Algebra 2 Interactive Chalkboard Copyright by The Mc

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 2 c– 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 3 a– 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(2 x – 1), replace m in b(m) with 2 x – 1. Original function Replace m with 2 x – 1. Evaluate 2(2 x – 1)2. Simplify.

To evaluate 3 b(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(2 x – 1) – 3 b(x). Replace b(2 x – 1) and 3 b(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 4 th degree polynomial function, it will have between 0 and 4 zeros, inclusive. x f (x) – 2 9 – 1 0 1 1 – 3 2 – 7 3 19 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) 0 110 1 2 3 4 109. 5 108. 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 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 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 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. or or Zero Product Property

Solve each equation. Answer: This equation has two real roots, – 2 and 2, and two imaginary roots, 2 i and – 2 i.

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, 3 i and – 3 i.

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 –

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 – 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 2 0 0 2 4 4 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 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 1 1 – 4 – 3 – 2 14 8 4 – 38 – 12 0 – 2 – 1 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.

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 i 2 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 x 4 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 10 1 1 2 3 6 12 – 8 – 5 2 112 – 1120 – 1125 – 1112 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 1 1 – 19 11 30 0 1 2 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 1 3 – 15 3 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 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 Composition of functions Replace g (x) with 2 x – 1. Substitute 2 x – 1 for x in f (x).

Evaluate (2 x – 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 Answer: b. Evaluate Answer: 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, 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, 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 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. a v 0 0 1 2 3 4 1. 41 2 2. 45 2. 83 5 3. 16 Make a table of values and 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 units 2

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 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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