Trig Help: Solve Y=2sec(3x-pi)

[nando94]

I have these practice trig problems for a test. I understand the simple graphs but its these more complicated ones.

y=2sec(3x-pi)

my approach:

I first find the period and phase shift. Since the period of a normal secant function which is 2pi. Then the period of this one is 2pi/3. The phase shift is pi/3. There is going to a be a vertical strech of 2.

After that I am stuck because I don't know how to exactly graph this even though I have all the information. Its mostly because of the odd period, 2pi/3. Really confused.

 

[SammyS]

I have these practice trig problems for a test. I understand the simple graphs but it's these more complicated ones.

y=2sec(3x-pi)

my approach:

I first find the period and phase shift. Since the period of a normal secant function which is 2pi. Then the period of this one is 2pi/3. The phase shift is pi/3. There is going to a be a vertical stretch of 2.

After that I am stuck because I don't know how to exactly graph this even though I have all the information. Its mostly because of the odd period, 2pi/3. Really confused.

I don't know about you, but I'd rather plot cosine rather than secant.

In addition to the information you have deduced, which all looks good, I like to look at the y-intercept, and find some points where the cosine is 1, where it's -1, and where it's zero.

For the y-intercept, set x=0 → y = cos(3(0)-π) = cos(-π) = ?

To find where cos(θ) = 1, of course an obvious place is where θ = 0, so that's 3x-π = 0. But in general cos(θ) = 1, where θ is an integer multiple of 2π. Of course you can generate these in terms of x, by using the easy case of the x which satisfies 3x-π = 0, and then using your information on period.

You can do similar for the the values of x which make cos(3x-π) equal to -1 and equal to zero. You can also generate these values from the information in the above paragraph, and then using the fact that the cosine is equal to -1 at 1/2 a period on either side of where it is +1. The zeros occur 1/4 of a period on either side of where it is +1.

Once you have cos(3x-π), then sketch sec(3x-π), then stretch that vertically to get y=2sec(3x-π).

 

[Hitchens]

A good way to check your graphs is this (after you've made an attempt of course!).

 

[nando94]

Thanks SammyS. I orginally planned to graph 1/cos which was the secant but it made it more confusing lol. Now I see what you mean by graph cos first.

My question is that can I also apply to secant and graph sine first or cotangent and graph tangent first? I am going to have all three of those on my test as well.

 

[Hitchens]

My question is that can I also apply to secant and graph sine first or cotangent and graph tangent first? I am going to have all three of those on my test as well.

Naturally, yes.

 

[eumyang]

My question is that can I also apply to secant and graph sine first or cotangent and graph tangent first? I am going to have all three of those on my test as well.

I don't think you really want to "apply to secant and graph sine first."

 

[nando94]

lol cosecant but thanks cause now I am going to ace this test.