Graph Sketching

[sweetliljenny]

I don't know if anyone will be able to help me out but for instance say i had
y = 3cos(1/2x - 90) - 1 on the interval 0 less than or equal to x less than or equal 360 degrees

y = a cos(b(x-h))+k
y = 3 cos(1/2(x - 180)) - 1

So,
a (amplitude) = 3
b: frequency = 1/2
Period: 2 x pi / b = 2 x pi / (1/2) = 4pi
k: vertical shift down 1 (midline y = -1)
h: horizontal shift to the right 180 degrees

Now my question is can anyone help me as far as graphing both sketches (before and after the shifts) ?

 

[wjm11]

I don't know if anyone will be able to help me out but for instance say i had
y = 3cos(1/2x - 90) - 1 on the interval 0 less than or equal to x less than or equal 360 degrees

y = a cos(b(x-h))+k
y = 3 cos(1/2(x - 180)) - 1

So,
a (amplitude) = 3
b: frequency = 1/2
Period: 2 x pi / b = 2 x pi / (1/2) = 4pi
k: vertical shift down 1 (midline y = -1)
h: horizontal shift to the right 180 degrees

Now my question is can anyone help me as far as graphing both sketches (before and after the shifts) ?

Good work, Jenny.

One way to approach this is to think of the “parent” cosine function (y = cos x) as a “U” shape, with the upper, left hand corner of the “U” starting at (0,1), the bottom of the U at (pi,-1), and the upper right hand corner of the U at (2pi,1).

Let’s consider just the “stretched” period and the vertical shift first, keeping the amplitude at “1” momentarily: Start by moving that “upper left hand corner” point down one position to (0,0). The “upper right hand corner” point will be at (4pi,0) due to the doubled period. The bottom of the U will be half way in between them at (2pi,-2). Does this make sense so far? All three points are one vertical unit away from the centerline you identified: y = -1.

Now just add the amplitude, moving all three points three units away from the centerline:
(0,0) goes up to (0,2)
(2pi,-2) goes down to (2pi,-4)
(4pi,0) goes up to (4pi,2)

I find the easiest way to do this sort of sketch is to first draw a dotted line where the centerline is, then draw two more dotted lines parallel to it, one above and one below, the distance equal to the size of the amplitude. Then sketch your sine or cosine curve between the upper and lower lines. So in this case, you’d draw dotted lines at
y = -1
y = 2
y = -4

Hope that helps.

 

[sweetliljenny]

after plugging cos(x) into y= and pressing graph i see how you arrived at (0,1) for the upper left hand corner of the "u" but i don't understand how you get that (pi, -1) is the bottom of the "u" and that (2pi,1) is the upper right hand corner of the "u"

is that standard to know?

 

[wjm11]

The pi, 2pi, and 4pi are if we're working in radians. They would be 180, 360, and 720 if we're working in degrees. You have to set either degrees or radians as the "mode" on your calculator, depending on which one you're working in.

Also, the period of the sine and cosine functions is 360 degrees or 2pi when the "b" value is 1. The shape of the curve gets stretched or compressed horizontally when b has values other than 1.