How do you graph # y=2+sin(1/2x)#?

1 Answer
jim p. · Stefan V.
Aug 14, 2017

Draw a sine wave with twice the wavelength. Then shift it UP two units.

Explanation:

You know what the graph of #sin(x)# is.

  • when #x# is #0, y = 0#
  • when #x# is #pi/2, y=1#
  • when #x# is #pi, y=0#
  • when #x# is #(3*pi)/2, y=-1#
  • when #x# is #2*pi, y=0#

graph{sin(x) [-10, 10, -5, 5]}

So now, for #y = sin(x/2)#:

  • when #x# is #0, x/2# is #0 , y= 0#
  • when #x# is #pi, x/2# is #pi/2, y= 1#
  • when #x# is #2 * pi, x/2# is #pi , y= 0#
  • when #x# is #3 * pi, x/2# is #(3*pi)/2 , y= -1#
  • when #x# is #4 * pi, x/2# is #2 * pi , y= 0#

So, it's the same sine wave, but stretched out.

Now, since your function is #2 + sin(x/2)#, simply add #2# to the value for #y# in the table above:

  • when #x# is #0, y= 2#
  • when #x# is #pi, y= 3#
  • when #x# is #2 * pi, y= 2#
  • when #x# is #3 * pi, y= 1#
  • when #x# is #4 * pi , y= 2#

graph{2 + sin(x/2) [-10, 10, -5, 5]}

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