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Lesson 7-1 Polynomial Functions

Lesson 7-2 Graphing Polynomial Functions

Lesson 7-3Solving 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.



Answer: This is not a polynomial in one variable. It contains two variables, a and b.



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.



Answer: This is a polynomial in one variable with degree of 4 and leading coefficient 1.

Rewrite the expression so the powers of y are in decreasing order.

State the degree and leading coefficient of each polynomial in one variable. If it is not a polynomial in one variable, explain why.

a.

b.

Answer: degree 3, leading coefficient 3

Answer: This is not a polynomial in one variable. It contains two variables, x and y.

Answer: degree 3, leading coefficient 1

Answer: This is not a polynomial in one variable. The term 3a–1 is not of the form ancn, where n is nonnegative.

c.

d.

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 forrespectively.

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 functiongives 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.

Simplify.Answer:

a. Find

b. Find

Answer:

Answer:

For the graph, describe the end behavior, determine whether it

represents an odd-degreeor 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.

..




Answer: It is an odd-degree polynomial function. The graph intersects the x-axis at one point,

so the function has one real zero.

..




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



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




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




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

Answer:

Graph by making a table of values.

x f(x)

–4 5

–3 –4

–2 –3

–1 2

0 5

1 0

2 –19

Answer:


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.

x f (x)

–3 –8

–2 1

–1 2

0 1

1 4

2 17


Answer:

Determine consecutive values of x between whicheach 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 signschange in signschange 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 whicheach 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 f (x)

–2 –19

–1 0

0 5

1 2

2 –3

3 –4

4 5

5 30

zero at x = –1

zero between x = 1 and x = 2

zero between x = 3 and x = 4

indicates a relative maximum

indicates a relative minimum

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 f (x)

–2 –19

–1 0

0 5

1 2

2 –3

3 –4

4 5

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.

Answer:n w(n)

0 110

1 109.5

2 108.4

3 107.3

4 106.8

5 107.5

6 110

7 114.9

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.

c. What trends in the amount of rainfall received by thetown does the graph suggest?

Answer: There is a relative maximum at Month 2, or May. For the end behavior, r (m) decreases as m increases.

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 in quadratic form, if possible.

Answer:

Answer: This cannot be written in quadratic form since



Answer:

Write each expression in quadratic form, if possible.

a.

b.

c.

d.

Answer: This cannot be written in quadratic form since

Answer:

Answer:

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 showsthat 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.

Simplify.

or

Answer: The solutions of the original equation are

Solve each equation.

a.

b.

Answer: –3, –1, 1, 3

Answer:

Solve

Original equation

Write the expression onthe left in quadratic form.

Zero Product Property

or


Isolate x on one side of the equation.

or

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.

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.

Zero Product Property

or


Solve

Answer: 25

Example 1 Synthetic Substitution

Example 2 Use the Factor Theorem

Example 3 Find All Factors of a Polynomial

If find f (4).

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

Answer: The remainder is 654. Thus, by using synthetic substitution, f (4) = 654.

Notice that there is no x term. A zero is placed in this position as a placeholder.

Method 2 Direct Substitution

Replace x with 4.

Answer: By using direct substitution, f (4) = 654.

Original function

Replace x with 4.

Simplify.or 654

If find f (3).

Answer: 34

Show that is a factor ofThen 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.


Answer: So,

Check You can see that the graph of the related functioncrosses the x-axis at 3, –6, and –1. Thus,

Show that is a factor ofThen find the remaining factors of the polynomial.

Answer:

1 6 5 0

So,Since

Geometry The volume of a rectangular prism is given byFind the missing measures.

The volume of a rectangularprism is You knowthat 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 byFind 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 2Find 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.


Original equation

Answer: This equation has two real roots, –8 and 6.

Factor.


Zero Product Propertyor


Original equation

Factor out the GCF.

Subtract 6 from each side.

Use the Zero Product Property.

or

Square Root Property

Answer: This equation has one real root at 0, and two imaginary roots at

Original equation


Factor differences of squares.

Factor differences of squares.

or or Zero ProductProperty


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.

b.

c.

Answer: This equation has exactly one root at –3.

Answer: This equation has exactly two roots, –3 and 4.

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.

no– to –

no– to –

yes– to +

yes+ to –

Since there are two sign changes, there are 2 or 0 negative real zeros. Make a chart of possible combinations.

x 1

Answer:

Number of Positive Real

Zeros

Number of Negative Real

Zeros

Number of Imaginary

ZerosTotal

2 2 2 6

0 2 4 6

2 0 4 6

0 0 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 yes no

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 andtwo imaginary zeros at and The graph of the function verifies that there is only one real zero.


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 thedifference of twosquares.

Square x – 4 and replace i2 with –1.

Simplify.

Multiply using the Distributive Property.

Combine liketerms.

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.

b.

Answer:

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 –8 –1120

1 1 3 –5 –1125

2 2 6 2 –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 Verify that the dimensions are correct.

Answer:

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:


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

0–15–13312

24–6–17211

3011–19110

3011–1911p

q

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

05613

–15–1331p

q

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.


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 offunctions

and

Simplify.Answer:

Given , find

Subtraction offunctions

and

Simplify.Answer:

Given find each function.

a.

b.

Answer:

Answer:

Givenfind

Product of functions

and

DistributiveProperty

DistributiveProperty

Simplify.Answer:

Given

find

Answer:and

Division of functions

Since 4 makes the denominator 0, it is excluded from

the domain of

Givenfind each function.

a.

b.

Answer:

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, is undefined for x = 2.

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 andand

Composition of functions

Replace g (x) with 2x – 1.

Substitute 2x – 1 for x in f (x).

Evaluate (2x – 1)2.

Simplify.

Compositionof 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.

Simplify.

Answer: So, and

Function from part a

Replace x with –2.

a. Find andand

b. Evaluate and x = 1.

Answer: and

Answer: and

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


If Tracie has her retirement deducted after taxes, then her net income is represented by



Answer: andThe difference is 1344 – 1340 or 4. So, her net income is $4 more if the retirement deduction is taken before taxes.

Examine 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 sideby –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 is the reflection for

over the line

Answer:

a. Find the inverse of

b. Graph the function and its inverse.

Answer:

Answer:

Determine whether and are inverse functions.

Check to see if the compositions of f (x) and g (x) are identity functions.

Answer: The functions are inverses since bothand equal x.

Determine whether and are inverse functions.

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.

2

3

2.836

2.555

2.234

1.873

1.142

0.711

0

yx

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, andx- 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 Make a table of values and graph the function.

a v

0 0

1 1.41

2 2

3 2.45

4 2.83

5 3.16

Answer:

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 equationSince the

boundary is not included, the graph should be dashed.

The domain includes values

for So the graph

is to the right of

Test (0, 0).

Shade the region that does not include (0, 0).false

Select a point and test its ordered pair.

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:

Explore online information about the information introduced in this chapter.

Click on the Connect button to launch your browser and go to the Algebra 2 Web site. At this site, you will find extra examples for each lesson in the Student Edition of your textbook. When you finish exploring, exit the browser program to return to this presentation. If you experience difficulty connecting to the Web site, manually launch your Web browser and go to www.algebra2.com/extra_examples.

Click the mouse button or press the Click the mouse button or press the Space Bar to display the answers.Space Bar to display the answers.

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