#7**+5 **

Thanks, heureka, for showing how to find that oblique asymptote....I was going to go through it but I was working on another problem......besides......I couldn't have presented it better, anyway!!!

I want to point out one more thing.....the remainder of (24)/(x-3) means that we actually don't have a perfectly linear asymptote when x is relatively "small." However, as heureka shows by limit analysis, this "fiddly" asymptotic behavior disappears as x aproaches "large" positive and negative values. Thus, we can ignore the remainder and write the aymptote as heureka has done !!!!

CPhill Jun 19, 2014

#1**+6 **

y=x^2+6x-3/x-3 horizontal asymptote?

$$y=\dfrac{x^{\textcolor[rgb]{1,0,0}{3}}+6x^2-3x^1-3}{x^\textcolor[rgb]{1,0,0}{1}}$$

degree top = 3

degree botton = 1

horizontal asymptote when degree top < degree botton

or degree top = degree botton

**no horizontal asymptote**! ( 3 < 1 no )

heureka Jun 19, 2014

#2**+5 **

I believe that the questioner might have also meant this (but I'm not sure)....... y = (x^2+6x-3)/(x-3)

heureka is correct.....there is no horizontal asymptote in a "high/low" rational function.......(I believe we have a "slant" asymptote, instead)

CPhill Jun 19, 2014

#3**0 **

Thank you Chris and Heureka,

I have wondered how to find 'slant' asumtotes (the proper name of these eludes me)

Can anyone show me?

Melody Jun 19, 2014

#4**+5 **

Actually...I believe the proper term is "oblique asymptote"....here's a page that might help:

http://www.purplemath.com/modules/asymtote4.htm

CPhill Jun 19, 2014

#6**+6 **

y = (x^2+6x-3)/(x-3) 'slant' asumtotes ?

$$y=\dfrac{x^{\textcolor[rgb]{1,0,0}{2}}+6x-3}{x^{\textcolor[rgb]{1,0,0}{1}}-3}$$

degree top = 2

degree botton = 1

'slant' asumtotes when **degree top = degree botton + 1**

**'slant' asumtote yes ! ( 2 = 1 + 1 )**

**$$\begin{array}{cccccc} (x^2&+&6x&-&3)& : (x-3)= \textcolor[rgb]{1,0,0}{x+9}+\dfrac{24}{x-3}\\ -(x^2&-&3x)\\ 0&+&9x&-&3 \\ &-&(9x&-&27)\\ &&0&+&24\\ \end{array}$$**

$$\lim\limits_{ x \to \pm\infty }(\frac{24}{x-3})=0$$

**'slant' asumtote** y = x + 9

heureka Jun 19, 2014

#7**+5 **

Best Answer

Thanks, heureka, for showing how to find that oblique asymptote....I was going to go through it but I was working on another problem......besides......I couldn't have presented it better, anyway!!!

I want to point out one more thing.....the remainder of (24)/(x-3) means that we actually don't have a perfectly linear asymptote when x is relatively "small." However, as heureka shows by limit analysis, this "fiddly" asymptotic behavior disappears as x aproaches "large" positive and negative values. Thus, we can ignore the remainder and write the aymptote as heureka has done !!!!

CPhill Jun 19, 2014