+101 The graph of \(\frac{4(x+1)}{x(x-4)}\) has a x intercept at -1 (plugging in 0 for y agrees with this and the graph does). Yet since the bottom power is greater than the top power, shouldn't the horizontal asymptote be zero and not allow an x intercept?
Could you explain why??
do x intercepts just not follow the horizontal asymptote rule???
Here is the function's graph: https://www.desmos.com/calculator/ullbutopnh
-Thanks
+118687 You have not quite graphed the given equation ChowMein.
dom6527,
You graph is correct.
AND The y value does approach 0 as \(x\rightarrow \pm \infty\)
However, you cannot divide by 0 which means that x cannot equal 0 or +4
There are vertical asymptotes when x =0 and when x=4
If -1
\(y=\frac{4(x+1)}{x(x-4)}\\ y=\frac{4(positive)}{neg*(negative)}\\ y=positive\)
BUT
if x<-1 then
\(If\;\; x<-1\\ y=\frac{4(x+1)}{x(x-4)}\\ y=\frac{4(negative)}{neg*(negative)}\\ y=negative\)
So the x axis has to be crossed at x=-1
I hope this helps but....
You probably need to think about it a lot.
+154 A horizontal asymptote is the line that a function approaches as x goes to infinity or negative infinity. So although the function crosses the x axis at (-1, 0), it's not what the function approaches as x goes to infinity or negative infinity. It's why something like this could happen. 
+118687 You have not quite graphed the given equation ChowMein.
dom6527,
You graph is correct.
AND The y value does approach 0 as \(x\rightarrow \pm \infty\)
However, you cannot divide by 0 which means that x cannot equal 0 or +4
There are vertical asymptotes when x =0 and when x=4
If -1
\(y=\frac{4(x+1)}{x(x-4)}\\ y=\frac{4(positive)}{neg*(negative)}\\ y=positive\)
BUT
if x<-1 then
\(If\;\; x<-1\\ y=\frac{4(x+1)}{x(x-4)}\\ y=\frac{4(negative)}{neg*(negative)}\\ y=negative\)
So the x axis has to be crossed at x=-1
I hope this helps but....
You probably need to think about it a lot.
