OpenStudy (anonymous):

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

OpenStudy (anonymous):

if it does...

OpenStudy (anonymous):

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

OpenStudy (anonymous):

and this has a horizantal asymptote of 1

OpenStudy (anonymous):

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

OpenStudy (anonymous):

It doesnt cross the horizantal asymptote my friend

OpenStudy (ash2326):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (ash2326):

oops , sorry :( :(

OpenStudy (zarkon):

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

OpenStudy (anonymous):

x=-4

OpenStudy (zarkon):

that is not a solution though

OpenStudy (anonymous):

could have sworn we did one just like this right?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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 :'(

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

okay :).

OpenStudy (ash2326):

$f(x)=\frac{x^2+4x}{x^2-16}$ 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