Question

how do you find the horizontal asymptote of an equation

Answer 1

what's the equation?

Answer 2

Take the limit as \(x\) goes to infinity.

Answer 3

yes ^ and divide by the highest power of x

Answer 4

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

Answer 5

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

Answer 6

how do you know

Answer 7

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

Answer 8

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.

Answer 9

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

Answer 10

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

Answer 11

so is anyone gonna respond orrrr

Answer 12

nvm gnight people

Answer 13

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

Answer 14

k thankss