AP Calculus I
Summer Packet
You will need to use your own paper to work the problems in this packet. You will turn in ALL
of your work and the attached completed answer sheet. Answers only will result in no credit.
I.
Symmetry:
x-axis
If a graph is symmetric to the x-axis for every point (x,y) on the graph the point (x,-y) is also on the
graph. To test for symmetry about the x-axis substitute in –y for y into the equation if this yields an
equivalent equation then it is symmetric.
y-axis
If a graph is symmetric to the y-axis for every point (x,y) on the graph the point (-x,y) is also on the
graph. To test for symmetry about the y-axis substitute in –x for x into the equation if this yields an
equivalent equation then it is symmetric.
Origin
If a graph is symmetric to the origin for every point (x,y) on the graph the point (-x,-y) is also on the
graph. To test for symmetry about the origin substitute in –x for x into the equation and substitute in –
y for y into the equation if this yields an equivalent equation then it is symmetric.
Example:
Test for symmetry with respect to each axis and the origin.
Given equation: xy  4  x 2  0
Solution:
x-axis (change all y to –y):
x(  y )  4  x 2  0
 xy  4  x 2  0 since there is no way to make this look like the original it is NOT symmetric to the x-
axis
y-axis: (change all x to –x)
 xy  4  ( x) 2  0
 xy  4  x 2  0 since there is no way to make this look like the original it is NOT symmetric to the y-
axis
origin: (change all x to –x and change all y to –y )
 x (  y )  4  (  x) 2  0
xy  4  x 2  0 since this does look like the original it is symmetric to the origin.
The figure to the right shows the graph of
It is symmetric only to the origin.
xy  4  x 2  0 .
Problem Set I
Test for symmetry with respect to each axis and the origin. Be sure to show your work.
1. y  4  x  3
2. y  x 3  x
x
y
x 1
4. y  x  3
3.
2
5. y  x 2  x
II.
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-axis. You can find the y-intercept by setting x=0.
Example:
Find the intercepts for y  ( x  3) 2  4
Solution:
x-intercept
0  ( x  3) 2  4
4  ( x  3)
2
 2  ( x  3)
 2  ( x  3) or
or
5  x
2  ( x  3)
1  x
set y=0
add 4 to both sides
take square root of both sides
Write as 2 equations
Subtract 3 from both sides
y-intercept
y  (0  3) 2  4
y 3 4
2
y 94
y5
set x=0
Add 0+3
Square 3
subtract
Problem Set II
Find the intercepts for each of the following.
1. 2 y  6 x  4
2. x  y 2  4
10
3. y  2
x 1
4. y  6  x
5. y  6  x
III.
Lines
The slope intercept form of a line is y=mx+b where m is the slope and b is the y-intercept. The point
slope form of a line is y  y1  m( x  x1 ) where m is the slope and ( x1, y1 ) is a point on the line. If two lines
are parallel then they have the same slope. If two lines are perpendicular then they have negative
reciprocal slopes.
Example:
Find the slope of the lines parallel and perpendicular to y   1 x  5
Solution: The slope of this line is
The parallel line has slope
m
1
3
3
1
m
3
and the perpendicular line has slope m  3 .
Example: Find the equations of (a) line parallel and (b) perpendicular to y   1 x  5 that contains the
3
point (-2,1)
Solution:
Part a (using slope from example above)
1
Using the slope-intercept form with m   1 and point (-2,1)
1
(2)  b
3
3
2
b
3
2
1  b
3
Multiply -1and -2
3 2
 b
3 3
1
b
3
Get a common denominator of 3
y
This is the equation of the line parallel to the given line that
contains (-2,1)
1
Subtract
2
from both sides
3
Combine like terms
1
1
x
3
3
Part b (using slope from example above)
1  3(2)  b
Using the slope-intercept form with m  3 and point (-2,1)
Multiply 3 and -2
Subtract -6 from both sides
subtract
This is the equation of the line perpendicular to the given line
that contains (-2,1)
1  6  b
1  6  b
7b
y  3x  7
Example:
Find the slope and y-intercept of 6 x  5 y  15
Solution: First you must get the line in slope-intercept form.
 5 y  15  6 x
Subtract 6x form both sides
Divide by -5
15  6 x
y
5
y
Simplify
6
x 3
5
The slope is m= 6 and the y-intercept is
5
-3
Example: Find the equation of the line that passes through the point (1,-2) and has slope m= -3.
Solution: Since we are given a point and slope it is easiest to use the point slope form of a line.
y  2  3( x  1)
Substitute into point slope form for
and m
Minus a negative makes + and
distribute -3
Subtract 2 from both sides
y  2  3x  1
y  3x  1
( x1, y1 )
Example: Find the equation of the line that passes through (-1,3) and (4,5).
Solution: You will need to find slope using m 
m
53 2

4  1 5
y 2  y1
x2  x1
choose one point to substitute back into either the point slope or slope-intercept form of a
line.
2
5  (4)  b
5
Using the slope-intercept form with m  2 and point
5
(4,5)
8
b
5
8
5  b
5
Multiply 2 and 4
25 8
 b
5 5
17
b
5
Get a common denominator of 5
5
Subtract
8
from both sides
5
Combine like terms
We know the equation of the line is
y
2
17
x
5
5
in slope intercept form.
Remember that when we graph lines that we plot the y-intercept first and then from that point we
graph the slope using rise over run.
Example: Graph the following equation: 3 y  3  x
Solution:
First you must get the equation in slope-intercept form:
3y  x  3
1
y  x 1
3
The slope =1/3 and the y-intercept = -1
Plot the point (0,-1). From the first point go up 1 and over 3 to the right to get a second point .
.
Now connect the two points to get the line.
Problem Set III
Find the equation of a line in slope intercept form:
1. contains (3,-4) and (5,2)
2. contains   1 , 2  and   3 , 1 
 2 3
3.
4.
5.
6.
7.
 4 6
contains (2,1) and m=0
contains (1,7) and m=-3
contains (-3,4) and m is undefined
contains (-2,-2) and m=2
contains (-2,4) and m   3
5
8. contains (0,0) and (2,6)
9. contains (-3,6) and (1,2)
10. contains (1,-2) and (3,-2)
11. x-intercept (2,0) and y-intercept (0,3)
12. x-intercept   2 ,0  and y-intercept (0,-2)
 3

13. contains (-3,4) and x-intercept (a,0) and y-intercept (0,a) and a≠0
Find the slope and y-intercept of the line:
14. x+5y = 20
15. 6x-5y=15
16. x=4
17. y=-1
Sketch a graph of the equation:
18. y=-3
19. x=4
20. y-1=3x+12
21. x+2y+6=0
Write an equation of a line through the point (a). parallel to the given line and (b) perpendicular to the
given line:
22. Point : (2,1)
line: 4x-2y=3
23. Point: (-3,2) : line: 3x+4y=7
IV.
Functions
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: Is x  y 2  2 a function?
The graph is below:
Solution: When a vertical line is drawn it will cross the graph more than one time so it is NOT a
function.
Definition:
Let f and g be functions. The function given by (f◦g)(x)=f(g(x)) is called the composite of f with g. The
domain of f◦g is the set of all x in the domain of g such that g(x) is in the domain of f.
Example: Given: f(x)=3x+5 and g(x)=2x-1
Find: f(g(2)), g(f(2)) and f(g(x))
Solution: To find f(g(2)) we must first find g(2): g(2)=2(2)-1 =4-1=3
Since g(2)=3 we can find f(g(2))=f(3)=3(3)+5=9+5=14
To find g(f(2)) we must first find f(2): f(2)=3(2)+5=6+5=11
Since f(2)=11 we can find g(f(2))=g(11)=2(11)-1=22-1=21
To find f(g(x)) we must put the function g(x) into f(x) equation in place of each x.
f(g(x))=f(2x-1)=3(2x-1)+5=6x-3+5=6x+2
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. If the domain were -1<x≤7 then in interval notation the domain would be (1,7]. Notice that the left side has a ( because it does not include -1 but the right side includes 7 so we
use a ]. When using interval notation we never use a [ or ] for infinity.
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 3 is included. If 3
was not to be included we would have used (3,∞). The smallest y value that the function can return is
0 so the range is (0,∞).
Problem Set IV
Sketch the graph of the equation and then use the vertical line test to determine whether it is a
function of x.
1. x 2  y  2
2. y  x 2  2 x
3. x  9  y 2
Let f(x)=2x+1 and g(x)=1-x2 find each of the following:
4. f(g(0))
5. g(g(2))
6. g(f(3))
7. g(f(x))
8. f(g(x))
Find the domain and range for each function give your answer using interval notation:
9. h(x)= 4  2 x
10. g(r)=cos r
11. f(x)= 2
x 1
12. h(x)=x2 +4
13. f(x)   x  1, x  1
 x  1, x  1
V.
Asymptotes and Holes
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.
Example: Find the holes 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 canceling: f ( x) 
x2
( x  1)( x  2)
1
but (x≠2) this restriction is from the original function before canceling. The graph of the
( x  1)
function f(x) will look identical to y  1 except for the hole at x=2.
( x  1)
f ( x) 
f ( x) 
x2
note the hole at x=2
x x2
2
y
1
( x  1)
Given a rational function if a number causes the denominator to be 0 but not the numerator to be 0
then there is a vertical asymptote at that x value.
Example: Find the vertical asymptotes for the function f ( x) 
x2
x x2
2
Solution: When x=-1 is substituted into f(x) then the numerator is -1 and the denominator is 0
therefore there is an asymptote at x=1. See the graphs above.
Given a rational function if a number causes the numerator to be 0 but not the denominator to be 0
then the value is an x-intercept for the rational function.
Example: Discuss the zeroes in the numerator and denominator f ( x)  x  3
2x
Solution: When x=-3 is substituted into the function the numerator is 0 and the denominator is -6 so
the value of the function is f(-3)=0 and the graph crosses the x-axis at x=-3. Also note that for x=0 the
numerator is 3 and the denominator is 0 so there is a vertical asymptote at x=0. The graph is below.
Example: Find the holes, vertical asymptotes and x-intercepts for the given function:
x 2  3x
f ( x)  2
3x  6 x
Solution: First we must factor to find all the zeroes for both the numerator and denominator:
x( x  3)
f ( x) 
3x( x  2)
Numerator has zeroes x=0 and x=3
Denominator has zeroes x=0 and x=-2.
x=0 is a hole
x=-2 is a vertical asymptote
x=3 is a x-intercept
Problem Set V
For each function below list all holes, vertical asymptotes and x-intercepts
1. f ( x)  ( x  3)( x  2)
( x  3)(2 x  1)
x2 1
2x 2  x  1
3
2
3. f ( x)  x 212 x  32 x
x  2x  8
2
4. g ( x)  x 2  9 x  14
x  3x  2
2. y 
VI.
Trig. Equations and Special Values
You are expected to know the special values for trigonometric functions. Fill in the table below and
study it.


(degrees) (radians)
0°
30°
45°
60°
90°
120°
135°
150°
180°
210°
225°
240°
270°
300°
315°
330°
360°
cos 
sin 
Quadrant
You should study the following trig identities and memorize them before school starts:
Reciprocal identities
1
csc x
1
csc x 
sin x
sin x 
1
sec x
1
sec x 
cos x
cos x 
1
cot x
1
cot x 
tan x
tan x 
Tangent Identities
sin x
cos x
tan x 
cot x 
cos x
sin x
Pythagorean Identities
sin 2 x  cos 2 x  1
tan 2 x  1  sec 2 x
cot 2 x  1  csc 2 x
Double angle Identities
sin 2 x  2 sin x cos x cos 2 x  cos 2 x  sin 2 x  2 cos 2 x  1  1  2 sin 2 x
Reduction Identities
sin( x)   sin x
cos( x)  cos x
tan( x)   tan x
We use these special values and identities to solve equations involving trig functions.
Example: Find all solutions to 2 sin 2 x  sin x  1
Solution:
2 sin 2 x  sin x  1
2 sin 2 x  sin x  1  0
(2 sin x  1)(sin x  1)  0
(2 sin x  1)  0 and (sin x  1)  0
1
and sin x  1
sin x 
2
x

Solve for x (these are special values)
 2k
6
5
x
 2k
6
Original Problem
Get one side equal to 0.
Factor
Set each factor equal to 0
Get the trig function by itself
and x 
3
 2k
2
Problem Set VI
Find all solutions to the equations. You should not need a calculator. (Hint: one of these has NO
solution.)
1. 4 cos 2 x  4 cos x  1
2. 2 sin 2 x  3 sin x  1  0
3. sin 2 x  2 sin x  0
4. 3 sin x  2 cos 2 x
5. 2 sin 2 x  3 sin x  2
6. cos 2 x  5 cos x  2 (hint: use double angle identity)
7. sin(cos x)  1
8. sin 2 x  2 sin x  3  0
VII. Absolute value and piecewise functions
In order to remove the absolute value sign from a function you must:
1. Find the zeroes of the expression inside of the absolute value.
2. Make 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: Rewrite the following equation without using absolute value.
f ( x)  2 x  4
Solution:
2x+4=0
Find where the expression is 0
2x=-4
Subtract 4
x=-4/2
Divide by 2
X=-2
Simplify
_-
+
Put in any value less than -2 into 2x+4
and you get a negative.
-2
Put in any value more than -2 into 2x+4
and you get a positive.
Write as a piecewise function. Be sure
to change the signs of each term for
any part of the graph that was negative
on the sign chart.
 2 x  4 x  2
f ( x)  

2 x  4 x  2
Example: Rewrite the following equation without using absolute value.
f ( x)  2 x 2  5 x  3
Solution:
2x2+5x-3 =0
(2x-1)(x+3)=0
2x-1=0 or x+3=0
X=1/2 or x=-3
+
Find where the expression is 0
factor
Set each factor equal to 0
Solve each equation
_+
-3
+
1/2
Put in any value less than -3 into (2x1)(x+3) and you get a positive number.
Put in any value more than -3 and less
than ½ into (2x-1)(x+3) and you get a
negative number.
Put in any value more than 1/2 into (2x1)(x+3) and you get a positive number.
2
x  3 and x  1 / 2


2 x  5 x  3
f ( x)  

2


 2 x  5 x  3  3  x  1 / 2

Write as a piecewise function. Be sure
to change the signs of each term for
any part of the graph that was negative
on the sign chart.
Example: Rewrite the following equation without using absolute value.
f ( x)  3 x  9  2
Solution:
3x-9=0
Find where the expression is 0 (For the
part in the absolute value only.)
Add 9
Divide by 3
3x=9
x=3
_-
+
3
 3x  9  2 x  3
f ( x)  

3x  9  2 x  3
 3x  11 x  3
f ( x)  

x  3
3x  7
Put in any value less than 3 into 3x-9
and you get a negative.
Put in any value more than 3 into 3x-9
and you get a positive.
Write as a piecewise function. Be sure
to change the signs of each term that is
inside the absolute value for any part of
the graph that was negative on the sign
chart.
Simplify.
Problem Set VII
Rewrite the following equation without using absolute value. Be sure to show your work, including a
sign chart:
1. f ( x)   5x  15
2. f ( x)  ( x  2)( x  4)
3. f ( x)  7 x  5  3
4. f ( x)  2 x 2  x  3
5. f ( x)  5x 2  13x  6  2
VIII. Exponents
A fractional exponent means you are taking a root. For example x1 / 2 is the same as
x.
Example: Write without fractional exponent: y  x 2 / 3
Solution: y  3 x 2 Notice that the root is the bottom number in the fraction and the power is the top
number in the fraction.
Negative exponents mean that you need to take the reciprocal. For example x 2 means 1 x 2 and
2 x 3 means 2x 3 .
Example: Write with positive exponents: y  2
Solution: y  2 x
4
5
5 x 4
Example: Write with positive exponents and without fractional exponents:
2
1/ 2
f ( x)  ( x  1) ( x  3)
(2 x  3) 1 / 2
Solution: f ( x)  x  3 2 x  3
( x  1) 2
When factoring, always factor out the lowest exponent for each term.
Example: y  3x 2  6 x  33x 1
Solution: The lowest exponent for x is -2 so 3x 2 can be factored from each term. Leaving
y  3x 2 (1  2 x 3  11x) . Notice that for the exponent for the 6x term we take 1- (-2) and get 3. For the
33x 1 term we take -1-(-2) and get 1 as our new exponent.
When dividing two terms with the same base, we subtract the exponents (numerator exponentdenominator exponent). 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: Simplify f ( x) 
(2 x) 3
x8
8x 3
Solution: First you must distribute the exponent. f ( x)  8 . Then since we have two terms with x
x
as the base we can subtract the exponents. Since 3-8 results in -5 we know that we will have x 5
8
in the denominator. f ( x)  5 .
x
Example: Simplify
f ( x) 
x 2  2x  1
( x 2  1)
( x  1) 2
. Then we
( x  1)( x  1)
can see that we have the term (x-1) in both the numerator and denominator. Subtracting exponents
( x  1)
we get 2-1=1 so the term will go in the numerator with 1 as it’s exponent. f ( x) 
.
( x  1)
Solution: First we must factor both the numerator and denominator. f ( x) 
Example: Factor and simplify f ( x)  4 x( x  3)1 / 2  x 2 ( x  3) 1 / 2
Solution: The common terms are x and (x-3). The lowest exponent for x is 1. The lowest exponent
for (x-3) is -1/2. So factor out x( x  3) 1 / 2 and obtain f ( x)  x( x  3) 1. / 2 [4( x  3)  x] . This will simplify
x(5 x  12)
to f ( x)  x( x  3) 1 / 2 [4 x  12  x] . Leaving a final solution of
.
x3
Problem Set VIII
Write without fractional exponents.
1. y  2x1 / 3
2. f ( x)  (16 x 2 )1 / 4
3. y  271 / 3 x 3 / 4
Write with positive exponents:
4. f ( x)  2 x 3
5. y  (4 x 2 ) 2
6. y  ( 2 4 ) 2
x
2
7. f ( x)  ( x  3)
(2 x  1) 3
Factor then simplify:
8. f ( x)  4 x 3  2 x  18x 2
9. f ( x)  5x 2 ( x  2) 1 / 2  ( x  2)1 / 2 3x
10. f ( x)  6 x(2 x  1) 1  4(2 x  1)
Simplify:
(4 x 2 ) 3
2x
(2 x  1)( x  3) 2
12. y 
( x  3) 4 (2 x  1)
11. f ( x) 
16 x 2  8 x  1
4 x 2  3x  1
x 2  25
14. y  2
x  10 x  25
13. f ( x) 
AP CALC 1 Summer Packet
Name _________________
Problem Set I
Problem Set II
1) symmetries ______________________
1) x-int ________ y-int ________
2) symmetries ______________________
2) x-int ________ y-int ________
3) symmetries ______________________
3) x-int ________ y-int ________
4) symmetries ______________________
4) x-int ________ y-int ________
5) symmetries ______________________
5) x-int ________ y-int ________
Problem Set III
1) y = _______________
10) y = _______________
2) y = _______________
11) y = _______________
3) y = _______________
12) y = _______________
4) y = _______________
13) y = _______________
5) x = _______________
14) m = _______
b = _______
6) y = _______________
15) m = _______
b = _______
7) y = _______________
16) m = _______
b = _______
8) y = _______________
17) m = _______
b = _______
9) y = _______________
18)
19)
20)
21)
22) parallel, y = ____________ and perpendicular,
y = ____________
23) parallel, y = ____________ and perpendicular,
y = ____________
Problem Set IV
1) yes or no
8) _________
2) yes or no
9) domain ___________ range ____________
3) yes or no
10) domain ___________ range ____________
4) _________
11) domain ___________ range ____________
5) _________
12) domain ___________ range ____________
6) _________
13) domain ___________ range ____________
7) _________
Problem Set V
1) holes _________ vertical asymptotes _________ x-int _________
2) holes _________ vertical asymptotes _________ x-int _________
3) holes _________ vertical asymptotes _________ x-int _________
4) holes _________ vertical asymptotes _________ x-int _________
Problem Set VI
1) x = ______________
2) x = ______________
x= _______________
x= _______________
3) x = ______________
4) x = ______________
x= _______________
x= _______________
5) x = ______________
6) x = ______________
x= _______________
x= _______________
7) x = ______________
8) x = ______________
Problem Set VII
1) f(x) =
2) f(x) =
_________________
3) f(x) =
_________________
4) f(x) =
_________________
5) f(x) =
_________________
_________________
Problem Set VIII
1)
3)
5)
y=
_____________
2) f (x) =
_____________
_____________
4) f (x) =
_____________
_____________
6) y =
_____________
_____________
8) f (x) =
_____________
_____________
10) f (x) =
_____________
_____________
12) y =
_____________
_____________
14) y =
_____________
y=
y=
7) f (x) =
9) f (x) =
11) f (x) =
13) f (x) =