## deepthiakella1 one year ago how do you find the horizontal asymptote of an equation

1. kari.lynee

what's the equation?

2. wio

Take the limit as $$x$$ goes to infinity.

3. kari.lynee

yes ^ and divide by the highest power of x

4. deepthiakella1

its for calc... so its like 4x^3-x^4

5. dumbcow

horizontal asymptote does not exist....f(x) -> neg infinity as x->infinity

6. deepthiakella1

how do you know

7. deepthiakella1

so when i graph it i can still find the relative extrema ?

8. wio

A horizontal asymptote is defined as a line $$y=L$$ such that $\lim_{x\to \infty }f(x)=L$Or$\lim_{x\to -\infty }f(x)=L$It's a definition. You can not argue with a definition.

9. dumbcow

yes there are still local max\min just not any asymptotes

10. deepthiakella1

ohhhh so can i still find the first der. and get the critical number then 2nd der. to find the point of inflection then do the number line thing to find the concavity then label the relative extrema? my only question is how do you find all of the asymptotes and where does the critical number go on the graph

11. deepthiakella1

so is anyone gonna respond orrrr

12. deepthiakella1

nvm gnight people

13. dumbcow

sorry...i think you are confusing local extrema with asymptotes f(x) = 4x^3 - x^4 f'(x) = 12x^2 -4x^3 = 0 --> x = 0,3 (critical points) f''(x) = 24x - 12x^2 = 0 --> x = 0 , 2 (inflection points) x=0 is inflection point x=3 is local max since f''(3) <0 indicating concave down

14. deepthiakella1

k thankss