review problems for exam 2 ce math 1050

14
Review problems for Exam 2 – CE Math 1050 This is a comprehensive set of review problems for Exam 2, CE Math 1050. The layout of the document is: Learning objectives are numbered and bolded. Sample problems for each learning objective are beneath the objective. Complete answers begin on page 8. For classes taking their exams as written by the UVU Math Department, the question types (not exact questions) for Exam 2 will be taken from a much smaller subset of this set. There is a new request on these problems, that some points be labeled/located on the graph. This means that either the points are labeled with their coordinates, or the points are clearly marked on the graph at obvious intersections of the graphing lines to make the location clear. For the answer key, we have chosen to label the points to make their positions clear. This exam will β€œcover” sections 3.2 to 4.3 from the Stewart text. Chapter 3: Polynomial and Rational Functions A student is able to: 2. Find the vertex and max/min values of a quadratic function. I. For the following functions: graph the function, indicate the location of the vertex, include the y-intercept and at least 3 other points well labeled/located on your graph. Then state the range of the function. a) 2 4 2 x x f b) 5 6 2 x x x f II. A ball is launched upward off a 200-foot cliff and then falls into the ocean below. The formula for height, in feet, t seconds after the ball is thrown is given by β„Ž() = βˆ’16 2 + 64 + 200. a) How many seconds after the ball is tossed does the ball reach its maximum height? [work expected] b) What is the maximum height the ball reaches? [work expected]

Upload: others

Post on 01-Feb-2022

9 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: Review problems for Exam 2 CE Math 1050

Review problems for

Exam 2 – CE Math 1050

This is a comprehensive set of review problems for Exam 2, CE Math 1050. The layout of the document is:

Learning objectives are numbered and bolded. Sample problems for each learning objective are beneath the

objective. Complete answers begin on page 8. For classes taking their exams as written by the UVU Math

Department, the question types (not exact questions) for Exam 2 will be taken from a much smaller subset of

this set.

There is a new request on these problems, that some points be labeled/located on the graph. This means that

either the points are labeled with their coordinates, or the points are clearly marked on the graph at obvious

intersections of the graphing lines to make the location clear. For the answer key, we have chosen to label the

points to make their positions clear.

This exam will β€œcover” sections 3.2 to 4.3 from the Stewart text.

Chapter 3: Polynomial and Rational Functions

A student is able to:

2. Find the vertex and max/min values of a quadratic function.

I. For the following functions: graph the function, indicate the location of the vertex, include the y-intercept

and at least 3 other points well labeled/located on your graph. Then state the range of the function.

a) 242 xxf

b) 562 xxxf

II. A ball is launched upward off a 200-foot cliff and then falls into the ocean below. The formula for height, in feet, t seconds after the ball is thrown is given by

β„Ž(𝑑) = βˆ’16𝑑2 + 64𝑑 + 200. a) How many seconds after the ball is tossed does the ball reach its maximum height? [work expected] b) What is the maximum height the ball reaches? [work expected]

Page 2: Review problems for Exam 2 CE Math 1050

3. Graph a polynomial, showing x- and y-intercepts and proper end behavior.

I. Sketch the graph of the polynomial 𝑃(π‘₯) = 1

2(π‘₯ + 3)(π‘₯ βˆ’ 2)2(π‘₯ βˆ’ 4)3.

What is the leading term? ___________ Describe the end behavior. ____________________________

____________________________ What is/are the x-intercept(s)? _________________________ What is/are the y-intercept(s)? _________________________

What are the zeros of the function and state their multiplicities? ____________________________________ Sketch:

II. Sketch the graph of the polynomial 𝑃(π‘₯) = βˆ’π‘₯3 + 4π‘₯2 βˆ’ π‘₯ βˆ’ 6 What is the leading term? __________ Describe the end behavior. ___________________________ ___________________________ What is/are the x-intercept(s)? ________________________ What is/are the y-intercept(s)? ________________________ What are the zeros of the function and state their multiplicities? ___________________________________ Sketch:

Page 3: Review problems for Exam 2 CE Math 1050

4. Perform long division on polynomials. (Also done in #13 below)

Divide p(x) by d(x) using long division.

a) p(x) = 2x3 βˆ’ 5x2 + 1; d(x) = x2 βˆ’ 3 b) p(x) = 10x3 + x2 + 18x + 2; d(x) = 2x2 βˆ’ x + 4

5. Perform synthetic division on polynomials. (Used in many objectives below.)

6. Use the Remainder Theorem to evaluate a polynomial.

Let 𝑃(π‘₯) = 2π‘₯4 βˆ’ 3π‘₯3 βˆ’ 5π‘₯2 + π‘₯ βˆ’ 6. Use synthetic division or the Remainder Theorem to evaluate P(-2).

7. Use the Factor Theorem to find a factor of a polynomial. (Used in #11 below, and for other

objectives.)

8. Construct a polynomial given the zeros and their multiplicities.

I. Write a formula for a polynomial which has degree 5; zeros 3, 2i, and -2 (with multiplicity 2). (For this problem, the polynomial need not be multiplied out.)

II. Find a polynomial in standard form with the given conditions. (Remember that Standard form is multiplied

out)

a) degree 3, zeros -2, 0, 3 b) degree 3, zeros 0 (multiplicity 2) and 1

III. Find the unique polynomial in standard form with the given condition.

a) degree 2, zeros -3 and -1, leading coefficient an = βˆ’3

Page 4: Review problems for Exam 2 CE Math 1050

9. Use the Rational Zeros Theorem, Descartes’ Rule of Signs, and the Upper and Lower Bounds

Theorem in finding zeros of polynomials.

I. Show that 4 is an upper bound of the real zeros of 𝑃(π‘₯) = π‘₯3 βˆ’ π‘₯2 + 4. Please, also explain what you see that indicates this is an upper bound.

II. Show that -4 is a lower bound of the real zeros of 𝑃(π‘₯) = π‘₯3 βˆ’ π‘₯2 + 4. Please, also explain what you see

that indicates this is a lower bound.

III. For the following function - Use the Rational Zero Theorem to determine all possible rational zeros - Use Descartes’s rule of signs to determine the number of positive and negative zeros - Factor the polynomial - Sketch the graph, showing intercepts and proper end behavior

P(x) = 6x3 + 17x2 + x – 10

IV. For the following function - Use the Rational Zero Theorem to determine all possible rational zeros (p/q’s) - Use Descartes’s rule of signs to determine the number of positive and negative zeros

a) P(x) = 2x6 - x5 - 13x4 + 13x3 + 19x2 - 32x + 12

b) 𝑃(π‘₯) = 4π‘₯7 βˆ’ π‘₯5 + 5π‘₯4 + π‘₯3 + 5.

10. Solve polynomial equations.

Solve the polynomial equation: 6x3 + x2 – 5x – 2 = 0

11. Factor a polynomial into linear and/or irreducible quadratic factors.

Factor the polynomial into linear factors and/or irreducible quadratic factors.

a) p(x) = x3 βˆ’ 3x2 + 4x – 2

b) p(x) = x3 + 5x2 + 17x + 13

c) p(x) = x4 + 4x3 βˆ’ 3x2 βˆ’ 10x + 8

Page 5: Review problems for Exam 2 CE Math 1050

12. Factor a polynomial completely into linear factors real and complex.

I. Given a zero of the polynomial, factor the polynomial into linear factors: p(x) = x4 βˆ’ 6x3 + 14x2 βˆ’ 24x + 40; π‘₯ = 2𝑖 is a zero.

II. Factor the following polynomial completely into linear factors: p(x) = 27x3 – 8

13. Find vertical and horizontal asymptotes of rational functions.

Find all the asymptotes (vertical and horizontal) of the following rational functions.

a) 𝑓(π‘₯) =π‘₯2βˆ’4

π‘₯3βˆ’2π‘₯2βˆ’3π‘₯

b) π‘Ÿ(π‘₯) =(2π‘₯βˆ’3)(π‘₯+1)

π‘₯(π‘₯βˆ’3)

14. Graph a rational function, showing intercepts and asymptotes.

I. Consider the rational function 𝑓(π‘₯) =π‘₯+2

π‘₯2βˆ’4π‘₯+3.

a) Show and label the vertical asymptote(s), if any.

b) Show and label the horizontal asymptote(s), if any.

c) Show and label the x and y intercepts, if any.

d) Write the domain in interval notation.

e) Graph the function.

II. Graph the rational function 𝑓(π‘₯) =π‘₯2+π‘₯βˆ’2

π‘₯3βˆ’π‘₯2βˆ’6π‘₯. Show and label all asymptotes, intercepts, and hole(s), if any.

Page 6: Review problems for Exam 2 CE Math 1050

Chapter 4: Exponential and Logarithmic Functions

A student is able to:

1. Graph exponential and logarithmic equations.

I. Starting with the basic function f (x) = log2x , use transformations, as needed, to answer the following

questions for the function )4(log1)( 2 xxg .

a) Identify the y-intercept of g(x) , if any:

b) State the equation of the asymptote of g(x) , if any:

c) Find the domain and the range of g(x)using

interval notation.

d) Sketch the graphs of 𝑓(π‘₯) and g(x) . (A complete

graph includes at least 3 labeled/located points on the graph.)

II. Starting with the basic function: f (x) = 2x , use transformations, as needed, to answer the following

questions for the function g(x) = 2x+2 -1.

a) Find the y-intercept of g(x) , if any:

b) Find the asymptote of g(x) , if any:

c) Sketch the graph of g(x) . (A complete graph includes

at least 3 labeled/located points on the graph.)

Page 7: Review problems for Exam 2 CE Math 1050

2. Solve interest problems using the compound interest formula:

A P 1r

n

nt

and the continuous

interest formula: A = Pert. I. If you put $3200 in a savings account that pays 2% a year compounded quarterly, how much will you have in

the account in 15 years? Use 𝐴 = 𝑃 (1 +π‘Ÿ

𝑛)

𝑛𝑑 or A = Pert whichever is appropriate. (Please, leave your

answer in a form ready to put into a calculator or you may enter it in your calculator OR round your answer to 2 decimal places.)

II. I want to have $50,000 in 7 years. How much should I invest now in an account earning 6% compounded

continuously? Use 𝐴 = 𝑃 (1 +π‘Ÿ

𝑛)

𝑛𝑑 or A = Pert whichever is appropriate. (Please, leave your answer in a

form ready to put into a calculator OR enter it in your calculator and round your answer to 2 decimal places.)

3. Switch between exponential and logarithmic forms using the definition of logarithm: π’š = 𝒃𝒙 β†”π’π’π’ˆπ’ƒπ’š = 𝒙.

I. Write log813 =1

4 in its equivalent exponential form.

II. Write ex = 6 in its equivalent logarithmic form.

III. Evaluate/Simplify each expression using the definition of logarithm

a) ln e3 = ___________________

b) 7log73βˆ’2 = _________________

c) log2√8 = _________________

d) log2 0.25 = ________________

IV. Solve for x:

a) log2 π‘₯ = βˆ’6

b) logπ‘₯ 81 = 4

Page 8: Review problems for Exam 2 CE Math 1050

ANSWERS Chapter 3: Polynomial and Rational Functions

2. Problem I:

a) Vertex (4,2); y-intercept (0,18); range [2, ∞)

b) Vertex (-3, 4); y-intercept (0, -5); range (-∞, 4]

Problem II:

a) 2 seconds b) 264 feet

Page 9: Review problems for Exam 2 CE Math 1050

3. Problem I:

The leading term is 1

2π‘₯6

End behavior: y β†’ ∞ as x β†’ ∞; y β†’ ∞ as x β†’ -∞, (The graph goes up on both ends)

x-intercepts: (-3, 0), (2, 0), (4, 0) y-intercept: (0, -384) zeros: x = -3 (mult 1), x = 2 (mult 2), x = 4 (mult 3)

Problem II: The leading term is βˆ’π‘₯3 End behavior: y β†’ -∞ as x β†’ ∞; y β†’ ∞ as x β†’ -∞,

(The graph goes up on the left and down on the right) x-intercepts: (-1, 0), (2, 0), (3, 0)

y-intercept: (0, -6) zeros: x = -1 (mult 1), x = 2 (mult 1), x = 3 (mult 1)

Page 10: Review problems for Exam 2 CE Math 1050

4.

a) The quotient is 2π‘₯ βˆ’ 5 and the remainder is 6π‘₯ βˆ’ 14 OR 𝑃(π‘₯)

𝑑(π‘₯)= 2π‘₯ βˆ’ 5 +

6π‘₯βˆ’14

π‘₯2βˆ’3

b) The quotient is 5π‘₯ + 3 and the remainder is π‘₯ βˆ’ 10 OR 𝑝(π‘₯)

𝑑(π‘₯)= 5π‘₯ + 3 +

π‘₯βˆ’10

2π‘₯2βˆ’π‘₯+4

5. No questions 6.

-2 2 -3 -5 1 -6

-4 14 -18 34

2 -7 9 -17 28 So P(-2) = 28 7. No questions 8.

Problem I: p(x) = (x – 3)(x – 2i)(x + 2i)(x + 2)2 OR if you insist on multiplying it out P(x) = x5 + x4 – 4x3 – 8x2 – 32x – 48. (Other answers are correct if they are a just a real number multiple

of P(x).) Problem II:

a) 𝑃(π‘₯) = π‘₯3 βˆ’ π‘₯2 βˆ’ 6π‘₯, (Other answers are correct if they are a just a real number multiple of P(x).) b) 𝑃(π‘₯) = π‘₯3 βˆ’ π‘₯2, (Other answers are correct if they are a just a real number multiple of P(x).)

Problem III: a) P(x) = -3x2 – 12x – 9

9.

Problem I: All the numbers in the last row (1, 3, 12, 52) are non-negative so 4 is an upper bound for the real zeros of P.

4 1 -1 0 4

4 12 48

1 3 12 52

Problem II: The numbers in the last row (1, -5, 20, -65) alternate between nonpositive and nonnegative so -4 is a lower bound for the real zeros of P.

-4 1 -1 0 4

-4 20 -80

1 -5 20 -76

Page 11: Review problems for Exam 2 CE Math 1050

Problem III: Possible rational zeros (p/q’s): Β±1, Β±2, Β±5, Β±10, Β±1/2, Β±1/3, Β±1/6, Β±2/3, Β±5/2, Β±5/3, Β±5/6, Β±10/3 Positive real zeros: one 𝑃(βˆ’π‘₯) = βˆ’6π‘₯3 + 17π‘₯2 βˆ’ π‘₯ βˆ’ 10 So Negative real zeros: two or zero P(x) = (2x + 5)(x + 1)(3x – 2)

Problem IV: a)

Possible rational zeros: Β±1, Β±2, Β±3, Β±4, Β±6, Β±12, Β±1/2, Β±3/2 Positive real zeros: four or two or zero

𝑃(βˆ’π‘₯) = 2π‘₯6 + π‘₯5 βˆ’ 13π‘₯4 + 19π‘₯2 + 32π‘₯ + 12 So Negative real zeros: two or zero

b) Possible rational zeros: Β±1, Β±5, Β±1/2, Β±5/2, Β±1/4, Β±5/4 Positive real zeros: two or zero

𝑃(βˆ’π‘₯) = βˆ’4π‘₯7 + π‘₯5 + 5π‘₯4 βˆ’ π‘₯3 + 5 So Negative real zeros: three or one 10. x = 1, x = -1/2, x = -2/3 11. a) p(x) = (x – 1)(x2 – 2x + 2)

b) p(x) = (x + 1)(x2 + 4x + 13)

c) p(x) = (x – 1)2(x + 2)(x + 4)

Page 12: Review problems for Exam 2 CE Math 1050

12. Problem I: p(x) = (x + 2i)(x – 2i)(x – (3 – i))(x – (3 + i))

Problem II: 𝑝(π‘₯) = (3π‘₯ βˆ’ 2) (π‘₯ βˆ’ (βˆ’1

3βˆ’

π‘–βˆš3

3)) (π‘₯ βˆ’ (βˆ’

1

3+

π‘–βˆš3

3))

= (3π‘₯ βˆ’ 2) (π‘₯ +1

3+

π‘–βˆš3

3) (π‘₯ +

1

3βˆ’

π‘–βˆš3

3) (Since the first term of the polynomial

we were factoring was 27π‘₯3, and ours would only be 3π‘₯3, we must multiply our answer by 9. We will multiply each of the last two factors by 3, which also serves to get

rid of the fractions, too.)

= (3π‘₯ βˆ’ 2)(3π‘₯ + 1 + π‘–βˆš3)(3π‘₯ + 1 βˆ’ π‘–βˆš3)

13. a) vert. asymptotes: x = 0, x = -1, x = 3 , horiz. Asymptote: y = 0 b) vert. asymptotes at x = 0 and x = 3, horizontal asymptote at y = 2 14. Problem I: Vertical asymptotes at x = 1 and x = 3 Horizontal asymptote at y = 0 x-intercept at (-2, 0) y-intercept at (0, 2/3) Domain (-∞, 1) u (1, 3) u (3, ∞)

Page 13: Review problems for Exam 2 CE Math 1050

Problem II: Vertical asymptotes at x = 0 and x = 3 Horizontal asymptote at y = 0 x-intercept at (1, 0) No y-intercept Hole at (-2, -3/10)

Chapter 4: Exponential and Logarithmic Functions

1. Problem I:

a) y-intercept: (0,3) b) Vertical asymptote at x = -4 c) Domain (-4, ∞), Range (-∞, ∞) d) Sketch:

Page 14: Review problems for Exam 2 CE Math 1050

Problem II:

a) y-intercept: (0,3) b) Horizontal Asymptote at y = -1

c) Sketch (the graph of f (x) = 2x need not be present for full credit.)

2.

Problem I: At the end of 15 years, there will be 3,200 (1 +0.02

4)

(4Γ—15) 𝑂𝑅 3,200(1.005)60 dollars in the

account OR approximately $4,316.32

Problem II: We will need to invest 50,000

𝑒(0.06Γ—7) 𝑂𝑅 50,000

𝑒(0.42) dollars initially OR approximately $32,852.34

3.

Problem I: 81(1 4⁄ ) = 3 Problem II: π‘™π‘œπ‘”π‘’6 = π‘₯ OR ln6 = x Problem III:

a) 3

b) πŸ‘βˆ’πŸ 𝑂𝑅 𝟏

πŸ—

c) log2 √8 = log2 23

2 = πŸ‘

𝟐

d) log2. 25 = log21

4= log2 2βˆ’2 = βˆ’πŸ

Problem IV:

a) π‘₯ =1

64

b) π‘₯ = 3