Graphing rational functions

[All†Fiction]

Hey guys,

Say, I were to sketch the graph of y=(x^3+2)/x, there would be a vertical asymptote with equation x=0, and a non-vertical asymptote with equation y=x^2. But as x approaches 0, doesn't the graph approach the asymptote y=2/x rather than x=0? Would it be necessary to draw this asymptote in the final sketch, and if so, over what domain (because the graph quickly leaves this asymptote as x gets bigger, tending to y=x^2 instead)?

Thanks in advance! 

[b^3]

You only get an asymptote when your function approaches a curve (or line) as it goes off to infinity (or negative infinity), excluding vertical asymptotes which are a different matter. As our function doesn't approach and so is not an asymptote for this function. But we still have the vertical asymptote.

But you will still need to do a light sketch of and to be able to draw the function in some cases.

https://www.desmos.com/calculator/ot3brtc46t

[lzxnl]

But you will still need to do a light sketch of and to be able to draw the function in some cases.

https://www.desmos.com/calculator/ot3brtc46t

For spesh, I disagree. I find that it is possible to determine the shape of the graph solely by looking at the function and its first derivative (second derivative may be needed for classifying stat points)

For y=x^2 + 2/x
y'=2x - 2/x^2
y"= 2 + 4/x^3
y'=0 when x=1 and y"> 0 here => local min
Also you have the asymptotes y=x^2 and x=0 (I think b^3 misread the function here; as x gets bigger so does y)
Now we have a 2/x term which dominates when the magnitude of x is small. You can see that where x is small and positive, the graph starts large and drops to the local minimum. Then it increases to approach the curved asymptote from above.
If x<0, initially y is very negative. There are no stat points in this part of the domain and the first derivative is continuous here (so no funny cusps like y=abs x) so the graph must be monotonically increasing or decreasing. When x is very negative, the x^2 term becomes a large positive number and the graph approaches that of y=x^2. As y=x^2 + 2/x, the 2/x term will be a tiny negative number, so the graph is always slightly under the asymptote. After around the x intercept at where x^3+2=0, the graph just keeps dropping and dropping.

If you sketch the graph on CAS, you'll find that the graph would have the features I described.  The point is, with some oractice you can work out the major features of graphs like asymptotes and shape just through logical reasoning.

[b^3]

For spesh, I disagree. I find that it is possible to determine the shape of the graph solely by looking at the function and its first derivative (second derivative may be needed for classifying stat points)

For y=x^2 + 2/x
y'=2x - 2/x^2
y"= 2 + 4/x^3
y'=0 when x=1 and y"> 0 here => local min
Also you have the asymptotes y=x^2 and x=0 (I think b^3 misread the function here; as x gets bigger so does y)
Now we have a 2/x term which dominates when the magnitude of x is small. You can see that where x is small and positive, the graph starts large and drops to the local minimum. Then it increases to approach the curved asymptote from above.
If x<0, initially y is very negative. There are no stat points in this part of the domain and the first derivative is continuous here (so no funny cusps like y=abs x) so the graph must be monotonically increasing or decreasing. When x is very negative, the x^2 term becomes a large positive number and the graph approaches that of y=x^2. As y=x^2 + 2/x, the 2/x term will be a tiny negative number, so the graph is always slightly under the asymptote. After around the x intercept at where x^3+2=0, the graph just keeps dropping and dropping.

If you sketch the graph on CAS, you'll find that the graph would have the features I described.  The point is, with some oractice you can work out the major features of graphs like asymptotes and shape just through logical reasoning.

Well yeah you can do it that way, it's just that I find visualising it quicker than having to go through all that effort above (although will come in handy if you need a particular location of a stationary point). You don't really have to draw them in if you can visualise it in your head (which comes with a little practice).

What I'm getting at is why put all this effort into working it out that way when you can get the same result without missing important details using a different method? (Although it may be because I'm a visual person that I find it easier, pretty much stick to the method that you find it easier, and are able to work out where the stationary point will lie, closer to one curve or the other, and note I did also say "some cases").

The second point to make is sometimes you won't be give the function but the shape of the function (e.g. multiple choice questions) and from that determine the shape of a function .

Also on the , yeah I misread that part. Fixed now.

[All†Fiction]

Thanks so much guys, that helped TONS!