Determine the Vertical Asymptotes (if any)

[Nicolas5150]

I have three problems in which I am having some trouble figuring out. The first one (if I understood correctly) I needed to get x alone, which I did and ended up with none as the answer due to a negative number within in a square root.
If anyone could help explain the process on how to solve the next two as well as verify the first as correct it would be much appreciated.

 

[Quaid]

Vertical asymptotes occur wherever the denominator becomes zero.

The first exercise has x^2 + 6 in the denominator.

You are correct; there is no Real value of x to make x^2 + 6 = 0


On the second exercise, use an identity, to rewrite the secant function in terms of cosine.

Now ask yourself, "what values of x cause the denominator to become zero?"


On the third exercise, use a common denominator to combine 5 + 6/x into a single ratio.

Now ask yourself, "what value of x causes the denominator to become zero?"


Cheers :cool:

 

[Nicolas5150]

Vertical asymptotes occur wherever the denominator becomes zero.

The first exercise has x^2 + 6 in the denominator.

You are correct; there is no Real value of x to make x^2 + 6 = 0


On the second exercise, use an identity, to rewrite the secant function in terms of cosine.

Now ask yourself, "what values of x cause the denominator to become zero?"


On the third exercise, use a common denominator to combine 5 + 6/x into a single ratio.

Now ask yourself, "what value of x causes the denominator to become zero?"


Cheers :cool:

Thank you for making sure the first question was correct.

The second one from what I gather should then deal with 1/cos(pi x) but not sure what I would be doing with it. I'm a little confused since there is no x in the denominator to begin the problem off.

As well, the third one so far got 5+(6/5) into (5x+6)/x I am not sure if that would be correct but if so then the x value could not equal 0?

 

[richardt]

Nicolas5150:

sec(pi*x/2) = 1 / cos(pi*x/2). Hence the function is undefined whenever the denominator, cos(pi*x/2), is equal to zero. The cosine function has value zero whenever the argument is an odd multiple of pi/2 (i.e., +/- pi/2, +/-3pi/2, +/-5pi/2, ...). That said, since the given denominator is pi*x/2 = x * pi/2, it follows that function, g, is undefined for all odd x.

As to f(x) = 5 + 6/x, f is defined for all real numbers, x, such that x not = 0. After all, substituting anything beside zero for x gives a real number; does it not?

Rich

 

[Quaid]

The second one from what I gather should then deal with 1/cos(pi x) but not sure what I would be doing with it.

Did you see the following (red highlight for emphasis)?

On the second exercise, use an identity, to rewrite the secant function in terms of cosine.

Now ask yourself, "what values of x cause the denominator to become zero?"

I'm a little confused since there is no x in the denominator to begin the problem off.

Rest assured, that is not an issue.

sec(A) = 1/cos(A) is an identity. This means that, for all radian measures A for which both sides are defined, those two expressions are interchangeable.

In other words, the cosine function is the reciprocal of the secant function, so your problem implicitly starts off with x in a denominator.

the third one so far got 5+(6/5) into (5x+6)/x

I am not sure if that would be correct

Your result is correct.

The statement in blue indicates a hole of concern in the prerequisite material.

so then the x value could not equal 0?

Is this question serious? That is, you actually do not know for certain whether x can equal zero (or not)?


The instructions in these exercises ask for vertical asymptotes.

Vertical asymptotes occur wherever the denominator becomes zero.

Now ask yourself, "what value of x causes the denominator to become zero?"

Are you enrolled in a calculus course? Are you preparing for a placement exam? How long has it been, since you studied math?

Thank you :cool: