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avatar+90 

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

dom6547  Mar 29, 2018

Best Answer 

 #2
avatar+94106 
+4

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.

 

 

Melody  Mar 29, 2018
edited by Melody  Mar 29, 2018
 #1
avatar+84 
+1

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. 

ChowMein  Mar 29, 2018
edited by Guest  Mar 29, 2018
 #2
avatar+94106 
+4
Best Answer

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.

 

 

Melody  Mar 29, 2018
edited by Melody  Mar 29, 2018
 #3
avatar+90 
0

Thank you very much I will think on this hard as suggested!

 

p.s. it's dom6547 not dom6527!

dom6547  Mar 29, 2018

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