Question

I need help finding the horizontal asymtopes of these 6 equation... I honestly don't get how to do it... f(x)=2x f(x)=0.5^x+3 f(x)=1/x f(x)=1/x^2 f(x)=(8x+5)/(2x) f(x)= (x^2+2x-1)/(x^2+x-2)

Answer 1

if you want to see where the slope is 0 , just get the derivations, f(x) = 2x so f'(x) = 2, which can't be 0 in the R(I mean the real numbers), about the others,
f(x) = 1/2^x+3, f'(x) = 1/2^x+ln 1/2 = 0 then you solve it for x, the same way for the others.

Answer 2

I think you need to know the concept first

Answer 3

this is the graph of e^x I suppose u know that.
in this case as the function goes to - infinity the graph get closer and closer to 0 ,though it will never reach the value 0.
so we express that in general form as :\[\lim_{x \rightarrow \pm \infty}f(x)=L \] such that L is real number

Answer 4

Is there a way to find them without using "lim"? The question in my book says to "study the graphs and rules of the given function to identify the equations of their horisontal asymtopes."

Answer 5

this is the definition of the horizontal asymptote , but If you allowed to use graphing tool , u can observe where does the function go for very large or very small values of x, and if it infinitily increase or decrease, there will be no HA for f(x)

Answer 6

It can be formed like this too, \[\lim_{x \rightarrow \pm \infty} [f(x)-(mx+b)] = 0\]

Answer 7

ain't that the oblique asymptote @MrWho ?

Answer 8

Thank you so much! This makes a bit of sense now.

Answer 9

you are welcome @kendra13 ,and also if you needed to know how to find it algebraically I'll be around ;)