# Finding the asymptotes

1. Sep 29, 2008

### Mentallic

1. The problem statement, all variables and given/known data
I am unsure how to find the asymptotes and thus, graph the function of -

$$f(x)=\frac{x+4}{x^{2}-4}$$

2. Relevant equations
Such functions that have the same degree polynomial in both the numerator and denominator can be simplified to change the numerator from a variable in x, to a constant which helps to find the asymptotes.

e.g.

$$y=\frac{x-2}{x+5}$$

$$y=\frac{(x+5)-7}{x+5}$$

$$y=1-\frac{7}{x+5}$$

Therefore it can be seen that there is an asymptote at y=1 - since the fraction $$\neq$$ 0 - and another at x=-5 since the denominator of the fraction $$\neq$$ 0

3. The attempt at a solution

The first asymptote is obvious, x=$$\pm$$2 since the denominator $$\neq$$ 0

Then I split the function into a similar form as shown with the e.g.

$$f(x)=\frac{(x+2)+2}{(x+2)(x-2)}$$

$$f(x)=\frac{1}{x-2}+\frac{2}{(x+2)(x-2)}$$

but from here I'm unsure how to find the y asymptotes. Could someone please show me the solution, or point me in the right direction?

2. Sep 29, 2008

### Defennder

Well first you need to be more specific as to what you mean by "asymptote". You want both the horizontal and vertical asymptotes and the means by which both are obtained are quite different.

As you have said, the vertical asymptotes are $$x = \pm 2$$. As for the horizontal asymptotes, note how the fraction is expressed. How is the degree of the numerator polynomial and that of the denominator polynomial related to the horizontal asymptote?

3. Sep 29, 2008

### Mentallic

Where the horizontal function approaches as $$x\rightarrow\pm\infty$$? Isn't this the definition of an asymptote? A line which cannot be cut. As for horizontal asymptotes, they occur because a fractional part in the function cannot = 0, thus there cannot be a y value there.

I think the vertical asymptotes are obtained in the same way but as for the horizontal asymptotes, I'm not too sure.

Sorry, I don't know how these polynomials of different degrees (quadratic and linear) affect the outcome of the horizontal asymptotes. You've already stated that they are obtained by another methods, but this is the only method I know of to tackle this new problem.

The attempt at a solution

Trying to use the same method to find the y asymptote as in my previous example of same degree polynomials in both numerator and denominator :

$$f(x)=\frac{1}{x-2}+\frac{2}{(x+2)(x-2)}$$

Each fractional part $$\neq$$ 0.
The first part, $$\frac{1}{x-2}$$, cannot = 0, but $$\rightarrow$$ 0 as $$x \rightarrow \infty$$. Therefore, this causes the second part $$\frac{2}{(x+2)(x-2)}$$ to approach 0 as well. So as $$x\rightarrow\pm\infty$$ : $$f(x)\rightarrow0$$

Does this mean there is a horizontal asymptote at y=0?

Although, the function crosses the x-axis at -4. Is this still in the form of an asymptote?

4. Sep 29, 2008

### Defennder

The former is for a horiztonal asymptote, the latter is for a vertical asymptote. You have to be clear about which ones you are talking about here.

See above.

You can arrive at this conclusion even without splitting up f(x). Just look at the degree of denominator polynomial and the numerator polynomial. If the degree of the former is greater than the latter, then when x approaches infinity, the bottom part of fraction would dominate and the function would tend to 0. Hence the horizontal asymptote at y = 0.

I don't know what you mean here. You mean instead the "y-axis"?

5. Sep 30, 2008

### Mentallic

Sorry I probably wasn't being very clear here. When I split up the function - $$y=\frac{x-2}{x+5}$$ : $$y=1-\frac{7}{x+5}$$ - The fraction part $$\neq$$ 0 so the range does not exist at y=1. This is how I came to terms with the horizontal asymptote at "that y value".

Yes thats true, thanks for the tip

No I still mean x-axis.

$$f(x)=\frac{x+4}{x^{2}-4}$$

when f(x)=0, x=-4. Therefore, the function intercepts the X-axis at (-4,0) and as $$x\rightarrow-\infty : f(x)\rightarrow0^{-}$$

Now, since the function cuts the x-axis, does the asymptote still exist at y=0? (I thought asymptotes were lines that are never crossed, even though the function approaches infinitely close to them)

Also, I've found another issue which I can't seem to find the answer to:
The range is all reals discluding $$\approx -\frac{4}{5}<f(x)<-\frac{1}{10}$$.
Is it possible to find the exact range?

6. Sep 30, 2008

### Defennder

Oh you're right. Apparently I misread. The x-axis is where y=0

Ok now this is where the definition of asymptote is involved. Does a line count as an asymptote if the function tends to it at infinity but crosses it at an earlier interval? I can't remember of the top of my head, but I'll say that you can consider it as an asymptote for a given interval.

7. Sep 30, 2008

### Mentallic

After a bit of scrounging, it seems that it is still asymptotic
No of course not
I much prefer exact answers to good estimates when theoretical mathematics is involved.

The answer to the asymptotes have been found : $$x=\pm2$$, $$y=0$$.

However, before I put this thread to rest, how could I go about finding the range of the function?

Thanks for all your help Defennder.

8. Sep 30, 2008

### Defennder

Well to find the range, just set the function = r for example. Then re-arrange it into a quadratic equation in x. Note that the quadratic has to satisfy some condition if x is restricted to the reals. Use that condition to find the range.

9. Sep 30, 2008

### Mentallic

I understood what you just said (how odd for me), but hopefully I can put it together -

$$f(x)=r$$

Therefore, $$r=\frac{x+4}{x^{2}-4}$$

$$rx^{2}-4r=x+4$$

$$rx^{2}-x-4r-4=0$$

$$x=\frac{1\pm\sqrt{1-4r(-4r-4)}}{2r}$$

So um...?

EDIT: The condition for the range that does not exist is when the discriminant < 0

Therefore, $$1-4r(-4r-4)<0$$

$$16r^{2}+16r+1<0$$

$$r=\frac{-16\pm\sqrt{16^{2}-4.16}}{2.16}$$

$$r=\frac{-2\pm\sqrt{3}}{4}$$

Since the parabola is concave up (a>0) and r is the roots, it is all values between the x values between the vertex and roots. i.e.

$$\frac{-2-\sqrt{3}}{4}<R<\frac{-2+\sqrt{3}}{4}$$

This looks right to me. Thanks defennder.