Question

Find where the function
\[\f(x)=\frac{x^2+4x}{x^2-16}\] crosses the horizontal asymptote.

Answer 1

if it does...

Answer 2

lol if there is any asymptote no line can touch it it can approach it but not touch it

Answer 3

and this has a horizantal asymptote of 1

Answer 4

Where does the graph cross the horizontal asymptote, if it does?

Answer 5

It doesnt cross the horizantal asymptote my friend

Answer 6

Let's simplify it
\[f(x)=\frac{x^2+4x}{x^2-16}=\frac{x(x+4)}{(x+4)(x-4)}\]
So we get now
\[f(x)=\frac{x}{x-4}\]
we can see that when \(x\to 4, y \to \infty\)
so x=4 is a vertical asymptote

Answer 7

@ash2326 he is asking if the graph crosses the asymptote or not

Answer 8

she... and how do I know that it doesn't cross the horizontal asymptote?

Answer 9

oops , sorry :( :(

Answer 10

try and solve\[\frac{x^2+4x}{x^2-16}=1\]

Answer 11

x=-4

Answer 12

that is not a solution though

Answer 13

could have sworn we did one just like this right?

Answer 14

no @satellite73, at least I don't think so... we were trying to get down definition and I still didn't understand

Answer 15

So it doesn't cross the horizontal asymptote when it's not a solution... if we get imaginary solution and when their is no solution? I think I said the same thing twice... but idk, I'm so confused :'(

Answer 16

zarkon gave it, same as last one
set it equal to zero and solve

Answer 17

if there is no solution, then no worries, it simply does not cross

Answer 18

You mean set it equal to the horizontal asymptote and solve?

Answer 19

exactly
horizontal asymptote in this case is \(y=1\) so set it equal to one and try and solve

Answer 20

Sigh, okay... Thanks so much everyone for all your help. I really do appreciate it...!!!

Answer 21

just as zarkon wrote above, and also the same as in your last question
\[\frac{x^2+4x}{x^2-16}=1\]

Answer 22

in this case there will be no solution, in the last case we got \(x=\frac{1}{2}\)

Answer 23

okay :).

Answer 24

\[f(x)=\frac{x^2+4x}{x^2-16}\]
as\(x\to \infty, y\to 1\)
y=1 is the horizontal asymptote, if you try to find x for which y=1 , you'd get x=-4, but that can't be a solution. Here no solution exist