Algebra 2 Semester 1 (#2221) - MathGuy.US · ALGEBRA 2 SEMESTER 1 INSTRUCTIONAL MATERIALS Course:...

ALGEBRA 2 SEMESTER 1 INSTRUCTIONAL MATERIALS Course: Algebra 2 Semester 1 (#2221) -

Released 8/22/16

Instructional Materials for WCSD Math Common Finals

The Instructional Materials are for student and teacher use and are aligned to the 2016-2017 Course Guides for the following course:

Algebra 2 Semester 1 (#2221)

When used as test practice, success on the Instructional Materials does not guarantee success on the district math common final.

Students can use these Instructional Materials to become familiar with the format and language used on the district common finals. Familiarity with standards and vocabulary as well as interaction with the types of problems included in the Instructional Materials can result in less anxiety on the part of the students. The length of the actual final exam may differ in length from the Instructional Materials.

Teachers can use the Instructional Materials in conjunction with the course guides to ensure that instruction and content is aligned with what will be assessed. The Instructional Materials are not representative of the depth or full range of learning that should occur in the classroom.

*Students will be allowed to use a Scientific or graphing calculator on

Algebra 2 Semester 1 and Algebra 2 Semester 2 final exams.

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Algebra 2 Semester 1 Test Reference Sheet

Sum of Two Cubes

Difference of Two Cubes

Note from Earl. Many of the solutions in this document use techniques presented in the Algebra Handbook, which is available on the www.mathguy.us website. If you have trouble following any of the techniques used, try looking in the handbook for pages that deal with the issue you are struggling with.

I solve the problems in this test using the quickest method available in most cases. Occasionally, I also make comments about some of the math involved in an effort to enhance your understanding of what is going on in the problem.

MultipleChoice:Identifythechoicethatbestcompletesthestatementoranswersthequestion.

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1. Solve the following equation for y: 7 2 5 7

A. 4 2 C. 5

2

B. 4 12 D. 52

To solve an equation in terms of a designated variable, perform the following steps:

Gather all terms containing the variable on one side of the equal sign, and all terms not

containing the variable on the other side of the equal sign.

Factor out the variable in question.

Divide both sides of the equation by the factor that does not contain the variable.

Here we go:

Original equation: 7 2 5 7 Subtract 7 from both sides: 7 7

Result: 2 5

Factor out : 2 5

Divide by 2 : 2 2

Result:

AnswerD

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2. Which graph represents the piecewise function?

12

1, 3

13

2, 3

A. C.

B.

D.

First, note that the breakpoint for the function occurs at 3. In terms of the above graphs,

this leaves out A (which has a gap) and D (which is not a function).

Next, notice that following about :

The piece of with a positive slope (since 1) is the piece to the

right of 3.

Of the remaining answers, B and C, this occurs in C and not in B.

AnswerC

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3. Which of the following piecewise functions represents the graph below?

A. 32

1, 2

2 3, 2 12, 1

B. 32

1, 2

2 3, 2 12, 1

C. 32

1, 2

2 3, 2 12, 1

D. 32

1, 2

2 3, 2 12, 1

Let’s look carefully at the pieces in the graph.

First, note that the breakpoints for the function occur at 2 and 1. The solid dots in the graph occur at the left endpoints of both the middle piece and the right piece. Solid dots

correspond to " " or " " signs in the notation given. This leaves out C and D, both of which have “<” signs on the left side of the middle piece.

Next, notice that the slopes on the left and middle pieces in the graph are positive and the slope of

the right piece is negative. Of, the remaining answers, A and B, this occurs in A and not in B.

AnswerA

Left Center Right

Piece Piece Piece

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4. Which of the following is a proper description of the range of the function shown:

A. : | 3 6 C. : | 2 1

B. : | 3 6 D. : | 2 1

Recall the following definitions:

The Domain of a function is the set of all values of for which it is possible to calculate a ‐

value.

The Range is the set of ‐values that function takes on for all in the function’s Domain.

To show the range, I like to locate a function’s bottom and/or top ‐values and draw a line

between them off to the side of the graph. This in indicated by the lines to the left of the figure

above. (Note: if the range is unlimited in either the up or down direction, I would show an arrow

in that direction instead of a line.)

Notice that I drew a solid magenta line for 1, indicating that there is a closed dot at the point 6, 1 . I drew a dashed magenta line for 2, indicating that there is an open dot at the point 3, 2 .

Based on this, the range is the set of all real numbers from 2 to 1, excluding 2 because of the open dot at the point 3, 2 .

In set notation, the range would be shown as: : |

In interval notation, the range would be shown as: ,

AnswerD

1

2

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5. Which of the following is the graph of over the

domains 3, 3 ∪ 5, 7 ?

A. C.

B. D.

Recall that the parent absolute value function is: | | , where , is the vertex.

What do we know about the equation given: for ∈ 3, 3 ∪ 5, 7 ?

It has vertex: , , or, in mixed numbers: 1 , 3 . All of the solutions have this

property. No help for us here.

It forms a “V” shape with a sharp point because it is an absolute value function. This

eliminates C.

It opens upside down because of the minus sign in front of the absolute value sign. This

eliminates D.

It does not have any values between 3 and 5 because must be in one of the

intervals 3, 3 or 5, 7 . All of the solutions have this property. No help for us here.

It has an open point at 3 because of the parenthesis in 3, 3 . It has a closed point

at 5 because of the bracket on the left of 5, 7 . This eliminates A.

The only solution left is B, and it fits everything we know about the function.

AnswerB

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6. Three students were chosen to show their solutions for solving the equation for x. Their work is shown below. Determine which students were correct.

Student #1 Student #2 Student #3

A. Student #1 and Student #2

B. Student #2 and Student #3

C. Student #1 and Student #3

D. All students were correct

Student 1 is correct.

Student 2 forgot to divide the constant by in step 2.

Student 3 is correct. (Note that the solution shown is

equivalent to that shown for Student 1.)

AnswerC

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7. Solve the following system for y:

35

2 3 2

A. 15

B. 4.6

C. 3

D. 2.5

Next, let’s re‐write the 2nd and 3rd equations, making 1. Note that, alternatively, we could use the first and 3rd equations. We just can’t use the first two, because we already used that pair

to find . We must use a different pairing of the original equations.

Equation 2: 5 becomes: 1 5, or, by subtracting 1: 4

Equation 3: 2 3 2 becomes: 1 2 3 2, or, by adding 1: 2 3 3

Let’s work with the two new equations to find the value of , as required by the problem. To

obtain , we need to eliminate . So let’s multiply the Revised Equation 2 by 3 and add it to the

Revised Equation 3.

4 3 3 3 12 2 3 3 1 2 3 3

5 15 so AnswerC

These can be a little tricky, so let’s look for a good way

to start. Notice that the first two equations have the

same ‐ and ‐terms. Let’s eliminate and to find

the value of . We can do this by multiplying the

second equation by 1 and adding it to the first equation:

First: 3 Second ∙ 1 : 5

2 2 so 1

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8. Which equation is obtained after the translation of the graph up 2 units and left 6 units?

A. | 2|

B. | | 2

C. | 2|

D. | | 2

The general form of an absolute value function is: | | , where , is the

vertex of the function.

The function shown has a vertex at 4, 2 , and a slope of 1, so it’s form would be:

| 4| 2.

Now, let’s translate it (see the arrows in the above chart).

Translation left 6 units means “subtract 6 from .” So, our new 4 6 .

Translation up 2 units means “add 2 to .” So, our new 2 2 .

These translations generate the equation | | , or | |

AnswerA

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9. Which of the following is the graph of over the domain 3, 3

A. C.

B. D.

Recall that the parent absolute value function is: | | , where , is the vertex.

What do we know about the equation given: over 3, 3 ?

It has vertex: , , or, in mixed numbers: 1 , 2 . All of the solutions

have this property.

It forms a “V” shape with a sharp point because it is an absolute value function. This eliminates A and C.

It opens upside down because of the minus sign in front of the absolute value sign. This eliminates D.

The only solution left is B, and it fits everything we know about the function.

AnswerB

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10. The graph of is vertically compressed by a factor of and translated to the

right three units and down one unit to produce the function . Which of the following equations represents ?

A. 1

23 1 C. 1

23 1

B. 12

1 3 D. 12

1 3

Recall that the parent quadratic function is: , where , is the vertex, is

the stretch or compression, and the sign of determines whether the function opens up or down.

Let’s look at what we are told:

There is no mention of a reflection, so is positive.

We have a compression factor of . So, .

The parent function is translated to the right three units and down one unit, so ,3, 1 .

The resulting equation, then, is:

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11. Compare the two functions represented below. Determine which of the following statements is true.

Function Function

6 4

This function:

Opens up,

Has vertex at (6, ‐4).

Is not stretched or compressed.

Is not reflected.

A. The functions have the same vertex.

Recall that the parent quadratic function is: , where , is the vertex. So, the vertex of the right function is 6, 4 . The vertex of the left function is 2, 4 from the graph. So, this statement is FALSE.

B. The minimum value of is the same as the minimum value of .

Both functions open up, so the minimum value of each is the ‐coordinate of its vertex (i.e., ). For the left function, this is 4 (from the graph), and for the right function, this is also 4. So, this statement is TRUE.

C. The functions have the same axis of symmetry.

The equation for the axis of symmetry is , where h is the ‐coordinate of the vertex. For the left function, the axis is 2, and for the right function, the axis is 6. So, this statement is FALSE.

D. The minimum value of is less than the minimum value of .

In B, above, we determined that the minimum values were the same, so this statement is FALSE.

AnswerB

Alternatively, graph on a calculator and read the required values off the graphs.

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12. If the function is translated left eight units and up ten units, how will the domain and range of the function change?

A. The domain will become : | 8 and the range will become : | 10 .

B. The domain will become : | 8 and the range will become : | 10 .

C. The domain will become : | 8 and the range will remain : | .

D. The domain will remain : | and the range will remain : | .

The Domain and Range of all polynomials with odd lead coefficients are All Real Numbers. Since

this function has degree 3, which is odd, translated is irrelevant; the Domain and Range remain All

Real Numbers.

AnswerD

13. Which of the following functions has a range of all real numbers?

I. 12

4 5

II. 12| 4| 5

III. 12

4 5

IV. 12

4 5

A. I only C. I and IV

B. II only D. II and III

Based on the above comments in blue, only and have unlimited ranges, that is, ranges

that are all real numbers. AnswerC

Linear function: always has range: .

Absolute value function: has a limited range because it has a vertex.

Quadratic function: has a limited range because it has a vertex.

Cubic function: always has range: .

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14. The function is graphed to

the right. Over which intervals of x is the graph positive?

A.

B.

C.

D.

15.

16.

Easy Peezy. Above the ‐axis is the

green part in the figure above.

Where does the curve exist in this

green area?

Answer: between 3 and 0, as well as to the right of 2.5. On the number line, all

intervals should be open.

AnswerA

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17.

18.

19. Simplify:

A. 2 2

B. 2 3

C. 6 22

D. 10 10

AnswerA

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You will need the quadratic formula for the next couple of questions:

√ 42

providessolutionstotheequation: 0

20. What are the solutions to the quadratic equation, 3 7 11 5 7?

A. 2 2 √11

3 C. 1 √11

3

B. 1 2 √113

D. √113

First, combine all terms on one side of the equation.

Original Equation: 3 7 11 5 7

Subtract 5 7 : 5 7 5 7

Result: 3 2 4 0

Next, use the quadratic formula with 3, 2, 4

2 2 4 3 42 3

2 √ 44

6

2 2 √116

√

AnswerC

21. What are the solutions to the quadratic equation, 2 9 5 ?

A. 3 3 √32

C. 3 3 √52

B. 3 3√52

D. 3 3√52

First, combine all terms on one side of the equation.

Original Equation: 2 9 5

Subtract 5 9 : 5 9 9 5

Result: 3 9 0

Next, use the quadratic formula with 1, 3, 9

3 3 4 1 92 1

3 √45

2

√

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22. Solve the following equation: 5 8

A. √23 C. 5

2√33

B. √43 D. 5 4√3

Original Equation: 5 8

Multiply by 6 to get rid of the fraction: ∙ 6 ∙ 6

Result: 5 48

Take square roots: 5 √48

Simplify: 5 4√3 Subtract 5: 5 5

Result: 5 4√3

AnswerD

23. The function is graphed below. What are the solutions to 0?

A. 8, 0

B. 3, 5

C. 4 2

D. 4

Real solutions are located where the curve crosses the ‐axis. Since this function does not

cross the x‐axis, it has no real solutions, which eliminates answers A and B.

Important point you may not know: The real portion of complex solutions is , the ‐

coordinate of the vertex. So the complex solutions have the form: ∙ something. Since this equation has complex solutions and 4, those solutions are of the form:

4 ∙ something. AnswerC

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24. Given the function, 4 3, determine which of the following statements is true.

A. The maximum value of is 3 and 0 has no real solutions.

B. The maximum value of is 3 and 0 has two real solutions.

C. The maximum value of is 4 and 0 has no real solutions.

D. The maximum value of is 4 and 0 has two real solutions.

Recall that the parent quadratic function is: , where , is the vertex. The

maximum (or minimum) of a quadratic function occurs at its vertex, and is the ‐coordinate of its

vertex (i.e., ).

This function opens down because of the negative sign in front of the quadratic term. Therefore,

it has a maximum value. Since 3, the maximum value is 3. Since the maximum is a

negative value, this function never crosses the ‐axis and so it has no real solutions.

AnswerA

Alternatively, graph the function on a calculator and read the required information off the graph.

25. Given the function, 2 7 find the minimum value.

A. Minimum value is 5

B. Minimum value is 6

C. Minimum value is 7

D. Minimum value is 8

This function opens up because of the positive sign in front of the quadratic term. Therefore, it

has a minimum value. The vertex of a quadratic in form occurs at:

2

22 ∙ 1

1

Substitute this into the original equation to find the minimum value of the function:

1 1 2 1 7

AnswerB

Alternatively, graph the function on a calculator and read the required information off the graph.

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26. Which of the following functions is equivalent to 6 11 ?

A. 12

6 7 C. 12

6 11

B. 12

6 14 D. 12

6 22

Looks like we need to complete the square to put in , form.

12

6 11

12

12 11

12

12 36 1112∙ 36

27.

Which function is represented by the graph?

A. 2 3

B. 2 3

C. 2 3

D. 2 3

The zeros of the function are 2 and 3. The signs on zeros reverse when in factor form,

so the factors are and . AnswerC

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28. Which of following functions does not represent the parabola with a vertex at 1, 4 and x-intercepts 1, 0 and 3, 0 .

A. 4 C. 2 3

B. 1 4 D. 1 3

Best way to approach this one may be with the answer that obviously has a vertex at 1, 4 –

answer B. So, answer B is not what we want. Let’s try multiplying it out and then factoring

to get the other two answers represent the same curve.

1 4

2 1 4

2 1 4 2 3, so answer C is not what we want.

Now, factor:

2 3

2 3 1 3 , so answer D is not what we want.

That leaves answer A as our solution.

AnswerA

Alternatively, graph each function on a calculator to see if it has the required vertex and x‐

intercepts.

29. Which of the following functions does not have zeros of 4 and 8 ?

A. 2 36 C. 4 32

B. 2 8 64 D. 4 8

A “zero” of a function gives a ‐value of zero when substituted into the equation. We could

substitute 4 and 8 into each equation to see if we get 0, but that would take a lot of time. Let’s try something else.

Inspecting the answers, we see that answer D is in intercept form, and that its zeros are 4 and 8. Since we want the equation that does not have 4 and 8 as zeros, we have found it by inspection. Whew! That was a close one!

AnswerD

Alternatively, graph each function and check its zeros (i.e., where it crosses the ‐axis).

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30. If 3 9 , then what is the value of ?

A. 36 C. 18

B. 27 D. 9

We could multiply 3 9 out and complete the square, but there is a simpler

approach. If you like this approach, use it, if not, multiply and complete the square (or graph

as indicated in the alternative below). Here are the steps:

We know that is the ‐coordinate of the vertex.

We also know that the ‐coordinate of the vertex is half way between the zeros of

the quadratic function.

The zeros are obviously at 3, 9 .

Halfway between these is 3.

Substitute 3 for in the original expression to get the value of :

3 3 3 3 9 6 6

AnswerA

Alternatively, graph the function on a calculator and read the value of off the graph.

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31. Compare the axis of symmetry and the minimum values for the two functions below. 2 3 7 4 21

Determine which of the following statements is correct. A. The functions and have the same axis of symmetry, but the minimum value of is

less than the minimum value of .

B. The functions and have the same axis of symmetry, but the minimum value of is greater than the minimum value of .

C. The functions and do not have the same axis of symmetry, and the minimum value of is less than the minimum value of .

D. The functions and do not have the same axis of symmetry, and the minimum value of is greater than the minimum value of .

Axis of Symmetry

We know that the axis of symmetry is half way between the zeros of a quadratic function.

For , the zeros are 3, 7 , so the axis of symmetry is 2.

The vertex of a quadratic in form occurs at . So, for , the

axis of symmetry is ∙

2. Therefore, the functions have the same axis of

symmetry. This eliminates answers C and D.

Minimum Values

Minimum values are determined at the vertex, and the vertex lies on the axis of symmetry.

So, for both functions, the minima will be found at the function values where 2.

2 2 2 3 2 7 2 ∙ 5 ∙ 5 50

2 2 4 ∙ 2 21 4 8 21 25

Therefore, the minimum value of is less than the minimum value of .

AnswerA

Alternatively, graph both functions on a calculator and compare the axes of symmetry and

minimum values of the functions from their graphs.

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32. Compare the functions, and , and explain how the graph of 4 4 is related to the graph of 4 2.

A. is vertically stretched to make

B. is translated 6 units left to make

C. is translated down 6 units to make

D. is compressed vertically to make

After the last couple of problems, it’s good to get an easy one. Let’s look at the two

equations:

4 4

4 2

They are pretty similar. The only difference between the two equations is that has a

constant term that is 6 less than .

Since the constant term indicates the vertical placement of the curve, is vertically 6 units below . Therefore, to get from , you would translate down 6 units.

AnswerC

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33. Translate 4 6 five units to the left. What is the graph obtained after the translation?

A.

C.

B. D.

Notice that the vertices in the graphs are at four different locations. So, if we can find the vertex

after the translation, we have our answer.

The easiest way to determine the vertex of this function is to complete the square:

4 6 4 ___ 6 ___

4 6

2 2

which has vertex , 2, 2

Finally, we need to translate the vertex 5 units to the left. Our new vertex becomes:

2 5, 2 , AnswerC

To get the missing constant term, divide

the coefficient of by 2, then square the

result. In this case,

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34. A company is planning on selling plastic widgets. The company has to pay $600 in production costs plus $50 per widget to produce widgets. The equation, 600 50 , models the total production costs. Based on sales of a similar product, the

company expects their revenue to be modeled by 400 . Use the

equation to find the maximum profit, , the company can expect to make.

A. $5,325 C. $15,025

B. $10,650 D. $40,650

12

400 600 50

20012

600 50

12

150 600

The maximum of a quadratic function (with a negative lead coefficient) will always occur at

the vertex. For an equation in form, this occurs at .

For the above equation, 150.

Finally, we need the ‐coordinate of the vertex:

15012150 150 150 600 10,650

AnswerB

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35. A local business wants to expand the size of their rectangular parking lot that currently measures 120 by 200 . The project will cost less if equal amounts are added to each side, as shown below. Zoning restrictions limit the total size of the parking lot to 35,000 . What is the maximum amount of distance that can be added on to the each side of the parking lot? Round your answer to the nearest foot.

A. 1 C. 105

B. 31 D. 187

The size of the new parking lot will be: 200 120 , which we want to be 35,000 square feet. This boils down to solving the equation: 200 120 35,000.

Original equation: 200 120 35,000

Expand the expression: 320 24,000 35,000 Subtract 35,000: 35,000 35,000

Result: 320 11,000 0

Use the quadratic formula:

Result: 351.3, 31.3

Only the positive solution makes sense in the context of this problem.

AnswerB

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36. A parabola has a vertex of 5, 6 and passes through the point 10, 4 . In the form of the parabola, what is the value of ?

A. 2 C. 25

B. 25

D. 245

We start with the form: , where , is the vertex. The vertex of the

left function is 5, 6 , so our equation becomes: 5 6. To find the value of ,

we must substitute the values of 10 and 4 (from our point 10, 4 ) into this

equation, and solve for .

Original equation: 5 6

Substitute values for and : 4 10 5 6

Result: 4 25 6 Subtract 6: 6 6

Result: 10 25

Divide by 25: 2525

Result: AnswerC

37. Which of the following systems of equations could a student use to write a quadratic function in standard form for the parabola passing through the points 1, 3 , 4, 6 , and 5, 9 ?

A. 3

4 65 9

C.3

16 625 9

B. 2 38 4 610 5 9

D.3

16 4 625 5 9

All of the solutions have three terms on the left, suggesting the use of the form:

. Creating a system of equations like this requires us to replace the values of and

in that form with the ‐ and ‐values of the points given. So, we are looking for equations

that do not have any ’s or ’s left in them. This eliminates answers A and C.

The multipliers of in the equations are the values of from the points, so we are looking

for coefficients of equal to 1 1, 4 16, and 5 25. In the remaining answers, B

and D, only D has these coefficients.

AnswerD

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38. A parabola has x-intercepts at 2 and 6 and goes through the point 8, 4 . What other point is on the parabola?

A. 2, 2 C. 3, 15

B. 4, 2 D. 1, 15

Knowing the zeros of a quadratic function tells us a lot about its equation. With zeros of 2 and 6, we know that the equation has the form: 2 6 , where is yet to be

determined.

Using the point on the line, 8, 4 , we can determine the value of by substituting 8 and 4 into the equation and solving for :

4 8 2 8 6

4 6 2

13

And, so our equation becomes: 2 6

To determine which point lies on the curve, substitute and of each point into this

equation to see which ones make the equation work(1):

A. 2, 2 2 2 2 2 6

2 4 4 FALSE

B. 4, 2) 2 4 2 4 6

2 2 2 FALSE

C. 3, 15) 15 3 2 3 6

15 5 9 15 TRUE

D. 1, 15 15 1 2 1 6

15 1 5 5/3 FALSE

AnswerC

Note from Earl: We would stop working on this problem after finding the true solution in answer

C. Unless I am missing something, this is a very time consuming problem and should be left until

after all other questions have been answered.

(1) Alternatively, graph the equation 2 6 and see which point is on the curve.

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39. Solve the following system for the y-coordinates of the system:

6 84 7

A. 13 19 C. 5 13

B. 3 5 D. 3 19

To get the ‐values, we need the ‐values, so let’s just solve the system.

Both equations are in the form " ", so let’s set them equal to each other and solve for .

Set the equations equal: 6 8 4 7 Add 4 7 : 4 7 4 7

Result: 2 15 0

Factor: 5 3 0

Solve for : 5, 3

Next, we substitute the ‐values into either of the original equations to obtain the ‐values.

5: 4 5 7

3: 4 3 7

AnswerA

40. What are the x-coordinates of the solution for the system given below?

10 6 8 132 3 7

A. 2 4 C. 3 3

B. 4 2 D. 3

We want to eliminate . Notice that the term in the first equation is double the ‐term in

the second equation, so let’s multiply the second equation by 2 and add it to the first.

10 6 8 13 1 10 6 8 13 2 3 7 2 4 6 14

6 8 1 Add 1 to both sides: 1 1

Result: 6 9 0

Factor: 3 3 0

Solve:

AnswerD

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41. Given 3 6 5 and 3 23, complete the tables below to find the x-values where .

3 6 5

3

2

1

0

1

2

3

4

5

6

7

3 23

3

2

1

0

1

2

3

4

5

6

7

A. C. 0, 1, 2 ,6 ,7

B. 14, 29 D. 2, 3

Earl’s advice is to ignore the method they tell you to use and just solve the system. In fact,

whenever you are told to use a particular method in a multiple choice test, treat it as a

suggestion that you are free to ignore!

Set the equations equal: 3 6 5 3 23 Add 3 23 : 3 23 3 23

Result: 3 3 18 0

Factor out a 3: 3 6 0

Factor the trinomial: 3 2 3 0

Solve for : ,

AnswerD

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42. Find the perimeter of the figure below:

A. 6 4 3 2 C. 2 4

B. 4 7 3 D. 4 8 2

To find the perimeter, add the lengths of the four sides of the rectangle:

Expression Side

2 Bottom

3 2 Left

2 Top (same as bottom)

3 2 Right (same as left)

4 8 2

AnswerD

43. Which expression must be subtracted from 4 6 2 to result in 5 9 3 ?

A. 9 15 1 C. 3 8

B. 3 5 D. 15 1

We want to find an “expression” such that:

4 6 2 expression 5 9 3

Adding “expression” to both sides and subtracting 5 9 3 from both sides of the

equation gives something we can work with:

4 6 2 5 9 3 expression

Then, remembering to change the signs of 5 9 3 and add, we get:

4 6 2 5 9 3

3 5 AnswerB

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44. Find the product: 1 5 3

A. 15 C. 3 13 15

B. 17 15 D. 7 7 15

First, let’s multiply two terms together, then multiply the result by the third term.

1 5 5 5 4 5

Then, multiply the result by 3

AnswerB

45. Multiply: 2 4 5 3 6

A. 2 2 29 9 30 C. 2 9 21 30

B. 2 10 19 9 30 D. 2 24 30

AnswerA

46. Factor completely: 13 36

A. 9 4 C. 3 3 2

B. 3 3 4 4 D. 3 3 2 2

13 36 4 9

2 2 3 3

3 3 2 2 after rearranging terms.

AnswerD

4 5

3

3 12 15

4 5

17 15

2 4 5

3 6

12 24 30

6 12 15

2 4 5

2 2 29 9 30

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47.

Factor completely: 64 27

A. 4 3 4 12 9 C. 4 3 16 12 9

B. 4 3 4 12 9 D. 4 3 16 12 9

To factor this expression, which is a difference of cubes, we must use the formula provided

on the second page of this document:

Let’s proceed to solve this problem:

64 27 4 3

4 3 ∙ 4 3 ∙ 4 3

4 3 16 12 9 AnswerC

48. Solve: 3 2 5 0

A. 1, 0,

53

C. 53, 0, 1

B. 3, 0, 5 D. 5, 0, 3

Alternatively, graph the function and see where it crosses the ‐axis.

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49. Solve: 3 10

A. 5, 2 C. √5, √2

B. 5, 2 D. √5, √2

50. Solve: 36 0

A. √6, 6 C. 6, 6

B. √6, √6 D. 6, √6

51. One way to factor 72 is to rewrite the expression as 72. What is the equivalent of ?

A. 9 C.

B. D.

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52. What is the remainder in the division 4 80 150 5 ?

A. 250 C. 1050

B. 650 D. 1150

53. Use synthetic or long division to find the quotient of 2 33 16 16 ?

A. 2 33

1616

B. 2 1

C. 2 13216

D. 2 1

For a full explanation of synthetic division, see pages 122‐123 of the Algebra Handbook or

use the synthetic division section of the Algebra App, both of which are available at

www.mathguy.us.

The easiest approach to this is to use synthetic division. First,

note that the root implied by the divisor 16 is 16.

162 3316

32 16

2 10

The result is: AnswerB

The easiest approach to this is to use synthetic division. First, note that the root implied

by the divisor 5 is 5. Also, don’t forget the 0 term in the first expression.

5 4080 150

20 100 100

4 20 20 AnswerA

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54. Given 9 is a factor of the polynomial, 16 144 81 729, what are the remaining factors?

A. 4 9 4 9

B. 16 81

C. 4 9

D. 4 9 4 9

55. What is the end behavior for the function, 4 3 3 6 ?

A. as → ∞, → ∞ and as → ∞, → ∞

B. as → ∞, → ∞ and as → ∞, → ∞

C. as → ∞, → ∞ and as → ∞, → ∞

D. as → ∞, → ∞ and as → ∞, → ∞

To determine end behavior, we need only know the lead term (the one with the highest

exponent). To get this term, multiply the terms with the highest exponents in each of the

factors given.

Lead term of is: ∙ 3 3 .

Notice that the sign of the lead term is positive and that the exponent is odd. So, this

functions will have end behavior the same as the function . That is:

As → ∞, → ∞ and as → ∞, → ∞

AnswerA

Check out the following table that describes the end behavior of any polynomial:

End Behavior of Polynomials

Positive lead coefficient

Acts like:

Negative lead coefficient

Acts like:

Odd exponent on lead term

Even exponent on lead term

If 9 is a factor, then a root is: 9.

916 144 81729

1440 729

16 0 81 0

The result is: 16x 81

Recognize a difference of squares: 4 9

Factor the expression: 4 9 4 9

AnswerA

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56..

Note: We typically report relative maxima and minima using ordered pairs, which is preferred in

higher level math classes. However, this problem is asking for the ‐values only. How do we

know this? Simply by looking at the form of the answers given.

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58.

Which polynomial is graphed below?

A. 1 3

B. 1 1 3

C. 3 1

D. 3 1

This function appears to be a cubic equation with

three real zeros – at 1, 0, 3 .

Each zero, , corresponds to a factor of .

The resulting equation has a factor for each zero, in particular: 1 3 .

Since the factors can appear in any order, we look through the answers to determine the

correct answer has the three factors slightly rearranged: 3 1 .

AnswerC

59. Which of the following is not a possible rational root of the function:

3 4 6 8 ? A. 1 C. 8

3

B. 43

D. 6

Possible roots are of the form where is a factor of the constant term, 8, and is a

factor of the lead coefficient, 3. The factors are as follows:

(factors of 8): 1, 2, 4, 8 (note: factors always have “ ” signs)

(factors of 3): 1, 3

So, all possible rational roots have form: ⇒ 1, 2, 4, 8, , , ,

Comparing these with the answers above, we see that answers A, B and C are in this list, but

answer D, 6, is not. AnswerD

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60. Given 5, use synthetic division to find the two remaining real solutions of 3 35 128 140 0.

A. 2,143

B. 6,143

C. 2,143

D. no other solutions

61. The equation 3 4 12 0 is graphed below. Use the graph to help solve the equation and find all the roots of the function.

A. 3, 2, 2

B. 12, 1, 3

C. 3, 2 , 2

D. 12,3 √7

2,3 √7

2

It is clear from the graph that one of the roots of the

equation is 3. We could use synthetic division on

the original equation to determine the quadratic that would

remain when we extract that root. However, we can be smarter and save ourselves some

work.

It appears that the only real root of this equation occurs at 3. Only answer C to this question that has a single real root equal to 3. Answers A and B have more than one real

root, and answer D does not have 3 as a root. AnswerC

Note that one root is given to us directly: 5.

53 35128 140

15 100140

3 20 28 0

The result is: 3x 20x 28 0

Split the middle term: 3x 6x 14x 28 0

Pair terms: 3x 6x 14x 28 0

Factor pairs: 3 2 14 2 0

Collect terms: 3 14 2 0

Break into separate equations: 3 14 0 2 0

Solutions for in the equation: , AnswerA

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62. A manufacturer is going to package their product in an open rectangular box made from a single flat piece of cardboard. The box will be created by cutting a square out from each corner of the rectangle and folding the flaps up to create a box. The original rectangular piece of cardboard is 20 long and 15 wide. Write a function that represents the volume of the box.

A. 35 300 C. 35 300

B. 4 70 300 D. 4 70 300

The length of the box is: 20 20 2

The width of the box is: 15 15 2

The height of the box is:

Volume of the box is: ∙ ∙ 20 2 ∙ 15 2 ∙

300 40 30 4 ∙

4 70 300 ∙

4 70 300

AnswerB

20 2

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63. Sketch the graphs of and on the same coordinate plane given the following information:

has zeros at 6, 1, 2 As → ∞, → ∞ and as → ∞, → ∞ has a local minimum at 1, 5 and a local maximum at 4, 9 5

How many real solutions exist when ?

A.

B. 1

C. 2

D. 3

Let’s see what each item tells us about the curves:

has zeros at 6, 1, 2. This suggests that is a cubic (it has 3 zeros).

As → ∞, → ∞ and as → ∞, → ∞. This tells us that

the degree of the curve is odd (this is consistent with the suggestion that it

is a cubic) and that its lead coefficient is positive. So it looks like this:

has a local minimum at 1, 5 and a local maximum at 4, 9 . This

is consistent with our shape; the minimum is to the right of the maximum.

5. This is a separate function, a horizontal line at 5.

This problem is asking us how many places the two functions intersect. A sketch of the two

curves is shown below. Notice that the line 5, being between the minimum 5 and

maximum 9 of , must intersect in three places if is a cubic.

AnswerD

Note: as described, could be 5th, 7th, or any other odd degree polynomial, in which

case could have more than three real solutions (points of intersection).

Therefore, a better answer to this question would be “at least 3 real solutions.”

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Algebra 2 Semester 1 (#2221) - MathGuy.US · ALGEBRA 2 SEMESTER 1 INSTRUCTIONAL MATERIALS Course:...

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