Solving Vertical Asymptotes: (x-1)(x-3) / y-1 or (x-1)/(x-1)(x-3)?

[cowgiljl]

The question is
fine a equation f(x) that has a vertical asymptotes of X=1and X=3 and a horizontal asymptotes of Y=1

A) (x-1)(x-3) / y-1 = x^2-4x+3 / x-1

OR

B) (x-1) /(x-1)(x-3) = just the oppsite of above

is any of these correct?
I think B is correct but right now I can't find any info on it in my book

 

[Pyrrhus]

I think it won't be correct, the Horizontal asymtopte for B will be y=0.

If i remember correctly the horizontal asymptote is

$$\lim_{x \rightarrow \infty} \frac{(x-1)}{(x-1)(x-3)}$$

The above case will give y=0. The numerator must of the form
$$x^2 + Bx +C$$ where B and C don't allow factors equal to those of the denominator.

To make myself clearer:

$$\lim_{x \rightarrow \infty} \frac{Ax^2 +Bx +C}{x^2-4x+3}$$

$$\lim_{x \rightarrow \infty} \frac{A +\frac{B}{x} +\frac{C}{x^2}}{1-\frac{4}{x}+\frac{3}{x^2}} = \frac{A}{1}$$

$$\frac{A}{1} = 1$$

because horizontal asymptote should be y = 1

 

[Chaotic42]

B gives you a removable discontinuity at x=1.

Try this:

f(x)=[(x+1)/(x-1)(x-3)]+1