Horizontal and vertical asymptotes

[ssmmss]

f(x)=(x^3)/3-(5x^2)/2+6x+1

The function is undefined since the denominator doesn't equal 0 (or at least that's my assessment). As a result, this function is defined for all values of x and as such will have no vertical asymptotes. Is this correct?

I am struggling with the horizontal asymptotes though. Kindly advise.

 

[ssmmss]

f(x)=(x^3)/3-(5x^2)/2+6x+1

The function is undefined since the denominator doesn't equal 0 (or at least that's my assessment). As a result, this function is defined for all values of x and as such will have no vertical asymptotes. Is this correct?

I am struggling with the horizontal asymptotes though. Kindly advise.

This is my attempt on the horizontal asymptote:
As X goes to infinity, f(x) equals infinity and as x goes to negative infinity, f(x) equals negative infinity, meaning this function does not have horizontal asymptotes. Thoughts please?

 

[stapel]

This is my attempt on the horizontal asymptote:

Are you saying that this is your first attempt on this exercise, or that you haven't worked with horizontal asymptotes before? If the latter, try here. If the former, it might help to convert the function to a common denominator, as this might make the computations and reasoning more clear.

Note: I am assuming that your function is as follows:

. . . . .\(\displaystyle f(x)\, =\, \dfrac{x^3}{3}\, -\, \dfrac{5x^2}{2}\, +\, 6x\, +\, 1\)

 

[ssmmss]

Are you saying that this is your first attempt on this exercise, or that you haven't worked with horizontal asymptotes before? If the latter, try here. If the former, it might help to convert the function to a common denominator, as this might make the computations and reasoning more clear.

Note: I am assuming that your function is as follows:

. . . . .\(\displaystyle f(x)\, =\, \dfrac{x^3}{3}\, -\, \dfrac{5x^2}{2}\, +\, 6x\, +\, 1\)

Thank you. Are my answers correct?

 

[Quaid]

f(x) = (1/3)x^3 - (5/2)x^2 + 6x + 1

The function is undefined since the denominator doesn't equal 0

Hi ssmmss:

I'm not sure what you're trying to say, above. Function f is a cubic polynomial; therefore, it has no denominator.

this function is defined for all values of x and as such will have no vertical asymptotes.

This is correct. The domain of all polynomial functions is the set of Real numbers. Polynomial functions do not have asymptotes.

But, you previously wrote that function f is undefined; this is why I do not understand your first statement above.

Cheers

 

[stapel]

Are my answers correct?

You are correct, in that there is no horizontal asymptote. Not being able to see how your book deals with this sort of thing (in the worked examples), I can't evaluate the correctness of your method.

 

[Quaid]

As X goes to infinity, f(x) equals infinity and as x goes to negative infinity, f(x) equals negative infinity, meaning this function does not have horizontal asymptote(s).

This is correct. The global behavior of any polynomial function is ±infinity. Therefore, these functions do not approach any number, as x becomes huge (positively or negatively).

Here are two notes about notation (see the red highlighting, above).

Symbols X and x represent different quantities. If you start out using symbol x to represent an independent variable, then stick with that symbol. Don't interchange X and x.

Be careful with the word "equals". It's not correct to say that a function equals infinity. Infinity is not a number; functions output only numbers. Instead, write "the function approaches infinity" or "the function is unbounded", as x becomes large in absolute value.

This situation with "equals" will come back, when you learn about limits. You will see people write that a particular limit equals infinity. That phrase "equals infinity" must be understood as an abbreviation that means "the limit does not exist" because the function is not approaching any fixed value; it means that the function "approaches infinity" or negative infinity. We could also say it means the function "increases without bound" or decreases without bound, as the case may be.

Point is, be careful with the word "equals"

:cool: