Sketching graphs of rational functions

[suskieanna]

Hello I am trying to do my schoolwork and the topic is about sketching graphs of rational functions like and I just have no idea about skeching this. I read the textbook but still have no idea  
Can someone explain me how to sketch this sort of function in detail and in an easy way?
Thanks in advance!

[fun_jirachi]

Okay first note that you can actually write this as

Note that there is an asymptote at x=0.
From there differentiate to find that

Note that this curve has no real stationary points, and also by noting that

that the function is monotonic increasing.
Noting from before that x=0, find the limiting behaviour of the curve as it approaches 0 from both sides, and positive and negative infinity.
You should find that it approaches y=x from beneath on the positive side, and from above on the negative side as it approaches positive and negative infinity respectively. SImilarly, as it approaches zero from the positive side, you get that it tends to negative infinity, while from the negative side it tends to positive infinity. You should get a graph like this (red is the function, blue is y=x (just to show that it is also an asymptote))

Spoiler

Hope this helps

[suskieanna]

Thanks a lot for helping!! It really helped and now I fully understand it

[suskieanna]

Okay first note that you can actually write this as

Note that there is an asymptote at x=0.
From there differentiate to find that

Note that this curve has no real stationary points, and also by noting that

that the function is monotonic increasing.
Noting from before that x=0, find the limiting behaviour of the curve as it approaches 0 from both sides, and positive and negative infinity.
You should find that it approaches y=x from beneath on the positive side, and from above on the negative side as it approaches positive and negative infinity respectively. SImilarly, as it approaches zero from the positive side, you get that it tends to negative infinity, while from the negative side it tends to positive infinity. You should get a graph like this (red is the function, blue is y=x (just to show that it is also an asymptote))

Spoiler

Hope this helps

Hello again, this time I am trying to plot and trying to find the equations of all asymptotes. I found that there's no vertical asymptotes. But I am not sure how to find horizontal asymptote. Do I just manually sub in values? It is too much work by hand... Is there any simpler way? Thanks in advance.

[fun_jirachi]

I'm not too sure on this one, but you know the rough shape right? Sorta like a sennheiser S near the origin, and at the tips of the S is a T.P/ inflexion point, where it then flattens out. I'm pretty sure that you find the limiting nature of the curve as it approaches infinity (positive and negative), because from the methods I've learned there aren't really any better ways to do this, since it only approaches the asymptote as it approaches really big or really small values of x. Subbing in values in general works, but you've got to sub in the right values

Hope this helps

[AlphaZero]

Hello again, this time I am trying to plot and trying to find the equations of all asymptotes. I found that there's no vertical asymptotes. But I am not sure how to find horizontal asymptote. Do I just manually sub in values? It is too much work by hand... Is there any simpler way? Thanks in advance.

Hey there, sketching this by hand is indeed yucky, but not completely impossible. I'll start you off.

First, let's define
I'm also going to define  \(g(x)=\dfrac{x}{x^2+9}\),  and note that  \(f(x)=1+4\,g\left[2\left(x-\dfrac{1}{2}\right)\right]\).

You should be able to sketch the graph of \(g\) relatively easily! Then just apply the transformations to get the graph of \(f\)

Some extra details you should note:
Indeed, there are no vertical asymptotes since the denominator of \(f(x)\) is never 0. Horizontal asymptotes occur by definition when either of  \(\lim_{x\to\infty}f(x)\)  or  \(\lim_{x\to-\infty}f(x)\)  exist. (In this case, you should find that  \(f(x)\rightarrow 1\)  as  \(x\rightarrow \pm\infty\)). You should also note that  \(g(x)\)  is an odd function, and so you only need to actually sketch half of it.

[suskieanna]

Hello I have a test today on this topic and I am still not sure about graphing. Some questions like has and as the asymptote. But not all questions have both things as the asymptote. For example, only has as its asymptote and not . Is there any particular rule for this? Really urgent I have a test today on period 5. Thanks to anyone who answers my question!

[Bri MT]

Snip

Hey, I didn't do spec & have never been assessed on this so take what a say with a grain of salt. But:

Looking at the first equation, we seperate the fraction out and get
 x^3/x + 2/x  . If we want x^2 we need 2/x to be zero. We can't just sub in a value of x to get that result, we need to take the limit as x approaches infinity. Similarly, if we want 2/x, we want x^2 to be zero. We can't just sub in a value of x ( 2/0 is undefined) so we take the limit as x approaches zero

In the second equation, if we want x-5, then 4/x must be zero. We can't just sub in a value of x to get that result, we need to take the limit as x approaches infinity. To get 4/x, we need x-5 to equal zero. That's simple enough, we'll just sub x=5 in and we're done.

I hope this somewhat helps

[S_R_K]

Hello I have a test today on this topic and I am still not sure about graphing. Some questions like has and as the asymptote. But not all questions have both things as the asymptote. For example, only has as its asymptote and not . Is there any particular rule for this? Really urgent I have a test today on period 5. Thanks to anyone who answers my question!

I realise this might be too late for your test, but in any event:

With respect to the first equation, the only asymptotes are x = 0 and y = x^2. y = 2/x is not an asymptote.

With respect to the second equation the asymptotes are x = 0 and y = x – 5.

In general, to find asymptotes of a rational function, use long division or equating coefficients to express it in the form P(x) + Q(x)/R(x), where P(x) is a polynomial (perhaps of degree 0, ie. just a constant), and where the degree of Q(x) is less than the degree of R(x). When your rational function is in this form, you will have the following asymptotes:

1. P(x) will be an asymptote. This is because as x approaches infinity, Q(x)/R(x) approaches zero. Hence, the only non-vanishing term in the function is P(x).

2. Any values of x for which R(x) = 0 will give a vertical asymptote. This is because the graph is undefined at these values (division by zero), and because Q(x)/R(x) approaches infinity as x approaches these asymptotes.