+295 let f(x)=(3x2+2x+14)/(1-5x2). The horizontal asymptote is given by y=?
round to 1 decimal place
+118608 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. :)
+128406 The hrizontal asymptote occurs at 3 / -5 = -3 / 5 = -0.6
Here's a graph : https://www.desmos.com/calculator/8mj0jllvgy
+128406 Hey......gimme' a break.....I was answering questions for 4 straight hours....!!!!...I can't give "expanded" explanations to every single one....!!!!!
+128406 LOL!!!.....it looks good....but.......you don't happen to have something a bit "stronger,' do you???
I think I'm gonna' need it......
+118608 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. :)
