AP Calculus AB Summer Math PacketName_____ Date_____ Section_____ AP Calculus AB Summer Math Packet...
Transcript of AP Calculus AB Summer Math PacketName_____ Date_____ Section_____ AP Calculus AB Summer Math Packet...
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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.
โ Every summer math packet will be due on MONDAY, AUGUST 19TH and
worth 75 POINTS.
โ 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 ๐ฆ = 0.
The y-intercept is where the graph
crosses the y-intercept. You can find
the y-intercept by setting ๐ฅ = 0
Example:
Find the intercepts for
๐ฆ = (๐ฅ + 3)2 โ 4
Solution
X-intercept
Set ๐ฆ = 0.
0 = (๐ฅ + 3)2 โ 4
Add 4 to both sides.
4 = (๐ฅ + 3)2
Take the square root of both sides
ยฑ2 = (๐ฅ + 3)
Write as two equations
โ2 = (๐ฅ + 3) ๐๐ 2 = (๐ฅ + 3)
Subtract 3 from both sides
โ5 = ๐ฅ ๐๐ โ 1 = ๐ฅ
Y-intercept
Set ๐ฅ = 0
๐ฆ = (0 + 3)2 โ 4
Add 0 + 3
๐ฆ = 32 โ 4
Square 3
๐ฆ = 9 โ 4
Add four to both sides
๐ฆ = 5
Find the intercepts for each of the following.
1. ๐ฆ = โ3๐ฅ + 2
2. ๐ฆ = ๐ฅ3 + 2
3. ๐ฆ =๐ฅ2+3๐ฅ
(3๐ฅ+1)2
4. ๐ฆ2 = ๐ฅ3 โ 4๐ฅ
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
โ7 โ6
๐ฅ + 15
๐ฅ + 1
=โ7 โ
6๐ฅ + 1
5๐ฅ + 1
โ๐ฅ + 1
๐ฅ + 1
=โ7๐ฅ โ 7 โ 6
5
=โ7๐ฅ โ 13
5
โ2๐ฅ +
3๐ฅ๐ฅ โ 4
5 โ1
๐ฅ โ 4
=โ
2๐ฅ +
3๐ฅ๐ฅ โ 4
5 โ1
๐ฅ โ 4
โ๐ฅ(๐ฅ โ 4)
๐ฅ(๐ฅ โ 4)
=โ2(๐ฅ โ 4) + 3๐ฅ(๐ฅ)
5(๐ฅ)(๐ฅ โ 4) โ 1(๐ฅ)
=โ2๐ฅ + 8 + 3๐ฅ2
5๐ฅ2 โ 20๐ฅ โ ๐ฅ
=3๐ฅ2 โ 2๐ฅ + 8
5๐ฅ2 โ 21๐ฅ
5. 25
๐โ๐
5+๐
6. 2โ
4
๐ฅ+2
5+10
๐ฅ+2
7. 4โ
12
2๐ฅโ3
5+15
2๐ฅโ3
III. System of Equations
Use substitution or elimination
method to solve the system of
equations.
Example:
๐ฅ2 + ๐ฆ2 โ 16๐ฅ + 39 = 0
๐ฅ2 โ ๐ฆ2 โ 9 = 0
Elimination Method
Add the two equations and you get
2๐ฅ2 โ 16๐ฅ + 30 = 0
๐ฅ2 โ 8๐ฅ + 15 = 0
(๐ฅ โ 3)(๐ฅ โ 5) = 0
๐ฅ = 3 ๐๐๐ ๐ฅ = 5
Plug ๐ฅ = 3 and ๐ฅ = 5 into an original
equation.
32 โ ๐ฆ2 โ 9 = 0
โ๐ฆ2 = 0
๐ฆ = 0
52 โ ๐ฆ2 โ 9 = 0
16 = ๐ฆ2
๐ฆ = ยฑ4
Points of intersection are
(5,4), (5, โ4), ๐๐๐ (3,0)
Substitution
Solve one equation for a variable
๐ฆ2 = โ๐ฅ2 + 16๐ฅ โ 39
Find the point(s) of intersection of the graphs
for the given equations.
8. ๐ฅ + ๐ฆ = 8, 4๐ฅ โ ๐ฆ = 7
9. ๐ฅ2 + ๐ฆ = 6, ๐ฅ + ๐ฆ = 4
Plug ๐ฆ2 into the other equation
๐ฅ2 โ (โ๐ฅ2 + 16๐ฅ โ 39) โ 9 = 0
2๐ฅ2 โ 16๐ฅ + 30 = 0
๐ฅ2 โ 8๐ฅ + 15 = 0
(๐ฅ โ 3)(๐ฅ โ 5) = 0
The rest is like the previous example
10. ๐ฅ2 โ 4๐ฆ2 โ 20๐ฅ โ 64๐ฆ โ 172 = 0,16๐ฅ2 + 4๐ฆ2 โ 320๐ฅ + 64๐ฆ + 1600 = 0
IV. Functions
To evaluate a function for a given
value, simply plug the value into the
function for x.
(๐ยฐ๐)(๐ฅ) = ๐(๐(๐ฅ)) ๐๐ ๐[๐(๐ฅ)]
read โf of g of xโ. Means to plug the
inside function (in this case ๐(๐ฅ) in
for x in the outside function (in this
case, ๐(๐ฅ)).
Example
Given ๐(๐ฅ) = 2๐ฅ2 + 1 and ๐(๐ฅ) = ๐ฅ โ
4 find ๐(๐(๐ฅ))
๐(๐(๐ฅ)) = ๐(๐ฅ โ 4) = 2(๐ฅ โ 4)2 + 1
= 2(๐ฅ2 โ 8๐ฅ + 16) + 1
= 2๐ฅ2 โ 16๐ฅ + 32 + 1
Let ๐(๐ฅ) = 2๐ฅ + 1 ๐๐๐ ๐(๐ฅ) = 2๐ฅ2 โ 1. Find
each
11. ๐(2) =
12. ๐(โ3) =
13. ๐(๐ก + 1)
๐(๐(๐ฅ)) = 2๐ฅ2 โ 16๐ฅ + 33
14. ๐(๐(โ2)) =
15. ๐(๐(๐ + 2)) =
Find ๐(๐ฅ+โ)โ๐(๐ฅ)
โ for the given function f
16. ๐(๐ฅ) = 9๐ฅ + 3
17. ๐(๐ฅ) = 5 โ 2๐ฅ
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 ๐(๐ฅ) =
โ๐ฅ โ 3
Solution
Since we can only take the square
root of positive numbers ๐ฅ โ 3 โฅ 0
which means that ๐ฅ โฅ 3. So we would
say the domain is [3, โ). 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, โ)
Find the domain and range.
18. โ(๐ฅ) = โ9 โ ๐ฅ2
19. โ(๐ฅ) = sin ๐ฅ
20. ๐(๐ฅ) =2
๐ฅโ1
VI. Inverses
To find the inverse of a function,
simply switch the x and the y and
solve for the new โyโ value.
๐(๐ฅ) = โ๐ฅ + 13
Rewrite ๐(๐ฅ) as y
๐ฆ = โ๐ฅ + 13
Switch x and y
๐ฅ = โ๐ฆ + 13
Solve for your new y.
๐ฅ3 = โ๐ฆ + 13 3
Find the inverse of each function
21. ๐(๐ฅ) = 2๐ฅ + 1
๐ฅ3 = ๐ฆ + 1
๐ฆ = ๐ฅ3 โ 1
Rewrite in inverse notation
๐โ1(๐ฅ) = ๐ฅ3 โ 1
To prove that one function is an
inverse of another function, you need
to show that ๐(๐(๐ฅ)) = ๐(๐(๐ฅ)) = ๐ฅ
22. ๐(๐ฅ) =๐ฅ3
3
Prove that f and g are inverse of each other
23. ๐(๐ฅ) =๐ฅ3
2
๐(๐ฅ) = โ2๐ฅ3
24. ๐(๐ฅ) = 9 โ ๐ฅ2, ๐ฅ โฅ 0
๐(๐ฅ) = โ9 โ ๐ฅ
VII. Symmetry
x-axis
substitute in โy 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 โx 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 โy 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 ๐ฅ โ ๐ฆ2 = 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. ๐ฆ = ๐ฅโ๐ฅ + 2
26. ๐ฆ = |6 โ ๐ฅ|
Example 2:
Test for symmetry with respect to
each axis and the origin given the
equation ๐ฅ๐ฆ โ โ4 โ ๐ฅ2 = 0
Solution:
x-axis
๐ฅ(โ๐ฆ) โ โ4 โ ๐ฅ2 = 0
โ๐ฅ๐ฆ โ โ4 โ ๐ฅ2 = 0 since there is no
way to make this look like the original
it is not symmetric to the x axis
y-axis
โ๐ฅ๐ฆ โ โ4 โ (โ๐ฅ)2 = 0
โ๐ฅ๐ฆ โ โ4 โ ๐ฅ2 = 0 since there is no
way to make this look like the original
it is not symmetric to the y axis
Origin
โ๐ฅ(โ๐ฆ) โ โ4 โ (โ๐ฅ)2 = 0
๐ฅ๐ฆ โ โ4 โ ๐ฅ2 = 0 since this does look
like the original it is symmetric to the
origin
27. ๐ฆ =๐ฅ
๐ฅ3+1
28. ๐ฆ = 3๐ฅ2 โ 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 (๐ฅ = #)
Determine the vertical asymptotes for the
function (it will be a line)
29. ๐(๐ฅ) =1
๐ฅ2
30. ๐(๐ฅ) =๐ฅ2
๐ฅ2โ4
31. ๐(๐ฅ) =2+๐ฅ
๐ฅ2(1โ๐ฅ)
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. ๐(๐ฅ) =(๐ฅโ3)(๐ฅ+2)
(๐ฅโ3)(2๐ฅ+1)
Example 1 Find the hole in the
following function
๐(๐ฅ) =๐ฅ โ 2
๐ฅ2 โ ๐ฅ โ 2
Solution: When ๐ฅ = 2 is substituted
into the function the denominator and
numerator both are 0.
Factoring and cancelling ๐(๐ฅ) =๐ฅโ2
(๐ฅ+1)(๐ฅโ2)
๐(๐ฅ) =1
๐ฅ+1 but ๐ฅ โ 2 this restriction
is from the original function before
canceling. The graph of the function
๐(๐ฅ) witll look identical to ๐(๐ฅ) =1
๐ฅ+1 except for the hole at ๐ฅ = 2
33. ๐(๐ฅ) =๐ฅ2โ1
2๐ฅ2+๐ฅโ1
X. Horizontal Asymptotes
Case 1: Degree of the numerator is
less than the degree of the
denominator. The asymptote is ๐ฆ = 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. ๐(๐ฅ) =๐ฅ2โ2๐ฅ+1
๐ฅ3+๐ฅโ7
35. ๐(๐ฅ) =3๐ฅ3โ2๐ฅ2+8
4๐ฅโ3๐ฅ3+5
degree of the denominator, then there
is a slant asymptote, which is
determined with long division.)
36. ๐(๐ฅ) =4๐ฅ5
๐ฅ2โ7
XI. Solving for indicated variables
37. ๐ฅ
๐+
๐ฆ
๐+
๐ง
๐= 1 ๐๐๐ ๐
38. ๐ = 2(๐๐ + ๐๐ + ๐๐)๐๐๐ ๐
39. 2๐ฅ โ 2๐ฆ๐ = ๐ฆ + ๐ฅ๐ ๐๐๐ ๐
40. 2๐ฅ
4๐+
1โ๐ฅ
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.
๐(๐ฅ) = |2๐ฅ + 4|
Solution:
Find where the expression is 0 for the part in the
absolute value
2๐ฅ + 4 = 0
2๐ฅ = โ4
๐ฅ = โ4
2
๐ฅ = โ2
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.
๐(๐ฅ) = {โ2๐ฅ โ 4 ๐ฅ < โ22๐ฅ + 4 ๐ฅ โฅ โ2
Write the following absolute value
expressions as piecewise
expressions (by remove the absolute
value):
41. ๐ฆ = |2๐ฅ โ 4|
42. ๐ฆ = |6 + 2๐ฅ| + 1
XIII. Exponents
A fractional exponent means you are
taking a root. For example ๐ฅ1
2 is the
same as โ๐ฅ
Example 1:
Write without fractional exponent:
๐ฆ = ๐ฅ2
3
Solution:
๐ฆ = โ๐ฅ23 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 ๐ฅโ2 means 1
๐ฅ2 and 2
๐ฅโ3 means
2๐ฅ3.
Example 2: Write with positive
exponents: ๐ฆ =2
5๐ฅโ4
Solution: ๐ฆ =2๐ฅ4
5
Example 3: Write with positive
exponents and without fractional
exponents: ๐(๐ฅ) =(๐ฅ+1)โ2(๐ฅโ3)
12
(2๐ฅโ3)โ12
Solution: ๐(๐ฅ) = โ๐ฅโ3โ2๐ฅโ3
(๐ฅ+1)2
Write without fractional exponents
43. ๐ฆ = 2๐ฅ1
3
44. ๐(๐ฅ) = (16๐ฅ2)1
4
45. ๐ฆ = 271
3๐ฅ3
4
46. 91
2 =
47. 641
3
48. 82
3 =
Write with positive exponents:
49. ๐(๐ฅ) = 2๐ฅโ3
50. ๐ฆ = (โ2
๐ฅโ4)
โ2
When factoring always factor out the
lowest exponent for each term.
Example 4: ๐ฆ = 3๐ฅโ2 + 6๐ฅ โ 33๐ฅโ1
Solution: ๐ฆ = 3๐ฅโ2(1 + 2๐ฅ3 โ 11๐ฅ)
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 ๐(๐ฅ) =(2๐ฅ)3
๐ฅ8
Solution: first you must distribute the
exponent. ๐(๐ฅ) =8๐ฅ3
๐ฅ8 . Then since we
have two terms with x as the base we
can subtract the exponents. Thus
๐(๐ฅ) =8
๐ฅ5
Example 6: Factor and simplify
๐(๐ฅ) = 4๐ฅ(๐ฅ โ 3)12 + ๐ฅ2(๐ฅ โ 3)โ
12
Solution: The common terms are x
and (x-3). The lowest exponent for x
is 1. The lowest exponents for (x-3) is
โ1
2. So factor out ๐ฅ(๐ฅ โ 3)โ
1
2 and
obtain
๐(๐ฅ) = ๐ฅ(๐ฅ โ 3)โ12[4(๐ฅ โ 3) + ๐ฅ]
This will simplify to
๐(๐ฅ) = ๐ฅ(๐ฅ โ 3)โ12[4๐ฅ โ 12 + ๐ฅ]
Leaving a final solution of ๐ฅ(5๐ฅโ12)
โ๐ฅโ3
Factor then simplify
51. ๐(๐ฅ) = 4๐ฅโ3 + 2๐ฅ โ 18๐ฅโ2
52. 5๐ฅ2(๐ฅ โ 2)โ1
2 + (๐ฅ โ 2)1
23๐ฅ = ๐ฆ
53. ๐(๐ฅ) = 6๐ฅ(2๐ฅ โ 1)โ1 โ 4(2๐ฅ โ 1)
XIV. Natural Logarithms
Recall that ๐ฆ = ln (๐ฅ) and ๐ฆ = ๐๐ฅ are
inverse to each other.
Properties of Natural Log:
ln(๐ด๐ต) = ln ๐ด + ln ๐ต
Example 1: ln(2) + ln(5) = ln (10)
ln (๐ด
๐ต) = ln ๐ด โ ln ๐ต
Example 2: ln 6 โ ln 2 = ln 3
ln ๐ด๐ = ๐ ln ๐ด
Example 3: ln ๐ฅ4 = 4 ln ๐ฅ
3ln 2 = ln 23 = ln 8
ln(๐๐ฅ) = ๐ฅ , ln ๐ = 1, ln 1 = 0, ๐0
= 1
Example 4: Use the properties of natural
logs to solve for x.
2 โ 5๐ฅ = 11 โ 7๐ฅ 5๐ฅ
7๐ฅ=
11
2
ln5๐ฅ
7๐ฅ= ln
11
2
ln 5๐ฅ โ ln 7๐ฅ = ln 11 โ ln 2
๐ฅ๐๐ 5 โ ๐ฅ ln 7 = ln 11 โ ln 2
๐ฅ(ln 5 โ ln 7) = ln 11 โ ln 2
๐ฅ =ln 11 โ ln 2
ln 5 โ ln 7
Express as a single logarithm:
54. 3 ln ๐ฅ + 2 ln ๐ฆ โ 4 ln ๐ง
Solve for x
55. 3 ln ๐ฅ = 1
56. ๐๐ฅโ3 = 7
57. 3๐ฅ = 5 โ 2๐ฅ
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 90ยฐ = 1
cos๐
2= 0
58. sin 180ยฐ
59. cos 270ยฐ
60. sin (โ90ยฐ)
61. cos(โ๐)
62. tan (๐
6)
63. cos (2๐
3)
64. sin (5๐
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 โค ๐ฅ < 2๐.
65. sin ๐ฅ = โ1
2
66. 2 cos ๐ฅ = โ3
67. 4 cos2 ๐ฅ โ 4 cos ๐ฅ = โ1
68. 2 sin2 ๐ฅ + 3 sin ๐ฅ + 1 = 0
69. 2 cos2 ๐ฅ โ 1 โ cos ๐ฅ = 0
XVII. Inverse Trig Functions
Inverse Trig Functions can be written
in one of two ways:
arcsin(๐ฅ) sinโ1(๐ฅ)
Inverse trig functions are defined only
in the quadrants as indicated below
due to their restricted domains.
Example 1:
Express the value of โyโ in radians
๐ฆ = arctanโ1
โ3
For each of the following, express the value for
โyโ in radians
70. ๐ฆ = arcsinโโ3
2
71. ๐ฆ = arccos(โ1)
Solution:
Draw a
reference
triangle
This means
the reverence angle is 30ยฐ or ๐
6. So
๐ฆ = โ๐
6 so it falls in the interval from
โ๐
2< ๐ฆ <
๐
2
Thus ๐ฆ = โ๐
6
Example 2: Find the value without a
calculator
cos (๐๐๐๐ก๐๐5
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 ๐ = (6
โ61)
72. ๐ฆ = tanโ1(โ1)
For each of the following give the value without
a calculator.
73. tan (arccos2
3)
74. sec (sinโ1 12
13)
75. sin (arctan12
5)
76. sin (sinโ1 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 ๐ฆ = ๐ฅ2 A. ๐ฆ = ๐ฅ2 โ 5 B. ๐ฆ = ๐ฅ2 + 3 C. ๐ฆ = (๐ฅ โ 10)2 D. ๐ฆ = (๐ฅ + 8)2 E. ๐ฆ = 4๐ฅ2 F. ๐ฆ = 0.25๐ฅ2 G. ๐ฆ = โ๐ฅ2 H. ๐ฆ = โ(๐ฅ + 3)2 + 6 I. ๐ฆ = (๐ฅ + 4)2 โ 8 J. ๐ฆ = โ2(๐ฅ + 1)2 + 4
1.
A
B
C
D
E
F
G
H
I j
2. Parent Function ๐ฆ = sin (๐ฅ) (set mode to radians) a. ๐ฆ = sin (2๐ฅ) b. ๐ฆ = sin(๐ฅ) โ 2 c. ๐ฆ = 2sin (๐ฅ) d. ๐ฆ = 2 sin(2๐ฅ) โ 2
2
a
b
c
d
3. Parent Function ๐ฆ = cos(๐ฅ) a. ๐ฆ = cos(3๐ฅ)
b. ๐ฆ = cos (๐ฅ
2)
c. ๐ฆ = 2cos(๐ฅ) + 2 d. ๐ฆ = โ2cos(๐ฅ) โ 1
3
a
b
c
d
4. Parent Function ๐ฆ = ๐ฅ3 a. ๐ฆ = ๐ฅ3 + 2 b. ๐ฆ = โ๐ฅ3 c. ๐ฆ = ๐ฅ3 โ 5 d. ๐ฆ = โ๐ฅ3 + 3 e. ๐ฆ = (๐ฅ โ 4)3 f. ๐ฆ = (๐ฅ โ 1)3 โ 4
4
a
b
c
d
e
f
5. Parent Function ๐ฆ = โ๐ฅ
a. ๐ฆ = โ๐ฅ โ 2
b. ๐ฆ = โโ๐ฅ
c. ๐ฆ = โ6 โ ๐ฅ
d. ๐ฆ = โโ๐ฅ
e. ๐ฆ = โโโ๐ฅ
f. ๐ฆ = โ๐ฅ + 2
g. ๐ฆ = โโ4 โ ๐ฅ 5
a
b
c
d
e
f
g
6. Parent Function ๐ฆ = ln ๐ฅ a. ๐ฆ = ln(๐ฅ + 3) b. ๐ฆ = ln ๐ฅ + 3 c. ๐ฆ = ln(๐ฅ โ 2) d. ๐ฆ = ln โ๐ฅ e. ๐ฆ = โln ๐ฅ f. ๐ฆ = ln(2๐ฅ) โ 4
6
a
b
c
d
e
f
7. Parent Function ๐ฆ = ๐๐ฅ a. ๐ฆ = ๐2๐ฅ b. ๐ฆ = ๐๐ฅโ2 c. ๐ฆ = ๐2๐ฅ + 3 d. ๐ฆ = โ๐๐ฅ e. ๐ฆ = ๐โ๐ฅ
7
a
b
c
d
E
8. Parent Function ๐ฆ = ๐๐ฅ a. ๐ฆ = 5๐ฅ b. ๐ฆ = 2๐ฅ c. ๐ฆ = 3โ๐ฅ
d. ๐ฆ =1
2
๐ฅ
e. ๐ฆ = 4๐ฅโ3
8
a
b
c
d
E
9. Resize your window to [0,1] ร [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 ๐ฅ > 1? a. ๐ฆ = ๐ฅ2 b. ๐ฆ = ๐ฅ3
c. ๐ฆ = โ๐ฅ
d. ๐ฆ = ๐ฅ2
3 e. ๐ฆ = |๐ฅ| f. ๐ฆ = ๐ฅ4
How do they compare?
(If you found any errors in the packet please let me know so I can correct it. Thanks!)
10. Given ๐(๐ฅ) = ๐ฅ4 โ 3๐ฅ3 + 2๐ฅ2 โ 7๐ฅ โ 11 Use your calculator to find all roots to the nearest 0.001
11. Given ๐(๐ฅ) = |๐ฅ โ 3| + |๐ฅ| โ 6 Use your calculator to find all the roots to the nearest 0.001
12. Find the points of intersection. a. ๐(๐ฅ) = 3๐ฅ + 2, ๐(๐ฅ) = โ4๐ฅ โ 2
b. ๐(๐ฅ) = ๐ฅ2 โ 5๐ฅ + 2, ๐(๐ฅ) = 3 โ 2๐ฅ