# Thread: Horizontal and vertical asymptotes

1. ## Horizontal and vertical asymptotes

I got problem with nr 39. If i understand right there shall be at limit x->2 because denominator. With other words if i understand correct i shall solve limit x->2 (2x+1)/(x-2) well i basicly got problem to solve that limit.

2. ## Re: Horizontal and vertical asymptotes

Originally Posted by Petrus
I got problem with nr 39. If i understand right there shall be at limit x->2 because denominator. With other words if i understand correct i shall solve limit x->2 (2x+1)/(x-2) well i basicly got problem to solve that limit.
1. A vertical asymptote occurs if the denominator of a quotient equals zero and the numerator unequals zero. With your example:

$y = \frac{2x+1}{x-1}$

you'll get a vertical asymptote at x = 1

2. Re-write the equation by synthetic division:

$y = \frac{2x+1}{x-1}~\implies~ \boxed{y = 2 + \frac3{x-1}}$

Since $\lim_{x \to \infty}\left(\frac3{x-1} \right) = 0$

you'll get

$\lim_{x \to \infty} y = 2$

3. The line with the equation y = 2 is the horizontal asymptote.

3. ## Re: Horizontal and vertical asymptotes

Thanks!
Just wanna correct u it was x-2 on denominator so vertical asymptot at x=2
so we rewrite the equation as 2+5/(x-2)
Edit: just a question. Should i also check lin x->-infinity aswell? In this problem it Will still be 2 but i should also check when lim x->-infinity? I am correct?

4. ## Re: Horizontal and vertical asymptotes

On nr 43. I get horizontal asymptotes at infinity and -infinity. Shall i write that?? I got vertical at x=5

5. ## Re: Horizontal and vertical asymptotes

Originally Posted by Petrus
On nr 43. I get horizontal asymptotes at infinity and -infinity. Shall i write that?? I got vertical at x=5
1. Factorize the denominator (and in this case the numerator too):

$y = \frac{x^3-x}{x^2-6x+5} = \frac{x(x^2-1)}{(x-1)(x-5)}= \frac{x(x-1)(x+1)}{(x-1)(x-5)}$

That means you'll get a vertical asymptote at x = 5 and a hole in the graph at x = 1.

2. Re-write the term of the function by synthetic division:

$y = \frac{x^3-x}{x^2-6x+5} ~\implies~y = \underbrace{x+5}_{\text{term of the asymptote}}+\frac{25x-25}{x^2-6x+5}$

That means your function has a slanted asymptote $y = x+5$

6. ## Re: Horizontal and vertical asymptotes

Originally Posted by earboth
1. Factorize the denominator (and in this case the numerator too):

$y = \frac{x^3-x}{x^2-6x+5} = \frac{x(x^2-1)}{(x-1)(x-5)}= \frac{x(x-1)(x+1)}{(x-1)(x-5)}$

That means you'll get a vertical asymptote at x = 5 and a hole in the graph at x = 1.

2. Re-write the term of the function by synthetic division:

$y = \frac{x^3-x}{x^2-6x+5} ~\implies~y = \underbrace{x+5}_{\text{term of the asymptote}}+\frac{25x-25}{x^2-6x+5}$

That means your function has a slanted asymptote $y = x+5$
Ty for taking ur time and thanks for the explain! all make sense now!
Greeting Petrus

7. ## Re: Horizontal and vertical asymptotes

IF limf(x)/x =a when x goes to infinity or minus infinity and lim(f(x)-ax)=b when x goes to infinity or - infinity then the line y=ax+b is a slant asymptote of f(x).
MINOAS

8. ## Re: Horizontal and vertical asymptotes

Originally Posted by Petrus
Ty for taking ur time and thanks for the explain! all make sense now!
Greeting Petrus
Ingen orsak!

9. ## Re: Horizontal and vertical asymptotes

Originally Posted by MINOANMAN
IF limf(x)/x =a when x goes to infinity or minus infinity and lim(f(x)-ax)=b when x goes to infinity or - infinity then the line y=ax+b is a slant asymptote of f(x).
MINOAS
I have not learned about "slant asymptote":P only learned horizontal and vertical yet but now i kinda know a litle ty

10. ## Re: Horizontal and vertical asymptotes

Some functions not all ...have asymptotes that are not vertical or horizontal .these are slant asymptotes...depending on your math carricculum you may or may not learnt them...the above theorem I mentioned is used to track them.
Minoas