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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.
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
2  ( x  3)
Write as 2 equations
5  x
1  x
Subtract 3 from both sides
or
set y=0
add 4 to both sides
take square root of both sides
y-intercept
y  (0  3) 2  4
set x=0
y  32  4
Add 0+3
y 94
Square 3
subtract
y5
Problem Set I
Find the intercepts for each of the following.
1.
2 y  6x  4
2.
x  y2  4
3.
y
4.
5.
10
x2 1
y  6 x
y  6 x
II.
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. In Calculus
the point slope form of a line is the preferred form. 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
3
Solution: The slope of this line is m   1
3
The parallel line has slope m   1 and the perpendicular line has slope
3
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 point-slope form with m   1 and point (-2,1)
𝑦−1 =
(𝑥 + 2)
3
3
Part b (using slope from example above)
𝑦 − 1 = 3(𝑥 + 2)
Using the point-slope form with m  3 and point (-2,1)
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
y
15  6 x
5
Divide by -5
y
6
x 3
5
Simplify
The slope is m= 6 and the y-intercept is -3
5
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
y 2  y1
x 2  x1
5  3 2 choose one point to substitute back into either the point slope or slope-intercept form of a

4  1 5
line.
2
𝑦 − 5 = (𝑥 − 4)
5
Using the point-slope form with m  2 and point (4,5)
5
Problem Set II
Find the equation of a line:
1. contains (3,-4) and (5,2)
3.
4.
contains   1 , 2  and   3 , 1 
 2 3
 4 6
contains (-3,4) and m is undefined
contains (-2,-2) and m=2
5.
x-intercept (2,0) and y-intercept (0,3)
2.
Find the slope and y-intercept of the line:
6. x+5y = 20
Sketch a graph of the equation:
7. y=-3
8. x=4
9. y-1=3(x+4)
Write an equation of a line through the point (a). parallel to the given line and (b) perpendicular to the given
line:
10. Point : (2,1)
line: 4x-2y=3
III. Functions
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
Example: Given 𝑓(𝑥) = 3𝑥 2 + √𝑥 find 𝑓(𝑥 + ℎ).
Solution: To find 𝑓(𝑥 + ℎ) we replace every instance of x with x+h.
Thus 𝑓(𝑥 + ℎ) = 3(𝑥 + ℎ)2 + √𝑥 + ℎ = 3(𝑥 2 + 2𝑥ℎ + ℎ2 ) + √𝑥 + ℎ = 3𝑥 2 + 6𝑥ℎ + 3ℎ2 + √𝑥 + ℎ
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 III
Let 𝑓(𝑥) = 2𝑥 + 1 and 𝑔(𝑥) = 1 − 𝑥2 find each of the following:
1. 𝑔(𝑓(𝑥))
2. 𝑓(𝑥 2 + 5)
3. 𝑔(𝑥 + ℎ) − 𝑔(𝑥)
Find the domain and range for each function give your answer using interval notation:
3. h(x)= 4  2 x
4.
f(x)=
2
x 1
IV. 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
x x2
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)
1 except for the hole at x=2.
function f(x) will look identical to y 
( x  1)
f ( x) 
f ( x) 
x2
note the hole at x=2
x x2
y
2
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:
f ( x) 
x( x  3)
3 x ( 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
Also, recall horizontal asymptotes can be found by comparing the degree of the numerator and denominator
of a rational function. If the denominator is of higher degree, the horizontal asymptote is 𝑦 = 0. If the
numerator is of higher degree, there is no horizontal asymptote. If the numerator and denominator are the
same degree, the horizontal asymptote is 𝑦 = 𝑘 where k is the ratio of the lead coefficients.
Example: Find the horizontal asymptote of 𝑓(𝑥)
=
2𝑥 2 −5𝑥+6
3𝑥 2 +7𝑥+8
Solution: Since the degrees of both polynomials is 2, we use the ratio of the lead coefficients to get an
asymptote of 𝑦 = 2⁄3.
Problem Set IV
For each function below list all holes, vertical asymptotes and x-intercepts, and horizontal asymptotes.
1. f ( x)  ( x  3)( x  2)
( x  3)( 2 x  1)
x2 1
2x 2  x  1
x 3  12 x 2  32 x
3. f ( x) 
x 2  2x  8
2
x  9 x  14
4. g ( x) 
x 2  3x  2
2.
y
V.
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)
0°
30°
45°
60°
90°
120°
135°
150°
180°
210°
225°
240°
270°
300°
315°
330°
360°
(radians)
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
1
cot x
1
cot x 
tan x
cos x 
tan x 
Tangent Identities
tan x 
sin x
cos x
cot x 
cos x
sin x
Pythagorean Identities
sin 2 x  cos 2 x  1
Reduction Identities
sin(  x)   sin x
tan 2 x  1  sec 2 x
cot 2 x  1  csc 2 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
Original Problem
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

Factor
Set each factor equal to 0
Get the trig function by itself
Solve for x (these are special values)
 2k
6
5
x
 2k
6
Get one side equal to 0.
and
x
3
 2k
2
Problem Set V
Find all solutions to the equations. You should not need a calculator.
1.
2.
3.
4.
4 cos 2 x  4 cos x  1
sin 2 x  2 sin x  0
3 sin x  2 cos 2 x
sin(cos x)  1
VI. Exponents
A fractional exponent means you are taking a root. For example
Example: Write without fractional exponent:
x1 / 2
is the same as
x.
y  x2/3
Solution: y  x Notice that the root is the bottom number in the fraction and the power is the top
number in the fraction.
2
3
Negative exponents mean that you need to take the reciprocal. For example
2 x
3
means
2x
3
y  2x
means
1 x2
and
.
Example: Write with positive exponents:
Solution:
x 2
y 2
5 x 4
4
5
Example: Write with positive exponents and without fractional exponents:
2
1/ 2
f ( x)  ( x  1) ( x  3)
Solution:
f ( x) 
(2 x  3) 1 / 2
x  3 2x  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
2
3 x 2 can be factored from each term.
Leaving
y  3x (1  2 x  11x) . Notice that for the exponent for the 6x term we take 1- (-2) and get 3.
33 x 1 term we take -1-(-2) and get 1 as our new exponent.
3
For the
When dividing two terms with the same base, we subtract the exponents (numerator exponent- denominator
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
Solution: First you must distribute the exponent.
8x 3
f ( x)  8
x
. Then since we have two terms with x as
the base we can subtract the exponents. Since 3-8 results in -5 we know that we will have
in the denominator.
f ( x) 
8
x5
.
x5
f ( x) 
Example: Simplify
x 2  2x  1
( x 2  1)
Solution: First we must factor both the numerator and denominator.
f ( x) 
( x  1) 2
.
( x  1)( x  1)
Then we
can see that we have the term (x-1) in both the numerator and denominator. Subtracting exponents we get
2-1=1 so the term will go in the numerator with 1 as it’s exponent.
Example: Factor and simplify
f ( x) 
( x  1)
.
( x  1)
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 (x3) is -1/2. So factor out
to
x( x  3) 1 / 2
and obtain
f ( x)  x( x  3) 1 / 2 [4 x  12  x] .
Leaving a final solution of
Problem Set VI
Write without fractional exponents.
1.
y  2x1 / 3
2. 𝑓(𝑥) = (16𝑥 3 )1⁄4
Write with positive exponents:
4.
f ( x)  2 x 3
y  (4 x 2 ) 2
5.
f ( x)  ( x  3)
3.
2
(2 x  1) 3
Factor then simplify:
6.
f ( x)  4 x 3  2 x  18 x 2
7.
f ( x)  5x 2 ( x  2) 1 / 2  ( x  2)1 / 2 3x
8.
f ( x)  6 x(2 x  1) 1  4(2 x  1)
Simplify:
9.
10.
11.
(4 x 2 ) 3
2x
(2 x  1)( x  3) 2
y
( x  3) 4 (2 x  1)
f ( x) 
y
x 2  25
x 2  10 x  25
f ( x)  x( x  3) 1. / 2 [4( x  3)  x] .
x(5 x  12)
x3
.
This will simplify
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.
AP CALC 1 Summer Packet
Name _________________
Problem Set I
1) x-int ________
y-int ________
2) x-int ________
y-int ________
3) x-int ________
y-int ________
4) x-int ________
y-int ________
5) x-int ________
y-int ________
Problem Set II
1) y = _______________
2) y = _______________
3) y = _______________
4) y = _______________
5) y = _______________
6) m = _____
7-9)
b = _________
(put on the same graph)
10) parallel, y = ____________ and perpendicular,
y = ____________
Problem Set III
1) f(g(x))= ___________________________
2) 𝑓(𝑥 2 + 5) = ____________________________________
3) 𝑔(𝑥 + ℎ) − 𝑔(𝑥) = ____________________________________
4) domain ___________ range ____________
5) domain ___________ range ____________
Problem Set IV
1) holes _________ vertical asymptotes _________ x-int _________ horizontal asymptote _________
2) holes _________ vertical asymptotes _________ x-int _________ horizontal asymptote _________
3) holes _________ vertical asymptotes _________ x-int _________ horizontal asymptote _________
4) holes _________ vertical asymptotes _________ x-int _________ horizontal asymptote _________
Problem Set V
1) x = ______________
2) x = ______________
3) x = ______________
4) x = ______________
Problem Set VI
1)
y=
2) f (x) =
_____________
3) f (x) =
_____________
4)
y=
_____________
5) f (x) =
_____________
6) f (x) =
_____________
7) f (x) =
_____________
8) f (x) =
_____________
9) f (x) =
_____________
10) y =
_____________
11) y =
_____________
_____________