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 !!!!
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 )
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)
Thank you Chris and Heureka,
I have wondered how to find 'slant' asumtotes (the proper name of these eludes me)
Can anyone show me?
Actually...I believe the proper term is "oblique asymptote"....here's a page that might help:
http://www.purplemath.com/modules/asymtote4.htm
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
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 !!!!