Let's first graph `y = sin 2x` (without the "+ 1") to get an idea of what phase shift means.

The amplitude is 1 and the period is `(2π)/2 = π.`

I have drawn a little more than 2 cycles, starting from `x = 0`.

The graph of `y=sin(2x)` for `0 ≤ x ≤ 6.5`

Now let's consider the phase shift. Using the formula above, we will need to shift our curve by:

Phase shift `=-c/b=-1/2=-0.5`

This means we have to shift the curve to the **left** (because the phase shift is negative) by `0.5`. Here is the answer (in green). I have kept the original *y* = sin 2*x * (in dotted gray) so you can see what's happening.

The graph of `y=sin(2x+1)` for `-0.5 ≤ x ≤ 6.7`

Note that the curve passes through:

- (−0.5, 0) on the
*x*-axis (because we shifted the curve to the left by 0.5) and - sin 1 = 0.84 on the
*y*-axis. This figure is obtained by substituting*x*= 0 into*y*= sin(2*x*+ 1); (in radians, of course).

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