How do I graph sinusoidal functions?

1 Answer
Aug 2, 2015

Answer:

See explanation.

Explanation:

We know what the basic graph of #y = sinx# look like.

graph{y = sinx [-12.11, 16.36, -6.92, 7.31]}

The graph goes up to a maximum of #1# and down to a minimum or #-1#. It is 'centered' on the #x# axis which is the line #y=0#.

Let's look at: #f(x) = D+Asin(Bx+C)#

The number #D# will shift the entire graph up (if #D > 0#) or down (#D < 0#) by the amount #absD#
If #D != 0#, the center line will move to the horizontal line: #y = D#

Let's leave that vertical shift out and just look at:

#y = Asin(Bx+C)#

If #B < 0#, use the fact that #sin(-u) = -sin(u)# to rewrite with a positve coefficient of #x#.

If #A# is negative, we will reflect the graph across the #x# axis.

Multiplying the sine by #A# makes the new graph have a maximum of #absA# and a minimum of #-absA#.
The number #absA# is called the Amplitude of the graph.

The period of the graph is the length of one complete cycle through the graph. For the basic graph, the period is #pi#.
We often think of and describe this by saying, "The first period of the graph starts when we take the sine of #0# and ends when we take the sine of #2pi#. That means the angle has gone once around the circle."
(This description is helpful, but can be a little misleading, too.)

Returning to #y = Asin(Bx+C)#

We will take the sine of #0# when #Bx+C = 0#
and we will take the sine of #2pi# when #Bx+C = 2pi#

So we will start the first period when #x = -C/B# (solve the first equation.)
The number #-C/B# is called the Phase (or Horizontal) Shift.

And we will finish the first period when #x = (2pi-C)/B# (Don't memorize that, we have another way of finding it.)

The "length of the period" is the end minus the start, or:

#((2pi-C)/B) - (-C/B)# And that simplifies to #(2pi)/B#

So that is the why, and here is the summary:

#y = Asin(Bx+C)# has

Amplitude: #" "absA" "# (That's how far above and below the center line we need to go.)

Period: #" "(2pi)/B" "#

Phase Shift: #" "# Is the solution to #Bx+C = 0#