How do you graph #f(x)=2sin(3x)#?

1 Answer

See below:

Explanation:

Let's start with the graph for #f(x)=sinx#:

graph{sinx [-6.25, 6.25, -3, 3]}

I've set the graph to be #-2pi<=x<=2pi#. The graph intersects the x-axis at #x=pi+npi# where n is an integer. The max is #y=1# and occurs at #x=pi/2+n2pi# and min at #y=-1# at #x=(3pi)/2+n2pi#.

So how does the graph change with #y=2sin(3x)#?

The first thing I'll look at is the 2. That will change the expansion along the y-axis by a factor of 2 - or in other words, each max and min will be 2 and #-2# respectively.

The other thing that happens is that the #3x# part of this will compress the graph along the x-axis by a factor of 3 - so what had been happening at #pi# now happens at #pi/3#.

When we put these two changes together, we get this:

graph{2sin(3x) [-6.25, 6.25, -3, 3]}

The graph is still set at #-2pi<=x<=2pi#. The graph intersects the x-axis at #x=pi/3+npi/3# where n is an integer. The max is #y=2# and occurs at #x=pi/6+n(2/3)pi# and min at #y=-2# at #x=(pi)/2+n(2/3)pi#.