Question

Give an example of a rational function that has a horizontal asymptote at y = 1 and a vertical asymptote at x = 4.

Answer 1

The horizontal asymptote is always 0 if the degree of the denominator is greater than that of the numerator. However, when the degrees of numerator/denominator are equal, we take the coefficients of the highest degree from the top and bottom and divide to get the horizontal asymptote. So you must make the degree of the top and bottom equal and make sure that their leading coefficients are also equal so that when you divide them, you get 1, which will give you the y = 1 horizontal asymptote.

If there is a V.A at x = 4, then the function is non-existent at x = 4. So make the denominator such that the degree and the leading coefficient are the same as the numerator, but the denominator restricts the domain in such a way such that the function doesn't exist at x = 4.

Get it?

@jtschellenberg

Answer 2

Yeah i get it thanks!

Answer 3

@genius12 that explanation of the horizontal asymptote is rather confusing.
I'm not quite sure I understand what you're saying.

I like to think of it more in these terms I guess:
Here is our principle function which will produce asymptotes at y=0 and x=0:
\[\large y=\frac{1}{x}\]

To get a horizontal asymptote at y=1, we simply shift the function up 1 unit,\[\large y=\frac{1}{x}+1\]

Then to shift our asymptote to the right 4 units, we subtract 4 from x,\[\large y=\frac{1}{x-4}+1\]


Maybe I could have given you an example instead of just giving the answer, sorry about that.

Answer 4

that helps too @zepdrix! thanks a lot but i do get it :)

Answer 5

Give an example of a rational function that has a horizontal asymptote at y = 1 and a vertical asymptote at x = 4.

For a horizontal asymptote at \(y=1\), we know \(y=1+f(x)\) where \(f(x)\to0\) for \(x\to\pm\infty\). For our vertical asymptote to be at \(x=4\), we require \(f(x)\) to have a singularity at \(x=4\) so \(f(x)=(x-4),(x-4)^2,\dots\) all should work.