and a sixth ring of 30 hexagons.
Show that the polynomial function
gives the total number of hexagons when
Find the values of f (4), f (5), and f (6).
Original function
Replace r with 4.
Simplify.
Original function
Replace r with 5.
Simplify.
Original function
Replace r with 6.
Simplify.
From the information given in Example 2 of your
textbook, you know that the total number of hexagons
for three rings is 19. So, the total number of hexagons
for four rings is 19 + 18 or 37, five rings is 37 + 24 or
61, and six rings is 61 + 30 or 91.
Answer: These match the function values for
respectively.
Nature Refer to Example 2 on page 347 of your
textbook. A sketch of the arrangement of hexagons
shows a fourth ring of 18 hexagons, a fifth ring of 24
hexagons, and a sixth ring of 30 hexagons.
Find the total number of hexagons in a honeycomb
with 20 rings.
Original function
Replace r with 20.
Answer:
Simplify.
Nature A sketch of the arrangement of hexagons
shows a seventh ring of 36 hexagons, an eighth ring
of 42 hexagons, and a ninth ring of 48 hexagons.
a. Show that the polynomial function
gives the total number of hexagons when
Recall that the total number of hexagons in
six rings is 91.
Answer: f (7) = 127; f (8) = 169; f (9) = 217; the total
number of hexagons for seven rings is 91 + 36 or 127,
eight rings is 127 + 42 or 169, and nine rings is 169 +
48 or 217. These match the functional values for r = 7,
8, and 9, respectively.
b. Find the total number of hexagons in a honeycomb
with 30 rings.
Answer: 2611
Find
Original function
Replace x with y 3.
Answer:
Property of powers
Find
To evaluate b(2x – 1), replace m in b(m) with 2x – 1.
Original function
Replace m
with 2x – 1.
Evaluate 2(2x – 1)2.
Simplify.
To evaluate 3b(x), replace m with x in b(m), then
multiply the expression by 3.
Original function
Replace m with x.
Distributive Property
Now evaluate b(2x – 1) – 3b(x).
Replace b(2x – 1)
and 3b(x) with
evaluated
expressions.
Answer:
Simplify.
a. Find
Answer:
b. Find
Answer:
For the graph,
 describe the end behavior,
 determine whether it
represents an odd-degree
or an even-degree function,
and
 state the number of real
zeros.
Answer:
.

.

 It is an even-degree polynomial function.
 The graph does not intersect the x-axis,
so the function has no real zeros.
For the graph,
 describe the end behavior,
 determine whether it
represents an odd-degree
or an even-degree function,
and
 state the number of real
zeros.
Answer:
.

.

 It is an odd-degree polynomial function.
 The graph intersects the x-axis at one point,
so the function has one real zero.
For the graph,
 describe the end behavior,
 determine whether it
represents an odd-degree
or an even-degree function,
and
 state the number of real
zeros.
Answer:
.

.

 It is an even-degree polynomial function.
 The graph intersects the x-axis at two points,
so the function has two real zeros.
For each graph,
a.
 describe the end
behavior,
 determine whether it
represents an odd-degree
or an even-degree function,
and
 state the number of real
zeros.
Answer:
.

.

 It is an even-degree polynomial function.
 The graph intersects the x-axis at two points,
so the function has two real zeros.
For each graph,
b.
 describe the end
behavior,
 determine whether it
represents an odd-degree
or an even-degree function,
and
 state the number of real
zeros.
Answer:
.

.

 It is an odd-degree polynomial function.
 The graph intersects the x-axis at three points,
so the function has three real zeros.
For each graph,
c.
 describe the end
behavior,
 determine whether it
represents an odd-degree
or an even-degree function,
and
 state the number of real
zeros.
Answer:
.

.

 It is an even-degree polynomial function.
 The graph intersects the x-axis at one point,
so the function has one real zero.
Example 1 Graph a Polynomial Function
Example 2 Locate Zeros of a Function
Example 3 Maximum and Minimum Points
Example 4 Graph a Polynomial Model
Graph
making a table of values.
Answer:
by
x
f(x)
–4
–3
–2
5
–4
–3
–1
0
1
2
2
5
0
–19
Graph
making a table of values.
Answer:
by
This is an odd degree
polynomial with a negative
leading coefficient, so
f (x)  + as x  – and
f (x)  – as x  +. Notice
that the graph intersects the
x-axis at 3 points indicating that
there are 3 real zeros.
Graph
by
making a table of values.
Answer:
x
f (x)
–3
–2
–1
–8
1
2
0
1
2
1
4
17
Determine consecutive values of x between which
each real zero of the function
is located. Then draw the graph.
Make a table of values. Since f(x) is a 4th degree
polynomial function, it will have between 0 and 4
zeros, inclusive.
x
f (x)
–2
9
–1
–1
0
1
1
–3
2
–7
3
19
change in signs
change in signs
change in signs
change in signs
Look at the value of f(x) to locate the zeros. Then use the
points to sketch the graph of the function.
Answer:
There are zeros between x = –2 and –1, x = –1
and 0, x = 0 and 1, and x = 2 and 3.
Determine consecutive values of x between which
each real zero of the function
is located. Then draw the graph.
Answer:
There are zeros between x = –1 and 0, x = 0
and 1, and x = 3 and 4.
Graph
Estimate the
x-coordinates at which the relative maximum and
relative minimum occur.
Make a table of values and graph the function.
x
–2
–1
0
1
2
3
4
5
f (x)
–19
0
5
2
–3
–4
5
30
zero at x = –1
indicates a relative maximum
zero between x = 1 and x = 2
indicates a relative minimum
zero between x = 3 and x = 4
Answer: The value of f(x) at x = 0 is greater than the
surrounding points, so it is a relative maximum. The value
of f(x) at x = 3 is less than the surrounding points, so it is
a relative minimum.
x
–2
–1
0
1
2
3
4
5
f (x)
–19
0
5
2
–3
–4
5
30
Graph
Estimate the
x-coordinates at which the relative maximum and
relative minimum occur.
Answer: The value of f(x) at
x = 0 is less than the
surrounding points, so it is a
relative minimum. The value
of f(x) at x = –2 is greater
than the surrounding points,
so it is a relative maximum.
Health The weight w, in pounds, of a patient during
a 7-week illness is modeled by the cubic equation
where n is the number of
weeks since the patient became ill.
Graph the equation.
Make a table of values for weeks 0 through 7. Plot the
points and connect with a smooth curve.
n
w(n)
0
1
110
109.5
2
3
4
108.4
107.3
106.8
5
6
7
107.5
110
114.9
Answer:
Describe the turning points of the graph and its
end behavior.
Answer: There is a
relative minimum at
week 4. For the end
behavior, w(n)
increases as n
increases.
What trends in the patient’s weight does the
graph suggest?
Answer: The patient
lost weight for each of
4 weeks after
becoming ill. After 4
weeks, the patient
started to gain weight
and continues to gain
weight.
Weather The rainfall r, in inches per month,
in a Midwestern town during a 7-month period
is modeled by the cubic equation
where m is the number of months after March 1.
a. Graph the equation.
Answer:
b. Describe the turning
points of the graph
and its end behavior.
Answer: There is a relative
maximum at Month 2, or
May. For the end behavior,
r(m) decreases as m
increases.
c. What trends in the amount of rainfall received by the
town does the graph suggest?
Answer: The rainfall increased for two months following
March. After two months, the amount of rainfall decreased
for the next five months and continues to decrease.
Example 1 Write an Expression in Quadratic Form
Example 2 Solve Polynomial Equations
Example 3 Solve Equations with Rational Exponents
Example 4 Solve Radical Equations
Write
Answer:
in quadratic form, if possible.
Write
Answer:
in quadratic form, if possible.
Write
in quadratic form, if possible.
Answer: This cannot be written in quadratic form
since
Write
Answer:
in quadratic form, if possible.
Write each expression in quadratic form, if possible.
a.
Answer:
b.
Answer:
c.
Answer: This cannot be
written in quadratic form
since
d.
Answer:
Solve
.
Original equation
Write the expression
on the left in
quadratic form.
Factor the trinomial.
Factor each
difference of
squares.
Use the Zero Product Property.
or
or
or
Answer: The solutions are –5, –2, 2, and 5.
Check
The graph of
shows
that the graph intersects the x-axis at –5, –2, 2,
and 5.
Solve
.
Original equation
This is the sum
of two cubes.
Sum of two cubes
formula with a = x and
b=6
or
Zero Product Property
The solution of the first equation is –6. The second
equation can be solved by using the Quadratic Formula.
Quadratic Formula
Replace a with 1, b
with –6, and c with 36.
Simplify.
or
Simplify.
Answer: The solutions of the original equation are
Solve each equation.
a.
Answer: –3, –1, 1, 3
b.
Answer:
Solve
Original equation
Write the expression on
the left in quadratic form.
Factor the trinomial.
or
Zero Product
Property
or
Isolate x on one side
of the equation.
Raise each side to
the fourth power.
Simplify.
Check Substitute each value into the original equation.
Answer: The solution is 81.
Solve
Answer: –8, –27
Solve
Original equation
Rewrite so that
one side is zero.
Write the expression
on the left side in
quadratic form.
Factor.
or
Zero Product
Property
Solve each
equation.
Answer: Since the square root of x cannot be negative,
the equation
has no solution. Thus, the only
solution of the original equation is 9.
Solve
Answer: 25
Example 1 Synthetic Substitution
Example 2 Use the Factor Theorem
Example 3 Find All Factors of a Polynomial
find f (4).
If
Method 1 Synthetic Substitution
By the Remainder Theorem, f(4) should be the
remainder when you divide the polynomial by x – 4.
3 10 41 164 654
Notice that there is no x
term. A zero is placed in
this position as a
placeholder.
Answer: The remainder is 654. Thus, by using synthetic
substitution, f (4) = 654.
Method 2 Direct Substitution
Replace x with 4.
Original function
Replace x with 4.
or 654
Simplify.
Answer: By using direct substitution, f (4) = 654.
If
Answer: 34
find f(3).
Show that
is a factor of
Then find the remaining factors of the polynomial.
The binomial x – 3 is a factor of the polynomial if 3 is a
zero of the related polynomial function. Use the factor
theorem and synthetic division.
1
7
6
0
Since the remainder is 0, (x – 3) is a factor of the
polynomial. The polynomial
can be factored as
The polynomial
is the depressed polynomial. Check to see if
this polynomial can be factored.
Factor the trinomial.
Answer: So,
Check
You can see that the graph of the related
function
crosses
the x-axis at 3, –6, and –1. Thus,
Show that
is a factor of
Then find the remaining factors of the polynomial.
Answer:
1
So,
Since
6
5
0
Geometry The volume of a
rectangular prism is given
by
Find the missing measures.
The volume of a rectangular
prism is
You know
that one measure is x – 2, so x – 2 is a factor of V(x).
1 9 20
0
The quotient is
. Use this to factor V(x).
Volume function
Factor.
Factor the trinomial
Answer: The missing measures of the prism
are x + 4 and x + 5.
Geometry The volume of a
rectangular prism is given
by
Find the missing measures.
Answer: The missing measures of the prism
are x – 2 and x + 5.
Example 1 Determine Number and Type of Roots
Example 2 Find Numbers of Positive and Negative Zeros
Example 3 Use Synthetic Substitution to Find Zeros
Example 4 Use Zeros to Write a Polynomial Function
Solve
State the number and type of roots.
Original equation
Add 10 to each side.
Answer: This equation has exactly one real root, 10.
Solve
of roots.
State the number and type
Original equation
Factor.
or
Zero Product Property
Solve each equation.
Answer: This equation has two real roots, –8 and 6.
Solve
of roots.
State the number and type
Original equation
Factor out the GCF.
Use the Zero Product Property.
or
Subtract 6 from
each side.
Square Root
Property
Answer: This equation has one real root at 0,
and two imaginary roots at
Solve
State the number and type of roots.
Original equation
Factor differences
of squares.
Factor differences
of squares.
or
or
Zero Product
Property
Solve each
equation.
Answer: This equation has two real roots, –2 and 2,
and two imaginary roots, 2i and –2i.
Solve each equation. State the number and type
of roots.
a.
Answer: This equation has exactly one root at –3.
b.
Answer: This equation has exactly two roots, –3 and 4.
c.
Answer: This equation has one real root at 0 and two
imaginary roots at
d.
Answer: This equation has two real roots, –3 and 3,
and two imaginary roots, 3i and –3i.
State the possible number of positive real zeros,
negative real zeros, and imaginary zeros of
Since p(x) has degree 6, it has 6 zeros. However, some of
them may be imaginary. Use Descartes Rule of Signs to
determine the number and type of real zeros. Count the
number of changes in sign for the coefficients of p(x).
yes
– to +
yes
+ to –
no
– to –
no
– to –
Since there are two sign changes, there are 2 or 0 positive
real zeros. Find p(–x) and count the number of sign
changes for its coefficients.
x
no
– to –
no
– to –
yes
– to +
1
yes
+ to –
Since there are two sign changes, there are 2 or 0 negative
real zeros. Make a chart of possible combinations.
Answer:
Number of
Positive Real
Zeros
Number of
Negative Real
Zeros
Number of
Imaginary
Zeros
Total
2
0
2
0
2
2
0
0
2
4
4
6
6
6
6
6
State the possible number of positive real zeros,
negative real zeros, and imaginary zeros of
Answer: The function has either 2 or 0 positive real
zeros, 2 or 0 negative real zeros, and 4, 2, or 0
imaginary zeros.
Find all of the zeros of
Since f (x) has degree of 3, the function has three zeros.
To determine the possible number and type of real zeros,
examine the number of sign changes in f (x) and f (–x).
yes
no
yes
no
no
yes
The function has 2 or 0 positive real zeros and exactly 1
negative real zero. Thus, this function has either 2 positive
real zeros and 1 negative real zero or 2 imaginary zeros
and 1 negative real zero.
To find the zeros, list some possibilities and eliminate
those that are not zeros. Use a shortened form of
synthetic substitution to find f (a) for several values of a.
x
1
–1
2
4
–3
1
–4
14
–38
–2
1
–3
8
–12
–1
1
–2
4
0
Each row in the table
shows the coefficients
of the depressed
polynomial and the
remainder.
From the table, we can see that one zero occurs at
x = –1. Since the depressed polynomial,
,
is quadratic, use the Quadratic Formula to find the roots of
the related quadratic equation
Quadratic Formula
Replace a with 1, b
with –2, and c with 4.
Simplify.
Simplify.
Answer: Thus, this function has one real zero at –1 and
two imaginary zeros at
and
The graph of
the function verifies that there is only one real zero.
Find all of the zeros of
Answer:
Short-Response Test Item
Write a polynomial function of least degree with
integer coefficients whose zeros include 4 and 4 – i.
Read the Test Item
• If 4 – i is a zero, then 4 + i is also a zero, according
to the Complex Conjugate Theorem. So, x – 4,
x – (4 – i), and x – (4 + i) are factors of the
polynomial function.
Solve the Test Item
• Write the polynomial function as a product of
its factors.
• Multiply the factors to find the polynomial function.
Write an equation.
Regroup terms.
Rewrite as the
difference of two
squares.
Square x – 4
and replace i2
with –1.
Simplify.
Multiply using
the Distributive
Property.
Combine like
terms.
Answer:
is a polynomial
function of least degree with integral coefficients whose
zeros are 4, 4 – i, and 4 + i.
Short-Response Test Item
Write a polynomial function of least degree with
integer coefficients whose zeros include 2 and 1 + i.
Answer:
Example 1 Identify Possible Zeros
Example 2 Use the Rational Zero Theorem
Example 3 Find All Zeros
List all of the possible rational zeros of
If
is a rational zero, then p is a factor of 4 and q is a
factor of 3. The possible factors of p are 1, 2, and 4.
The possible factors of q are 1 and 3.
Answer: So,
List all of the possible rational zeros of
Since the coefficient of x4 is 1, the possible zeros must be
a factor of the constant term –15.
Answer: So, the possible rational zeros are 1, 3, 5,
and 15.
List all of the possible rational zeros of each function.
a.
Answer:
b.
Answer:
Geometry The volume of a rectangular solid is 1120
cubic feet. The width is 2 feet less than the height and
the length is 4 feet more than the height. Find the
dimensions of the solid.
Let x = the height, x – 2 = the width,
and x + 4 = the length.
Write the equation for volume.
Formula for volume
Multiply.
Subtract 1120.
The leading coefficient is 1, so the possible integer zeros
are factors of 1120. Since length can only be positive, we
only need to check positive zeros.
The possible factors are 1, 2, 4, 5, 8, 10, 14, 16, 20, 28,
32, 35, 40, 56, 70, 80, 112, 140, 160, 224, 280, 560,
and 1120. By Descartes’ Rule of Signs, we know that
there is exactly one positive real root. Make a table and
test possible real zeros.
p
1
2
1
1
2
2
3
6
–8
–5
2
–1120
–1125
–1112
10
1
12
112
0
So, the zero is 10. The other dimensions are
10 – 2 or 8 feet and 10 + 4 or 14 feet.
Check
Answer:
Verify that the dimensions are correct.
Geometry The volume of a rectangular solid is 100
cubic feet. The width is 3 feet less than the height and
the length is 5 feet more than the height. Find the
dimensions of the solid.
Answer:
Find all of the zeros of
From the corollary to the Fundamental Theorem of
Algebra, we know there are exactly 4 complex roots.
According to Descartes’ Rule of Signs, there are 2 or 0
positive real roots and 2 or 0 negative real roots.
The possible rational zeros are 1, 2, 3, 5, 6, 10,
15, and 30.
Make a table and test some possible rational zeros.
p
q
0
1
2
p
q
1
1
–19
11
30
1
1
1
1
2
3
–19
–17
–13
11
–6
–15
30
24
0
Since f (2) = 0, you know that x = 2 is a zero.
The depressed polynomial is
Since x = 2 is a positive real zero, and there can only be
2 or 0 positive real zeros, there must be one more
positive real zero. Test the next possible rational zeros on
the depressed polynomial.
p
q
3
p
q
1
3
–13
–15
1
6
5
0
There is another zero at x = 3. The depressed polynomial
is
Factor
Write the depressed
polynomial.
Factor.
or
Zero Product Property
There are two more real roots at x = –5 and x = –1.
Answer: The zeros of this function are –5, –1, 2,
and 3.
Find all of the zeros of
Answer: –5, –3, 1, and 3
Example 1 Add and Subtract Functions
Example 2 Multiply and Divide Functions
Example 3 Evaluate Composition of Relations
Example 4 Simplify Composition of Functions
Example 5 Use Composition of Functions
Given
find
,
Addition of
functions
and
Answer:
Simplify.
Given
find
,
Subtraction of
functions
and
Answer:
Simplify.
Given
find each function.
a.
Answer:
b.
Answer:
Given
find
Product of
functions
and
Distributive
Property
Distributive
Property
Answer:
Simplify.
Given
find
Division of functions
Answer:
and
Since 4 makes the denominator 0, it is excluded from
the domain of
Given
find each function.
a.
Answer:
b.
Answer:
If f (x) = {(2, 6), (9, 4), (7, 7), (0, –1)} and
g(x) = {(7, 0), (–1, 7), (4, 9), (8, 2)}, find
and
To find
, evaluate g(x) first. Then use the range of g
as the domain of f and evaluate f(x).
Answer:
To find
evaluate f(x) first. Then use the range of f
as the domain of g and evaluate g(x).
is undefined.
Answer: Since 6 is not in the domain of g,
undefined for x = 2.
is
If f (x) = {(1, 2), (0, –3), (6, 5), (2, 1)} and
g(x) = {(2, 0), (–3, 6), (1, 0), (6, 7)}, find
and
Answer:
Find
and
and
Composition
of functions
Replace g(x)
with 2x – 1.
Substitute
2x – 1 for x
in f(x).
Evaluate
(2x – 1)2.
Simplify.
Composition
of functions
Replace f (x)
with
Substitute
for x in g(x).
Simplify.
Answer: So,
and
Evaluate
and
x = –2.
Function from part a
Replace x with –2.
Simplify.
Function from part a
Replace x with –2.
Simplify.
Answer: So,
and
a. Find
and
and
and
Answer:
b. Evaluate
Answer:
and
and
x = 1.
Taxes Tracie Long has $100 deducted from every
paycheck for retirement. She can have this deduction
taken before state taxes are applied, which reduces
her taxable income. Her state income tax is 4%. If
Tracie earns $1500 every pay period, find the
difference in her net income if she has the retirement
deduction taken before or after state taxes.
Explore Let x = her income per paycheck, r(x) = her
income after the deduction for retirement,
t(x) = her income after tax.
Plan
Write equations for r(x) and t(x).
$100 is deducted for retirement.
The tax rate is 4%.
Solve
If Tracie has her retirement deducted before
taxes, then her net income is represented by
Replace x with
1500 in
Replace x with
1400 in
If Tracie has her retirement deducted after
taxes, then her net income is represented by
Replace x with 1500 in
Replace x with
1440 in
Answer:
Examine
and
The difference is 1344 – 1340 or 4. So, her
net income is $4 more if the retirement
deduction is taken before taxes.
The answer makes sense. Since the taxes are
being applied to a smaller amount, less taxes
will be deducted from her paycheck.
Taxes Brandi Smith has $200 deducted from every
paycheck for retirement. She can have this deduction
taken before state taxes are applied, which reduces
her taxable income. Her state income tax is 10%. If
Brandi earns $2200 every pay period, find the
difference in her net income if she has the retirement
deduction taken before or after state taxes.
Answer: Her net income is $20 more if she has the
retirement deduction taken before her state
taxes.
Example 1 Find an Inverse Relation
Example 2 Find an Inverse Function
Example 3 Verify Two Functions are Inverses
Geometry The ordered pairs of the relation {(1, 3),
(6, 3), (6, 0), (1, 0)} are the coordinates of the vertices
of a rectangle. Find the inverse of this relation and
determine whether the resulting ordered pairs are also
the coordinates of the vertices of a rectangle.
To find the inverse of this relation, reverse the coordinates
of the ordered pairs. The inverse of the relation is {(3, 1),
(3, 6), (0, 6), (0, 1)}.
Answer: Plotting the points shows that the ordered pairs
also describe the vertices of a rectangle. Notice that the
graph of the relation and the inverse are reflections over
the graph of y = x.
Geometry The ordered pairs of the relation {(–3, 4),
(–1, 5), (2, 3), (1, 1), (–2, 1)} are the coordinates of the
vertices of a pentagon. Find the inverse of this relation
and determine whether the
resulting ordered pairs are
also the coordinates of the
vertices of a pentagon.
Answer: {(4, –3), (5, –1),
(3, 2), (1, 1), (1, –2)}
These ordered pairs also
describe the vertices of
a pentagon.
Find the inverse of
Step 1
Replace f(x) with y in the original equation.
Step 2
Interchange x and y.
Step 3
Solve for y.
Inverse
Multiply each side
by –2.
Add 2 to each side.
Step 4
Replace y with f –1(x).
Answer: The inverse of
is
Graph the function and its inverse.
Graph both functions on the coordinate plane. The graph
of
over the line
is the reflection for
Answer:
a. Find the inverse of
Answer:
b. Graph the function
and its inverse.
Answer:
Determine whether
are inverse functions.
and
Check to see if the compositions of f(x) and g(x) are
identity functions.
Answer: The functions are inverses since both
and
equal x.
Determine whether
are inverse functions.
and
Answer: The functions are inverses since both
compositions equal x.
Example 1 Graph a Square Root Function
Example 2 Solve a Square Root Problem
Example 3 Graph a Square Root Inequality
Graph
State the domain, range, and
x- and y-intercepts.
Since the radicand cannot be negative, identify the domain.
Write the expression inside the
radicand as  0.
Solve for x.
The x-intercept is
Make a table of values to graph the function.
x
2
3
y
0
1
2
3
4
5
6
0.71
1.14
1.87
2.23
2.55
2.83
Answer:
From the graph, you can
see that the domain is
and the range is
y  0. The x-intercept is
There is no y-intercept.
Graph
State the domain, range, and
x- and y-intercepts.
Answer:
domain: x  1
range: y  0
x-intercept: 1
y-intercept: none
Physics When an object is spinning in a circular path
of radius 2 meters with velocity v, in meters per
second, the centripetal acceleration a, in meters per
second squared, is directed toward the center of the
circle. The velocity v and acceleration a of the object
are related by the function
Graph the function. State the domain and range.
The function is
graph the function.
Make a table of values and
Answer:
a
v
0
1
2
3
4
5
0
1.41
2
2.45
2.83
3.16
The domain is a  0 and
the range is v  0.
What would be the centripetal acceleration of an
object spinning along the circular path with a velocity
of 4 meters per second?
Original equation
Replace v with 4.
Square each side.
Divide each side by 2.
Answer: The centripetal acceleration would be
8 meters per second squared.
Geometry The volume V and surface area A of a soap
bubble are related by the function
a. Graph the function. State the domain and range.
Answer:
domain: A  0
range: V  0
b. What would the surface area be if the volume were
94 cubic units?
Answer: 100 units2
Graph
Graph the related equation
Since the
boundary is not included,
the graph should be
dashed.
The domain includes values
for
So the graph
is to the right of
Select a point and test
its ordered pair.
Test (0, 0).
false
Shade the region that does not include (0, 0).
Graph
Graph the related equation
The domain includes values for
Select a point and test its ordered pair.
Test (4, 1).
true
Shade the region that
includes (4, 1).
a. Graph
Answer:
b. Graph
Answer:
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