1
π
To graph one period of y = − cos 4x −
− 1, I begin by calculating some basic facts
2
8
about the function:
1
• Amplitude :
2
• Vertical Shift: −1
• Period:
2π
π
=
4
8
π
π
• Phase Shift: 8 =
4
32
Our fundamental interval is
h
π π
φ 2π
φ
π π
πi
π 16
π 17π
=
,
+
=
,
+
,
+
=
,
ω ω
ω
32 32 2
32 32 2 16
32 32
I need to divide the fundamental interval into four pieces of equal length and find the endpoints. In this case we are dealing with annoying fractions. Our four subintervals are
5π 9π
9π 13π
13π 17π
π 5π
,
,
,
,
,
,
,
32 32
32 32
32 32
32 32
Now I will explain how I got those numbers in
class. I know that the whole fundamental interval
1 π π
π
π
long, so each subinterval has width
= . My first key point is at x =
, and I
is
2
4 2
8
32
π
π
4 π
4π
get each new point by moving an additional
units to the right. I rewrite
as
=
8
8
4 8
32
because then it is easier to do the addition if the denominators of the fractions match:
π
• Key Point 1: x =
32
• Key Point 2: x =
π
4π
5π
+
=
32 32
32
• Key Point 3: x =
5π 4π
9π
+
=
32
32
32
• Key Point 4: x =
9π 4π
13π
+
=
32
32
32
13π 4π
17π
+
=
32
32
32
Now I need to calculate the y-values that go with these x-values.
• Key Point 5: x =
1
2
x
π
32
5π
32
9π
32
13π
32
17π
32
1
π
− cos 4x −
−1
2
8
π π
1
1
1
3
− cos 4
−
− 1 = − cos (0) − 1 = − (1) − 1 = −
2
32
8
2
2
2
π 1
5π
π
1
1
− cos 4
−
− 1 = − cos
− 1 = − (0) − 1 = −1
2
32
8
2
2
2
π
1
1
9π
1
1
−
− 1 = − cos (π) − 1 = − (−1) − 1 = −
− cos 4
2
32
8
2
2
2
π
1
1
1
13π
3π
−
− 1 = − cos
− 1 = − (0) − 1 = −1
− cos 4
2
32
8
2
2
2
1
π
1
17π
1
− cos 4
−
− 1 = − cos (0) − 1 = − (1) − 1
2
32
8
2
2
So we plot the five points
π
3
,−
32 2
5π
9π 1
13π
17π 3
,
, −1 ,
,−
,
, −1 ,
,−
32
32 2
32
32
2
Note that because of the negative sign the coefficient A, our graph is upside down. So we
connect the five points with an upside down cosine curve.
3
Another way to find the y-values is to recall that the x-values for my key points for will
π
3π
correspond to the first five quadrantal angles 0, , π, , 2π. The y-values of the cosine function
2
2
at these angles are 1, 0, −1, 0, and 1. So in our transformed graph, we can obtain the y-value of
each key point by first multiplying by A and then adding the vertical shift B. So our y-values
are obtained as follows:
4
Quadrantal Angle Y = cos(Quadrantal Angle) Key Point x
0
π
2
π
cos(0) = 1
π cos
=0
2
3π
2
cos(π) = 1
3π
=0
cos
2
2π
cos(2π) = 1
π
32
5π
32
9π
32
13π
32
17π
32
A(Y ) + B = f (x)
1
3
− (1) − 1 = −
2
2
1
− (0) − 1 = −1
2
1
1
− (−1) − 1 = −
2
2
1
− (0) − 1 = −1
2
1
3
− (1) − 1 = −
2
2
You can compare this table with the first table and see how the y values correspond to both
the key points and the quadrantal angles.