TI-83 Problem: Finding the Domain of a Secant Graph

[Sabellic]

TI-83 problem...Secant Graphs

Homework Statement

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

Homework Equations

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

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?

 

[HallsofIvy]

Homework Statement

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

Homework Equations

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

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.

Do you mean RANGE rather than domain here?

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

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 [itex]-2(x- 90)= (2n+1)90[/itex]. Solve that for x.

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?

 

[Sabellic]

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

Thank you, HallsofIvy!