Question from Patel's textbook... (1 Viewer)

[aud]

Page 10 -
y = x/(x+2)(x-1)

The graph's drawn, but two of the questions are:
Why is the RIGHT (TOP) branch approaching y = 1 from below?
Why is the LEFT (TOP) branch approaching y = 1 from above?

I get the second on, but not the first one... why does it cut through the asymptote?

 

[Giant Lobster]

oblique? i dont see any on that graph

i cant explain "why" but if u find the derivative, u will find 2 stationary points. one at x = 0, one at x = 4. the one at x = 4 is a minimum point, so obviously it must approach x-> infinite from the bottom.

 

[kimmeh]

its something to do with the limits.. my idiot froiend tok my 4 unit book full of curve sketching with all my notes i'll post the reason why when i find it. it has something to do with as x -> positive infinty , then y -> 1- (negative being under) .. not too sure

 

[turtle_2468]

see kimmeh...

It's to do with limits as x -> infinity.
y=(x^2)/(x^2-x+2)=1/(1+(1/x)-(2/x^2)) on dividing by the highest factor.

Now observe that 1/x-2/x^2 has roots 0,2. Also, from that it is less than 0 for x < -2 and > 0 for x > 2.

As x goes to infinity, the 2/x^2 term becomes less important than 1/x (only dividing by a really big number once rather than twice). So as x -> pos inf, the denominator is larger than zero hence approaching from above
but as x -> neg inf, denom less than zero (as 1/x < 0) so approaching y=1 from below.

The reason it "cuts the asymptote" is that you're mixing terms Technically, an asymptote is just a line that the graph never cuts. So while the graph tends to that line on both sides it also crosses it...

For another one of these graphs that cut the line of the limit going to inifinity (y=1 in this case), think about sin(x)/x.

PS I think markers wouldn't really mind if you named it an asymptote... check with your teacher, I might even be wrong with the defn...