Question

Determine the equations of the asymptotes of the graph for the function f(x)=x/5x+2

Answer 1

va = -2/5

ha = 1/5

Answer 2

assuming that 5x+2 is the denominator....

Answer 3

yes it is , thank you ! could you show me how you achieved that. Im clueless.

Answer 4

va..vertical asymptote...is whatever turns the bottom to zero;

5x+2 =0
5x = -2
x = -2/5 is the va

Answer 5

the ha is found by dividing top and bottom by the highest power of x in the bottom; in this case that is just 'x'

x/x 1
-------- = -------
5x/x + 2/x 5 + (2/x)

Answer 6

with it like this; we see that when x get very large;

1
------------ = .0000...00001 - is very tiny
10000...0000

Answer 7

so 2/x gets close to zero and fades away... in fact, anything with an x left on bottom goes to zero and we are left with:

1/5

Answer 8

ohh right , so to find the the ha you always have to take the variable x into consideration

Answer 9

yes

Answer 10

The Rule for horizontals is that it depends on the degree of the top as compared to the degree of the bottom.

If the degree of the top is smaller, then you have a horizontal asymptote on the x-axis.

If the degrees are the same, then you have a horizontal asymptote on the line y=a/b where a is the leading coefficient on the top and b is the leading coefficient on the bottom.

If the degree of the bottom is smaller, then you don't have a horizontal asymptote - you have an oblique one instead. I teach my students to find those using synthetic division and ignoring the remainder but you can just use regular polynomial long-division

Answer 11

and if you want to read a novel..there it is :)

Answer 12

thank you both for your help !

Answer 13

:) youre welcome.

Answer 14

sorry for butting in - someone told me about this site and I was looking and saw this question. I am a teacher and gave a final exam that had this question on it yesterday so thought I'd try to help. Cool site - I'm going to tell my students about it!