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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 20TH 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 am very excited to see you all in August.
Good Luck!!
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. You
should not need a caluclator. (hint one of these
has NO solution)
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
For each of the following, express the value for
“y” in radians
70. 𝑦 = arcsin−√3
2
71. 𝑦 = arccos(−1)
𝑦 = arctan−1
√3
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