Need Help with knowing when a graph approaches an asymptote from ABOVE or BELOW (1 Viewer)

[blackops23]

Hi guys, this is the question I had to do:

Q. Draw careful sketches of the following, clearly showing asymptotes. DO NOT USE CALCULUS.

(g). y= (2x^2 + 4x + 3)/(x^2 - 1)

So for this question, there's an asymptote at y=2, and x=1 and x=-1, no x-intercepts, and passes through (0,-3).

The guide graph shows that when x< -1, graph IS ABOVE x-axis
For -1<(x)<1 graph is BELOW X-AXIS
For x> 1, graph is ABOVE X-AXIS
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So it seemed like a fairly simple graph with the graph being greater than 2, at x<1 and x>1 and below 0 for -1<(x)<1

So I drew the graph approaching the asymptotes FROM ABOVE y=2.
However to my surprise, for x<1, the graph cut y=2 and approached the asymptote from below as x--> -inf. (so there was a minima at (-2,1))
(If anyone has graphing software, sketch the graph to know what I mean)...

So how am I supposed to know whether a graph cuts the asymptotes and approaches it from the other side? Normally graphs don't act that weirdly, they approach the asymptotes like normal graphs. Is there any special feature about these particular graphs that make it cross the asymptote and approach it from the other side??

Or should I just sub in 10^10 and -10^10 for every single graph I come across?

Advice would be immensely appreciated.

 

[cutemouse]

So it seemed like a fairly simple graph with the graph being greater than 2, at x<1 and x>1 and below 0 for -1<(x)<1

Maple seems to show these properties...

 

[khfreakau]

So I drew the graph approaching the asymptotes FROM ABOVE y=2.
However to my surprise, for x<1, the graph cut y=2 and approached the asymptote from below as x--> -inf. (so there was a minima at (-2,1))
(If anyone has graphing software, sketch the graph to know what I mean)...

I'm not entirely sure what you meant here, I got completely lost...

But in answer to your question, to find out where the graph cuts the asymptotes for equations in the form P(x)/A(x), you should polynomially divide them through to split them into two parts like so:
y=Q(x)+R(x)/A(x), which, for this question, is
y=2 + (4x+5)/(x^2-1)

Q(x) is the horizontal or oblique asymptote. Solve A(x)=0 to find vertical asymptotes and solve R(x)=0 to find the POINTS AT WHICH the curve cuts the horizontal/oblique asymptote. In this instance, x=-5/4, which makes sense, since as you approach the asymptote, it cuts it at -5/4, and has a minimum at (-2,1), at which point it begins to curve back towards the asymptote.

Hope I cleared that up for you.

 

[blackops23]

and solve r(x)=0 to find the points at which the curve cuts the horizontal/oblique asymptote. In this instance, x=-5/4, which makes sense, since as you approach the asymptote, it cuts it at -5/4, and has a minimum at (-2,1), at which point it begins to curve back towards the asymptote.

Hope i cleared that up for you.

omg, thanks man!