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

Guest Jun 19, 2014

Best Answer 

 #7
avatar+78754 
+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
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7+0 Answers

 #1
avatar+18715 
+5

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
avatar+78754 
+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
avatar+91051 
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
avatar+78754 
+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
 #5
avatar+91051 
0

Thanks Chris

Melody  Jun 19, 2014
 #6
avatar+18715 
+5

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
avatar+78754 
+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

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