# MitPY 11: Asymptotes and Wolfram Alpha

This MitPY comes from frequent commenter, John Friend:

Dear colleagues,

I figured this was as good place as any to ask for help. I’m writing a small test on rational functions. One of my questions asks students to consider the function where and to find the values of for which the function intersects its oblique asymptote.

The oblique asymptote is so they must first solve … (1)

for . The solution is and there are no restrictions along the way to getting this solution that I can see. So obviously .

It can also be seen that if then equation (1) becomes which has no solution. So obviously .

When I solve equation (1) using Wolfram Alpha the result is also . But here’s where I’m puzzled:

Wolfram Alpha gives the obvious restriction but also the restriction . emerges naturally (and uniquely) from this second restriction and I really like that this happens as a natural part of the solution process. BUT ….

I cannot see where this second restriction comes from in the process of solving equation (1)! Can anyone see what I cannot?

Thanks.

## 5 Replies to “MitPY 11: Asymptotes and Wolfram Alpha”

1. Grant Cairns says:

The solution is fine as long as it doesn’t happen to equal one of the points where isn’t defined; namely, . A little exploration shows that this happens when .

2. Anonymous says:

5a^3 + 4a^2 + a = 0, when solved over the reals, gives only a = 0.

If you type this into wolfram alpha: “(x^3 + x)/(x^2 + ax – 2a) = x – a, reals” , you won’t get the fancy second restriction.

As for why 5a^3 + 4a^2 + a =/= 0 over the complex numbers, when you sub x=2a^2 / (a+1)^2 into x^2 + ax – 2a, you get an expression that has the same complex zeros as 5a^3 + 4a^2 + a.

Wolfram alpha presumably converts a complex (pun intended) restriction into a polynomial where possible, so as not to have something like “(4 a^4)/(a + 1)^4 + (2 a^3)/(a + 1)^2 – 2 a =/= 0”.

1. John Friend says:

Thanks very much GC and Anonymous. Much appreciated. I’d substituted x into all of f(x), not just the denominator and after simplifying a got something different. Of course, now that it’s been pointed out, it’s obvious to simply require the denominator of f(x) to not be zero.

Thanks again.

3. Steve R says:

Hi,

GC and Anonymous have highlighted the shortcomings of Mathematica .

IMO it would not hurt to add a warning that 5a^2 + 4a + 1 = 0 has no real solutions as discriminant is negative

Steve R

1. John Friend says:

Thanks, Steve. I agree. I think Wolfram Alpha has assumed x and a are complex (this is the default assumption in Mathematica). Things look different when I specify reals:

https://www.wolframalpha.com/input/?i=Solve+%28x%5E3+%2B+x%29%2F%28x%5E2+%2B+a*x+-+2a%29+%3D+x-a+where+x+is+a+real+number

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