# TI-83 problem Secant Graphs

1. Oct 13, 2008

### Sabellic

TI-83 problem....Secant Graphs

1. The problem statement, all variables and given/known data

What are the properties of:
y= 2 sec(-2x + 90deg) + 3

2. Relevant equations

sec (x) = (1/cos (x))

3. The attempt at a solution

I have a problem with finding the Domain of y= 2 sec(-2x + 180deg) + 3.

First of all, I have to put the equation in a neater form:

y= 2 [sec -2(x - 90deg)] + 3

Now, if I want to find the domain, I need to find what sec CAN'T equal. That is the vertical asymptotes. Now asymptotes can be found where the graph of the inverse of sec (which is cos) crossed the x-axis.

So therefore I look at the corresponding cos function:
y= 2 [cos -2(x - 90deg)] + 3

But the problem is: this cos function does NOT cross the x-axis. If it does not cross the x-axis then how can any vertical asymptotes appear?

I later typed in "y= 2 (1/cos (-2x + 180deg)) + 3" into the TI-83. But it DID show vertical asymptotes. Does anyone know where these asymptotes came from???

2. Oct 13, 2008

### HallsofIvy

Staff Emeritus
Re: TI-83 problem....Secant Graphs

Do you mean RANGE rather than domain here?

No. You are not concerned about where this function "crosses the x-axis". The secant is undefined where cosine itself is 0. cos(x) is 0 when x is an odd multiple of 90: x= (2n+1)90 for n any integer. Here you want $-2(x- 90)= (2n+1)90$. Solve that for x.

3. Oct 13, 2008

### Sabellic

Re: TI-83 problem....Secant Graphs

You know, that's starting to make sense. I will try this.

Thank you, HallsofIvy!!