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Cards (34)
Midpoint Formula
(
x
1
+
x
2
2
,
y
1
+
y
2
2
)
( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} )
(
2
x
1
+
x
2
,
2
y
1
+
y
2
)
Distance Formula
d
(
A
,
B
)
=
d(A,B)=
d
(
A
,
B
)
=
\sqrt{(x_2-x_1)^2+
(y_2-y_1)^2}
Equation of a
circle
(
x
−
h
)
2
+
(x-h)^2+
(
x
−
h
)
2
+
(
y
−
k
)
2
=
(y-k)^2=
(
y
−
k
)
2
=
r
2
r^2
r
2
standard form of a
circle
x
2
+
x^2+
x
2
+
y
2
=
y^2=
y
2
=
r
2
r^2
r
2
Domain
all possible
x-values
Range
all possible
y-values
what does [0,25] U [35,40] mean?
that [] is a
specific interval
,
stops
, then
continues
[]
Shifts
upwards
f
(
x
)
+
f(x)+
f
(
x
)
+
c
c
c
shifts
downward
f
(
x
)
−
c
f(x)-c
f
(
x
)
−
c
Shifts to the
right
f
(
x
−
c
)
f(x-c)
f
(
x
−
c
)
shifts to the
left
f
(
x
+
c
)
f(x+c)
f
(
x
+
c
)
Flips from
up
to
down
y
=
y=
y
=
−
f
(
x
)
-f(x)
−
f
(
x
)
flips from
right
to
left
y
=
y=
y
=
f
(
−
x
)
f(-x)
f
(
−
x
)
y
=
y=
y
=
c
f
(
x
)
cf(x)
c
f
(
x
)
if c>1, it will
stretch vertically
by a
factor
of c
if 0<c<1, it will
shrink vertically
by a factor of c
f is
even
if f(-x) = f(x) for all x in the domain of f
f is
odd
if f(-x)= -f(x) for all x in the domain of f
Terminal point
are determined by
t
=
t =
t
=
π
2
,
π
,
3
π
2
,
2
π
\frac{\pi}{2} , \pi, \frac{3\pi}{2}, 2\pi
2
π
,
π
,
2
3
π
,
2
π
c
o
s
(
x
)
=
cos(x)=
cos
(
x
)
=
x
x
x
s
i
n
(
x
)
=
sin(x) =
s
in
(
x
)
=
y
y
y
t
a
n
(
x
)
=
tan(x)=
t
an
(
x
)
=
y
x
\frac{y}{x}
x
y
c
s
c
(
x
)
=
csc(x) =
csc
(
x
)
=
1
s
i
n
(
x
)
\frac{1}{sin(x)}
s
in
(
x
)
1
s
e
c
(
x
)
=
sec(x) =
sec
(
x
)
=
1
c
o
s
(
x
)
\frac{1}{cos(x)}
cos
(
x
)
1
c
o
t
(
x
)
=
cot(x) =
co
t
(
x
)
=
1
t
a
n
(
x
)
\frac{1}{tan(x)}
t
an
(
x
)
1
t
a
n
(
x
)
=
tan(x) =
t
an
(
x
)
=
s
i
n
(
x
)
c
o
s
(
x
)
\frac{sin(x)}{cos(x)}
cos
(
x
)
s
in
(
x
)
c
o
t
(
x
)
=
cot(x) =
co
t
(
x
)
=
c
o
s
(
x
)
s
i
n
(
x
)
\frac{cos(x)}{sin(x)}
s
in
(
x
)
cos
(
x
)
s
i
n
2
x
+
sin^2x+
s
i
n
2
x
+
c
o
s
2
=
cos^2=
co
s
2
=
1
1
1
pythagorean identity
t
a
n
2
x
+
tan^2x+
t
a
n
2
x
+
1
=
1=
1
=
s
e
c
2
x
sec^2x
se
c
2
x
c
o
t
2
x
+
cot^2x+
co
t
2
x
+
1
=
1=
1
=
c
s
c
2
x
csc^2x
cs
c
2
x
pythagorean identity
Graph of
sin
(x)
Graph of sin
(
X
)
Domain
: (-∞, ∞)
range
[
-1
,
1
]
period is 2π
goes through the
origin
odd
function
Graph of
cos
(x)
Graph of cos(x)
Domain
(-∞,∞)
range
[-1,1]
period is 2π
even
function
a =
amplitude
,(vertical stretch or shrink)
k =
period
, (changes
length
of
wavelength
)
y
=
y=
y
=
a
s
i
n
k
x
a \:sin \:kx
a
s
in
k
x
a =
amplitude
,(vertical stretch or shrink)
k =
period
, (changes
length
of
wavelength
)
y
=
y=
y
=
a
c
o
s
k
x
a\:cos\: kx
a
cos
k
x
How to find period of y= a
sin
kx and y= a
cos
kx?
2
π
k
\frac{2\pi}{k}
k
2
π