+0  
 
+5
115
9
avatar+295 

let f(x)=(3x2+2x+14)/(1-5x2). The horizontal asymptote is given by y=?

round to 1 decimal place

Deathstroke_rule  Feb 26, 2017

Best Answer 

 #8
avatar+90633 
+5

let f(x)=(3x2+2x+14)/(1-5x2). The horizontal asymptote is given by y=?

round to 1 decimal place

 

\(f(x)=\frac{3x^2+2x+14}{1-5x^2}\\ \text{The horizontal asymptotes is }\\ f(x)=\displaystyle\lim_{x\rightarrow \pm \infty}\frac{3x^2+2x+14}{1-5x^2}\\ f(x)=\;-\;\displaystyle\lim_{x\rightarrow \pm \infty}\frac{3x^2+2x+14}{5x^2-1}\\ \text{Now the numerator and the denominator tend to infinity}\\ \text{so use l'Hopital's theorum}\\ f(x)=\;-\;\displaystyle\lim_{x\rightarrow \pm \infty}\frac{6x+2}{10x}\\ f(x)=\;-\;\displaystyle\lim_{x\rightarrow \pm \infty}(\frac{6}{10}+\frac{2}{10x})\\ f(x)=\;-0.6 \)

 

You can also do this using algebraic division.  It is almost as easy. :)

Melody  Feb 27, 2017
Sort: 

8+0 Answers

 #1
avatar+77149 
+5

The hrizontal asymptote occurs at  3 / -5    =  -3 / 5   =  -0.6

 

Here's a graph : https://www.desmos.com/calculator/8mj0jllvgy

 

 

cool cool cool

CPhill  Feb 26, 2017
 #3
avatar+90633 
+5

There is not much explanation here Chris     frown

Melody  Feb 27, 2017
 #4
avatar+77149 
+5

Hey......gimme' a break.....I was answering questions for 4 straight hours....!!!!...I can't give "expanded" explanations to every single one....!!!!!

 

 

cool cool cool

CPhill  Feb 27, 2017
 #5
avatar+90633 
+5

ok ok keep your hat on!     :D

Here, take a break and drink this.  I will make you feel better :)

 

Melody  Feb 27, 2017
 #6
avatar+77149 
+5

LOL!!!.....it looks good....but.......you don't happen to have something a bit "stronger,' do you???

 

I think I'm gonna' need it......

 

 

cool cool cool

CPhill  Feb 27, 2017
 #7
avatar+90633 
+5

Mmm ok, here is some Turkish coffee,

It should be strong enough!

 

Melody  Feb 27, 2017
 #8
avatar+90633 
+5
Best Answer

let f(x)=(3x2+2x+14)/(1-5x2). The horizontal asymptote is given by y=?

round to 1 decimal place

 

\(f(x)=\frac{3x^2+2x+14}{1-5x^2}\\ \text{The horizontal asymptotes is }\\ f(x)=\displaystyle\lim_{x\rightarrow \pm \infty}\frac{3x^2+2x+14}{1-5x^2}\\ f(x)=\;-\;\displaystyle\lim_{x\rightarrow \pm \infty}\frac{3x^2+2x+14}{5x^2-1}\\ \text{Now the numerator and the denominator tend to infinity}\\ \text{so use l'Hopital's theorum}\\ f(x)=\;-\;\displaystyle\lim_{x\rightarrow \pm \infty}\frac{6x+2}{10x}\\ f(x)=\;-\;\displaystyle\lim_{x\rightarrow \pm \infty}(\frac{6}{10}+\frac{2}{10x})\\ f(x)=\;-0.6 \)

 

You can also do this using algebraic division.  It is almost as easy. :)

Melody  Feb 27, 2017
 #9
avatar+1508 
0

You guys are hilarious!

MysticalJaycat  Feb 27, 2017

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