How do you graph $y=-2 - 3 sin (x - pi)$ ?

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Answer 1 · 2016-11-18T04:45:21

See the inserted graph and the explanation.

$y=-2-3 sin (x-pi)=-2+3 sin (pi-x)=-2+ 3 sin x$

The period $2pi$ for sin x is the period of y..

The graph is symmetrical about (0. -2).

So, a Table for the half period $x in [0, pi]$ is sufficient for extending

it for one period $x in [-pi. pi]$ .

$(x, y): (0, -2) (pi/6, -3/2) (pi/3, 3sqrt3/2-2) (pi/2, 1)$

$(2/3pi, 3sqrt3/2) (5/6pi, =3/2) (pi, -2)$

The crests of f the periodic wave in the graph are at

$(-3/2pi, 1), (pi/2, 1) and (5/2pi, 1)$ .

graph{y=3 sinx -2 [-10, 10, -5, 5]}

How do you graph y=-2 - 3 sin (x - pi)y=-2 - 3 sin (x - pi)?