Rules of the game:

When given a function of the form:

y = asink(x - b) + c, or y = acosk(x - b) + c

then we have the following by the basic transformation rules:

amplitude = |a|....this is how high and how low the graph goes from the resting point

Period = 2pi/|k|.....this is how long the graph is, how long is one complete oscillation

Phase shift = b......this is how much we shift to the left or right, note that we change the sign

vertical shift = c....this is how much we shift up or down, if c is positive, it's up, if c is negative, it's down

appropriate graphing interval = [b, b + 2pi/|k|]....this is the appropriate interval to draw the graph between. that is, we start where we shifted to and continue for one period

I will do the first one for you. try the second one and tell me if you have any problems, the two are very similar.

Okay, so they want us to graph this over 2 cycles, that is we will graph two periods.Graph each function over 2 cycles:

1) y = 2cos (x+5pi/3) - 2

now, from the rules above, we have:

amplitude = |2| = 2

period = 2pi/|1| = 2pi

phase shift = - 5pi/2

vertical shift = -2, that is 2 units down

appropriate graphing interval = [-5pi/3, -5pi/3 + 2pi] = [-5pi/3, pi/3]

this would be the appropriate period to graph, however, they don't want this, they want 2 periods, so we will graph in the interval [-5pi/3, 7pi/3]

Now remember what the cosine graph looks like. in one period, it starts at the peak of the amplitude, goes down and then back up. halfway between the beginning and the end it hits its minimum point. halfway between its begining and it's minimum point it hits the resting position on its way down. halfway between its minimum point and its end, it hits the resting position on its way up.

so, for the first period, between -5pi/3 and pi/3, this is what our graph will look like, as described by words.

we have a vertical shift of 2 units down, so the resting position is the line, y = -2. the amplitude is 2, so the graph will go 2 units above this (that is to the x-axis) and then 2 units below this (that is to the line y = -4). the period is 2pi, just like a regular cosine graph, so no problems here, we're on familiar territory. halfway between -5pi/3 and pi/3 we have -2pi/3, this is the x-value for which we will hit the first minimum point, our minimum point will be (-2pi/3, -4). halfway between the begining and the minimum point we have x = -7pi/6. so we will pass through (-7pi/6, -2) halfway on our way to the minimum point. halfway between the min point and the end of the first period is x = -pi/6. so we will pass through (-pi/6, -2) halfway on our way up to the end. the second period repeats this pattern.

here's what the graph will look like:

I drew a grid for the first period. tell me if you don't get what it means. the lines of course are markers. each vertical line except the first and last, represent halfway between two points, i told you which points above