Thread: asymptotes and discontinuitites

1. asymptotes and discontinuitites

$\displaystyle (x^2 +5x+4)/(x^3 +x^2 -9x -9)$

just trying to find the vertical and horizontal asymptotes of this function...

hortizontal at 0.

the vertical asymptotes at -1, 3 and -3..just wondering if this is correct..

and how do i go on about finding where the discontinuities are in this function by looking at the graph...how would i work it out??

thanks

2. Originally Posted by b00yeah05
$\displaystyle (x^2 +5x+4)/(x^3 +x^2 -9x -9)$

just trying to find the vertical and horizontal asymptotes of this function...

hortizontal at 0. Mr F says: Strictly speaking, y = 0.

the vertical asymptotes at -1, 3 and -3.. Mr F says: Strictly speaking, x = -1, x = 3 and x = -3.

just wondering if this is correct.. Mr F says: Now you know.

and how do i go on about finding where the discontinuities are in this function by looking at the graph...how would i work it out??

thanks
Imagine trying to run a pen along the graph. Every time you have to lift your pen off the page, you've met a discontinuity. Clearly the function is discontinuous at the vertical asymptotes, that is, at x = -1, 3 and -3.

I spoke too soon. There's a mistake in the above .... See my next reply.

3. Originally Posted by b00yeah05
$\displaystyle (x^2 +5x+4)/(x^3 +x^2 -9x -9)$

just trying to find the vertical and horizontal asymptotes of this function...

hortizontal at 0. Mr F says: Yes, at y = 0.

the vertical asymptotes at -1, 3 and -3..just wondering if this is correct.. Mr F says: NO! There's vertical asymptotes at x = -3 and x = 3. But NOT x = -1. See my main reply below.

and how do i go on about finding where the discontinuities are in this function by looking at the graph...how would i work it out??

thanks
$\displaystyle \frac{x^2 +5x+4}{x^3 +x^2 -9x -9} = \frac{(x + 1)(x + 4)}{(x + 1)(x + 3)(x - 3)} = \frac{x+4}{(x+3)(x - 3)}$, provided $\displaystyle x \neq -1$ .....

So the function is undefined at x = 3 and x = -3 => vertical asympotes. But the function does not exist (different from undefined) at x = -1. At x = -1 you have an open circle (to indicate the point is NOT included).

The function is still discontinuous though at x = -3, x = 3 AND x = -1 ..... you still have to left your pen off the page at x = -1 ......