AP Calculus AB Summer Math PacketName_____ Date_____ Section_____ AP Calculus AB Summer Math Packet This assignment is to be done at you leisure during the summer. It is meant to help

Name____________________________________Date_______________Section__________

AP Calculus AB

Summer Math Packet

This assignment is to be done at you leisure during the summer. It is meant to help you

practice mathematical skills necessary to be successful in Calculus AB. All of the skills

covered in this packet are skills from Algebra 2 and Pre-Calculus. If you need to use

reference materials please do so.

While graphing calculators will be used in class the majority of this packet should be done

without one. If it says to you use one then please do otherwise please refrain.

As you know AP Calculus AB is a fast paced course that is taught at the college level. There

is a lot of material in the curriculum that must be covered before the AP exam in May. The

better you know the prerequisite skills coming into the class the better the class will go for

you. Spend some time with this packet and make sure you are clear on everything covered.

If you have questions please contact me via email and I will be glad to help. (If you take a

picture of your work and the questions it usually makes things go faster)

This assignment will be collected and graded as your first test. Be sure to show all

appropriate work. In addition, there may be a quiz on this material during the first quarter.

All questions must be complete with the correct work.

― Every summer math packet will be due on MONDAY, AUGUST 19TH and

worth 75 POINTS.

― This Summer Math Packet is a Summative Assessment.

Please email the math department with any questions, [email protected] or any of

the math teachers.

I. Intercepts

The x-intercept is where the graph

crosses the x-axis. You can find the x-

intercept by setting 𝑦 = 0.

The y-intercept is where the graph

crosses the y-intercept. You can find

the y-intercept by setting 𝑥 = 0

Example:

Find the intercepts for

𝑦 = (𝑥 + 3)2 − 4

Solution

X-intercept

Set 𝑦 = 0.

0 = (𝑥 + 3)2 − 4

Add 4 to both sides.

4 = (𝑥 + 3)2

Take the square root of both sides

±2 = (𝑥 + 3)

Write as two equations

−2 = (𝑥 + 3) 𝑜𝑟 2 = (𝑥 + 3)

Subtract 3 from both sides

−5 = 𝑥 𝑜𝑟 − 1 = 𝑥

Y-intercept

Set 𝑥 = 0

𝑦 = (0 + 3)2 − 4

Add 0 + 3

𝑦 = 32 − 4

Square 3

𝑦 = 9 − 4

Add four to both sides

𝑦 = 5

Find the intercepts for each of the following.

1. 𝑦 = −3𝑥 + 2

2. 𝑦 = 𝑥3 + 2

3. 𝑦 =𝑥2+3𝑥

(3𝑥+1)2

4. 𝑦2 = 𝑥3 − 4𝑥

II. Complex Fractions When simplifying fractions, multiply

by a fraction equal to 1 which has a

numerator and denominator

composed of the common

denominator of all the denominators

in the complex fraction.

Example

−7 −6

𝑥 + 15

𝑥 + 1

=−7 −

6𝑥 + 1

5𝑥 + 1

∙𝑥 + 1

𝑥 + 1

=−7𝑥 − 7 − 6

5

=−7𝑥 − 13

5

−2𝑥 +

3𝑥𝑥 − 4

5 −1

𝑥 − 4

=−

2𝑥 +

3𝑥𝑥 − 4

5 −1

𝑥 − 4

∙𝑥(𝑥 − 4)

𝑥(𝑥 − 4)

=−2(𝑥 − 4) + 3𝑥(𝑥)

5(𝑥)(𝑥 − 4) − 1(𝑥)

=−2𝑥 + 8 + 3𝑥2

5𝑥2 − 20𝑥 − 𝑥

=3𝑥2 − 2𝑥 + 8

5𝑥2 − 21𝑥

5. 25

𝑎−𝑎

5+𝑎

6. 2−

4

𝑥+2

5+10

𝑥+2

7. 4−

12

2𝑥−3

5+15

2𝑥−3

III. System of Equations

Use substitution or elimination

method to solve the system of

equations.

Example:

𝑥2 + 𝑦2 − 16𝑥 + 39 = 0

𝑥2 − 𝑦2 − 9 = 0

Elimination Method

Add the two equations and you get

2𝑥2 − 16𝑥 + 30 = 0

𝑥2 − 8𝑥 + 15 = 0

(𝑥 − 3)(𝑥 − 5) = 0

𝑥 = 3 𝑎𝑛𝑑 𝑥 = 5

Plug 𝑥 = 3 and 𝑥 = 5 into an original

equation.

32 − 𝑦2 − 9 = 0

−𝑦2 = 0

𝑦 = 0

52 − 𝑦2 − 9 = 0

16 = 𝑦2

𝑦 = ±4

Points of intersection are

(5,4), (5, −4), 𝑎𝑛𝑑 (3,0)

Substitution

Solve one equation for a variable

𝑦2 = −𝑥2 + 16𝑥 − 39

Find the point(s) of intersection of the graphs

for the given equations.

8. 𝑥 + 𝑦 = 8, 4𝑥 − 𝑦 = 7

9. 𝑥2 + 𝑦 = 6, 𝑥 + 𝑦 = 4

Plug 𝑦2 into the other equation

𝑥2 − (−𝑥2 + 16𝑥 − 39) − 9 = 0

2𝑥2 − 16𝑥 + 30 = 0

𝑥2 − 8𝑥 + 15 = 0

(𝑥 − 3)(𝑥 − 5) = 0

The rest is like the previous example

10. 𝑥2 − 4𝑦2 − 20𝑥 − 64𝑦 − 172 = 0,16𝑥2 + 4𝑦2 − 320𝑥 + 64𝑦 + 1600 = 0

IV. Functions

To evaluate a function for a given

value, simply plug the value into the

function for x.

(𝑓°𝑔)(𝑥) = 𝑓(𝑔(𝑥)) 𝑂𝑅 𝑓[𝑔(𝑥)]

read “f of g of x”. Means to plug the

inside function (in this case 𝑔(𝑥) in

for x in the outside function (in this

case, 𝑓(𝑥)).

Example

Given 𝑓(𝑥) = 2𝑥2 + 1 and 𝑔(𝑥) = 𝑥 −

4 find 𝑓(𝑔(𝑥))

𝑓(𝑔(𝑥)) = 𝑓(𝑥 − 4) = 2(𝑥 − 4)2 + 1

= 2(𝑥2 − 8𝑥 + 16) + 1

= 2𝑥2 − 16𝑥 + 32 + 1

Let 𝑓(𝑥) = 2𝑥 + 1 𝑎𝑛𝑑 𝑔(𝑥) = 2𝑥2 − 1. Find

each

11. 𝑓(2) =

12. 𝑔(−3) =

13. 𝑔(𝑡 + 1)

𝑓(𝑔(𝑥)) = 2𝑥2 − 16𝑥 + 33

14. 𝑓(𝑔(−2)) =

15. 𝑔(𝑓(𝑚 + 2)) =

Find 𝑓(𝑥+ℎ)−𝑓(𝑥)

ℎ for the given function f

16. 𝑓(𝑥) = 9𝑥 + 3

17. 𝑓(𝑥) = 5 − 2𝑥

V. Domain and Range

The domain of a function is the set of

x values for which the function is

defined. The range of a function is the

set of y values that a function can

return. In Calculus we usually write

domains and ranges in interval

notation.

Example:

Find the domain and range for 𝑓(𝑥) =

√𝑥 − 3

Solution

Since we can only take the square

root of positive numbers 𝑥 − 3 ≥ 0

which means that 𝑥 ≥ 3. So we would

say the domain is [3, ∞). Note that we

have used a [ to indicate that 2 is

included. If 3 was not to be included

we would have used a (. The smallest

y value that the function can return is

0 so the range is [0, ∞)

Find the domain and range.

18. ℎ(𝑥) = √9 − 𝑥2

19. ℎ(𝑥) = sin 𝑥

20. 𝑓(𝑥) =2

𝑥−1

VI. Inverses

To find the inverse of a function,

simply switch the x and the y and

solve for the new “y” value.

𝑓(𝑥) = √𝑥 + 13

Rewrite 𝑓(𝑥) as y

𝑦 = √𝑥 + 13

Switch x and y

𝑥 = √𝑦 + 13

Solve for your new y.

𝑥3 = √𝑦 + 13 3

Find the inverse of each function

21. 𝑓(𝑥) = 2𝑥 + 1

𝑥3 = 𝑦 + 1

𝑦 = 𝑥3 − 1

Rewrite in inverse notation

𝑓−1(𝑥) = 𝑥3 − 1

To prove that one function is an

inverse of another function, you need

to show that 𝑓(𝑔(𝑥)) = 𝑔(𝑓(𝑥)) = 𝑥

22. 𝑓(𝑥) =𝑥3

3

Prove that f and g are inverse of each other

23. 𝑓(𝑥) =𝑥3

2

𝑔(𝑥) = √2𝑥3

24. 𝑓(𝑥) = 9 − 𝑥2, 𝑥 ≥ 0

𝑔(𝑥) = √9 − 𝑥

VII. Symmetry

x-axis

substitute in –y for y into the

equation. If this yields an equivalent

equation then the graph has x-axis

symmetry. If this is the case, this is

not a function as it would fail the

vertical line test.

y-axis

substitute in –x for x into the

equation. If this yields an equivalent

equation then the graph has y-axis

symmetry. A function that has y-axis

symmetry is called an even function.

Origin

Substitute in -x for x into the equation

and substitute –y for y into the

equation. If this yields an equivalent

equation then the graph has origin

symmetry. If a function has origin

symmetry it is called an odd function.

In order for a graph to represent a

function it must be true that for every

x value in the domain there is exactly

one y value. To test to see if an

equation is a function we can graph it

and then do the vertical line test.

Example 1

Is 𝑥 − 𝑦2 = 2 a function?

Solution: this is not a function

because it does not pass the vertical

line test.

Test for symmetry with respect to each axis and

the origin.

25. 𝑦 = 𝑥√𝑥 + 2

26. 𝑦 = |6 − 𝑥|

Example 2:

Test for symmetry with respect to

each axis and the origin given the

equation 𝑥𝑦 − √4 − 𝑥2 = 0

Solution:

x-axis

𝑥(−𝑦) − √4 − 𝑥2 = 0

−𝑥𝑦 − √4 − 𝑥2 = 0 since there is no

way to make this look like the original

it is not symmetric to the x axis

y-axis

−𝑥𝑦 − √4 − (−𝑥)2 = 0

−𝑥𝑦 − √4 − 𝑥2 = 0 since there is no

way to make this look like the original

it is not symmetric to the y axis

Origin

−𝑥(−𝑦) − √4 − (−𝑥)2 = 0

𝑥𝑦 − √4 − 𝑥2 = 0 since this does look

like the original it is symmetric to the

origin

27. 𝑦 =𝑥

𝑥3+1

28. 𝑦 = 3𝑥2 − 1

VIII. Vertical Asymptotes

To find the vertical asymptotes, set

the denominator equal to zero to find

the x-value for which the function is

undefined. That will be the vertical

asymptote. Vertical asymptotes are

always lines (𝑥 = #)

Determine the vertical asymptotes for the

function (it will be a line)

29. 𝑓(𝑥) =1

𝑥2

30. 𝑓(𝑥) =𝑥2

𝑥2−4

31. 𝑓(𝑥) =2+𝑥

𝑥2(1−𝑥)

IX. Holes (Points of Discontinuity)

Given a rational function if a number

causes the denominator and the

numerator to be 0 then both the

numerator and denominator can be

factored and the common zero can be

cancelled out. This means there is a

hole in the function at this point.

For each function list find the holes

32. 𝑓(𝑥) =(𝑥−3)(𝑥+2)

(𝑥−3)(2𝑥+1)

Example 1 Find the hole in the

following function

𝑓(𝑥) =𝑥 − 2

𝑥2 − 𝑥 − 2

Solution: When 𝑥 = 2 is substituted

into the function the denominator and

numerator both are 0.

Factoring and cancelling 𝑓(𝑥) =𝑥−2

(𝑥+1)(𝑥−2)

𝑓(𝑥) =1

𝑥+1 but 𝑥 ≠ 2 this restriction

is from the original function before

canceling. The graph of the function

𝑓(𝑥) witll look identical to 𝑓(𝑥) =1

𝑥+1 except for the hole at 𝑥 = 2

33. 𝑓(𝑥) =𝑥2−1

2𝑥2+𝑥−1

X. Horizontal Asymptotes

Case 1: Degree of the numerator is

less than the degree of the

denominator. The asymptote is 𝑦 = 0

Case 2: Degree of the numerator is the

same as the degree of the

denominator. The asymptote is the

ratio of the lead coefficients.

Case 3: Degree of the numerator is

greater than the degree of the

denominator. There is no horizontal

asymptote. The function increases

without bound. (If the degree of the

numerator is exactly 1 more than the

Determine the horizontal asymptotes using the

three cases.

34. 𝑓(𝑥) =𝑥2−2𝑥+1

𝑥3+𝑥−7

35. 𝑓(𝑥) =3𝑥3−2𝑥2+8

4𝑥−3𝑥3+5

degree of the denominator, then there

is a slant asymptote, which is

determined with long division.)

36. 𝑓(𝑥) =4𝑥5

𝑥2−7

XI. Solving for indicated variables

37. 𝑥

𝑎+

𝑦

𝑏+

𝑧

𝑐= 1 𝑓𝑜𝑟 𝑎

38. 𝑉 = 2(𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎)𝑓𝑜𝑟 𝑎

39. 2𝑥 − 2𝑦𝑑 = 𝑦 + 𝑥𝑑 𝑓𝑜𝑟 𝑑

40. 2𝑥

4𝜋+

1−𝑥

2= 0 for x

XII. Absolute Value and Piecewise Functions

In order to remove the absolute value sign from a

function you must:

1) Find the zeros of the expression inside of the absolute value.

2) Make a sign chart of the expression inside the absolute value

3) Rewrite the equation without the absolute value as a piecewise function. For each interval where the expression is positive we can write that interval by just dropping the absolute value. For each interval that is negative we must take the opposite sign.

Example 1

Rewrite the following equation without using

absolute value symbols.

𝑓(𝑥) = |2𝑥 + 4|

Solution:

Find where the expression is 0 for the part in the

absolute value

2𝑥 + 4 = 0

2𝑥 = −4

𝑥 = −4

2

𝑥 = −2

Put in any value less than -2 into 2x+4 and you

get a negative. Put in any value more than -2 and

you get a positive.

Write as a piecewise function. Be sure to change

the sign of each term for any part of the graph

that was negative on the sign chart.

𝑓(𝑥) = {−2𝑥 − 4 𝑥 < −22𝑥 + 4 𝑥 ≥ −2

Write the following absolute value

expressions as piecewise

expressions (by remove the absolute

value):

41. 𝑦 = |2𝑥 − 4|

42. 𝑦 = |6 + 2𝑥| + 1

XIII. Exponents

A fractional exponent means you are

taking a root. For example 𝑥1

2 is the

same as √𝑥

Example 1:

Write without fractional exponent:

𝑦 = 𝑥2

3

Solution:

𝑦 = √𝑥23 Notice that the index is the

denominator and the power is the

numerator.

Negative exponents mean that you

need to take the reciprocal. For

example 𝑥−2 means 1

𝑥2 and 2

𝑥−3 means

2𝑥3.

Example 2: Write with positive

exponents: 𝑦 =2

5𝑥−4

Solution: 𝑦 =2𝑥4

5

Example 3: Write with positive

exponents and without fractional

exponents: 𝑓(𝑥) =(𝑥+1)−2(𝑥−3)

12

(2𝑥−3)−12

Solution: 𝑓(𝑥) = √𝑥−3√2𝑥−3

(𝑥+1)2

Write without fractional exponents

43. 𝑦 = 2𝑥1

3

44. 𝑓(𝑥) = (16𝑥2)1

4

45. 𝑦 = 271

3𝑥3

4

46. 91

2 =

47. 641

3

48. 82

3 =

Write with positive exponents:

49. 𝑓(𝑥) = 2𝑥−3

50. 𝑦 = (−2

𝑥−4)

−2

When factoring always factor out the

lowest exponent for each term.

Example 4: 𝑦 = 3𝑥−2 + 6𝑥 − 33𝑥−1

Solution: 𝑦 = 3𝑥−2(1 + 2𝑥3 − 11𝑥)

When dividing two terms with the

same base, we subtract the exponents.

If the difference is negative then the

term goes in the denominator if the

difference is positive then the term

goes in the numerator.

Example 5: Simplify 𝑓(𝑥) =(2𝑥)3

𝑥8

Solution: first you must distribute the

exponent. 𝑓(𝑥) =8𝑥3

𝑥8 . Then since we

have two terms with x as the base we

can subtract the exponents. Thus

𝑓(𝑥) =8

𝑥5

Example 6: Factor and simplify

𝑓(𝑥) = 4𝑥(𝑥 − 3)12 + 𝑥2(𝑥 − 3)−

12

Solution: The common terms are x

and (x-3). The lowest exponent for x

is 1. The lowest exponents for (x-3) is

−1

2. So factor out 𝑥(𝑥 − 3)−

1

2 and

obtain

𝑓(𝑥) = 𝑥(𝑥 − 3)−12[4(𝑥 − 3) + 𝑥]

This will simplify to

𝑓(𝑥) = 𝑥(𝑥 − 3)−12[4𝑥 − 12 + 𝑥]

Leaving a final solution of 𝑥(5𝑥−12)

√𝑥−3

Factor then simplify

51. 𝑓(𝑥) = 4𝑥−3 + 2𝑥 − 18𝑥−2

52. 5𝑥2(𝑥 − 2)−1

2 + (𝑥 − 2)1

23𝑥 = 𝑦

53. 𝑓(𝑥) = 6𝑥(2𝑥 − 1)−1 − 4(2𝑥 − 1)

XIV. Natural Logarithms

Recall that 𝑦 = ln (𝑥) and 𝑦 = 𝑒𝑥 are

inverse to each other.

Properties of Natural Log:

ln(𝐴𝐵) = ln 𝐴 + ln 𝐵

Example 1: ln(2) + ln(5) = ln (10)

ln (𝐴

𝐵) = ln 𝐴 − ln 𝐵

Example 2: ln 6 − ln 2 = ln 3

ln 𝐴𝑝 = 𝑝 ln 𝐴

Example 3: ln 𝑥4 = 4 ln 𝑥

3ln 2 = ln 23 = ln 8

ln(𝑒𝑥) = 𝑥 , ln 𝑒 = 1, ln 1 = 0, 𝑒0

= 1

Example 4: Use the properties of natural

logs to solve for x.

2 ∙ 5𝑥 = 11 ∙ 7𝑥 5𝑥

7𝑥=

11

2

ln5𝑥

7𝑥= ln

11

2

ln 5𝑥 − ln 7𝑥 = ln 11 − ln 2

𝑥𝑙𝑛 5 − 𝑥 ln 7 = ln 11 − ln 2

𝑥(ln 5 − ln 7) = ln 11 − ln 2

𝑥 =ln 11 − ln 2

ln 5 − ln 7

Express as a single logarithm:

54. 3 ln 𝑥 + 2 ln 𝑦 − 4 ln 𝑧

Solve for x

55. 3 ln 𝑥 = 1

56. 𝑒𝑥−3 = 7

57. 3𝑥 = 5 ∙ 2𝑥

XV. Trig. Equations and special values

You are expected to know the special

values for trigonometric functions.

Fill in the table to the right and study

it. (Please)

You can determine sine or cosine of a

quadrantal angle by using the unit

circle. The x-coordinate is the cosine

and the y-coordinate is the sine of the

angle.

Example:

sin 90° = 1

cos𝜋

2= 0

58. sin 180°

59. cos 270°

60. sin (−90°)

61. cos(−𝜋)

62. tan (𝜋

6)

63. cos (2𝜋

3)

64. sin (5𝜋

4)

XVI. Trig. Identities

You should study the following trig

identities and memorize them before

school starts (we use them a lot)

Find all the solutions to the equations. DO NOT

use a Calculator.

Find all angle values 0 ≤ 𝑥 < 2𝜋.

65. sin 𝑥 = −1

2

66. 2 cos 𝑥 = √3

67. 4 cos2 𝑥 − 4 cos 𝑥 = −1

68. 2 sin2 𝑥 + 3 sin 𝑥 + 1 = 0

69. 2 cos2 𝑥 − 1 − cos 𝑥 = 0

XVII. Inverse Trig Functions

Inverse Trig Functions can be written

in one of two ways:

arcsin(𝑥) sin−1(𝑥)

Inverse trig functions are defined only

in the quadrants as indicated below

due to their restricted domains.

Example 1:

Express the value of “y” in radians

𝑦 = arctan−1

√3

For each of the following, express the value for

“y” in radians

70. 𝑦 = arcsin−√3

2

71. 𝑦 = arccos(−1)

Solution:

Draw a

reference

triangle

This means

the reverence angle is 30° or 𝜋

6. So

𝑦 = −𝜋

6 so it falls in the interval from

−𝜋

2< 𝑦 <

𝜋

2

Thus 𝑦 = −𝜋

6

Example 2: Find the value without a

calculator

cos (𝑎𝑟𝑐𝑡𝑎𝑛5

6)

Solution

Draw the reference triangle in the

correct quadrant fits. Find the missing

side using the Pythagorean Theorem.

Find the ratio of the cosine of the

reference triangle.

cos 𝜃 = (6

√61)

72. 𝑦 = tan−1(−1)

For each of the following give the value without

a calculator.

73. tan (arccos2

3)

74. sec (sin−1 12

13)

75. sin (arctan12

5)

76. sin (sin−1 7

8)

XVIII. Transformations of a Graph

Graph the parent function of each set, try not to use your calculator. Draw a quick sketch

on your paper of each additional equation in the family. Check your sketch with the

graphing calculator.

1. Parent Function 𝑦 = 𝑥2 A. 𝑦 = 𝑥2 − 5 B. 𝑦 = 𝑥2 + 3 C. 𝑦 = (𝑥 − 10)2 D. 𝑦 = (𝑥 + 8)2 E. 𝑦 = 4𝑥2 F. 𝑦 = 0.25𝑥2 G. 𝑦 = −𝑥2 H. 𝑦 = −(𝑥 + 3)2 + 6 I. 𝑦 = (𝑥 + 4)2 − 8 J. 𝑦 = −2(𝑥 + 1)2 + 4

1.

A

B

C

D

E

F

G

H

I j

2. Parent Function 𝑦 = sin (𝑥) (set mode to radians) a. 𝑦 = sin (2𝑥) b. 𝑦 = sin(𝑥) − 2 c. 𝑦 = 2sin (𝑥) d. 𝑦 = 2 sin(2𝑥) − 2

2

a

b

c

d

3. Parent Function 𝑦 = cos(𝑥) a. 𝑦 = cos(3𝑥)

b. 𝑦 = cos (𝑥

2)

c. 𝑦 = 2cos(𝑥) + 2 d. 𝑦 = −2cos(𝑥) − 1

3

a

b

c

d

4. Parent Function 𝑦 = 𝑥3 a. 𝑦 = 𝑥3 + 2 b. 𝑦 = −𝑥3 c. 𝑦 = 𝑥3 − 5 d. 𝑦 = −𝑥3 + 3 e. 𝑦 = (𝑥 − 4)3 f. 𝑦 = (𝑥 − 1)3 − 4

4

a

b

c

d

e

f

5. Parent Function 𝑦 = √𝑥

a. 𝑦 = √𝑥 − 2

b. 𝑦 = √−𝑥

c. 𝑦 = √6 − 𝑥

d. 𝑦 = −√𝑥

e. 𝑦 = −√−𝑥

f. 𝑦 = √𝑥 + 2

g. 𝑦 = −√4 − 𝑥 5

a

b

c

d

e

f

g

6. Parent Function 𝑦 = ln 𝑥 a. 𝑦 = ln(𝑥 + 3) b. 𝑦 = ln 𝑥 + 3 c. 𝑦 = ln(𝑥 − 2) d. 𝑦 = ln −𝑥 e. 𝑦 = −ln 𝑥 f. 𝑦 = ln(2𝑥) − 4

6

a

b

c

d

e

f

7. Parent Function 𝑦 = 𝑒𝑥 a. 𝑦 = 𝑒2𝑥 b. 𝑦 = 𝑒𝑥−2 c. 𝑦 = 𝑒2𝑥 + 3 d. 𝑦 = −𝑒𝑥 e. 𝑦 = 𝑒−𝑥

7

a

b

c

d

E

8. Parent Function 𝑦 = 𝑎𝑥 a. 𝑦 = 5𝑥 b. 𝑦 = 2𝑥 c. 𝑦 = 3−𝑥

d. 𝑦 =1

2

𝑥

e. 𝑦 = 4𝑥−3

8

a

b

c

d

E

9. Resize your window to [0,1] × [0,1] Graph all of the following functions in the same window. List the functions from the highest graph to the lowest graph in the table. How do they compare for values of 𝑥 > 1? a. 𝑦 = 𝑥2 b. 𝑦 = 𝑥3

c. 𝑦 = √𝑥

d. 𝑦 = 𝑥2

3 e. 𝑦 = |𝑥| f. 𝑦 = 𝑥4

How do they compare?

(If you found any errors in the packet please let me know so I can correct it. Thanks!)

10. Given 𝑓(𝑥) = 𝑥4 − 3𝑥3 + 2𝑥2 − 7𝑥 − 11 Use your calculator to find all roots to the nearest 0.001

11. Given 𝑓(𝑥) = |𝑥 − 3| + |𝑥| − 6 Use your calculator to find all the roots to the nearest 0.001

12. Find the points of intersection. a. 𝑓(𝑥) = 3𝑥 + 2, 𝑔(𝑥) = −4𝑥 − 2

b. 𝑓(𝑥) = 𝑥2 − 5𝑥 + 2, 𝑔(𝑥) = 3 − 2𝑥