Tarantulafudge | TRWBW: I'm confused with your answer TRWBW: how do I get it in y=a*x^2 form |

TRWBW | Tarantulafudge: just saying that once you solved this first example, you can solve any other example by rotating and tranlating it to be in the form y=a*x^2 |

Tarantulafudge | TRWBW: oh okay |

Adlai | what's the directrix again? |

Dovetailed | Hey, don't suppose anyone knows a free piece of statistical software which can be used to compute robust correllation matrices? |

demute__ | |Steve|: isn't it possible to do a newline if i use displaymath? |

SeveredCross | Directrix... |

Tarantulafudge | I'm so lost -.- |

SeveredCross | http://en.wikipedia.org/wiki/Directrix |

TRWBW | Tarantulafudge: or start with wikipedia article on conic section even |

|Steve| | demute__: Why don't you tell me what you are really trying to do and I can offer a suggestion for how to go about that. If q(x) is a polynomial over some field and we define the derivative q'(x) in the obvious way without using limits, is there some simple way to consider the gcd(q(x), q'(x))? I wrote out the first iteration of the algorithm, but I don't think it's going to get any simpler. |

Kasadkad | Isn't it just q(x) with the multiplicity of each zero reduced by 1? |

|Steve| | I have no idea. What makes you think that? |

TRWBW | |Steve|: you can show that (qp)'=qp'+q'p |Steve|: so if you have p=(x-r)^kq, then (x-r)^(k-1) divides p' |Steve|: ignoring things like k=0 in the field |

|Steve| | Okay. Do I need to do anything like look at it in the splitting field of the polynomials so that I have linear factors? |

Tarantulafudge | Okay I read some of the wikipedia stuff but it really didn't help much I'm not sure how to find the coordinates of the vertex because my book has an example but in y=ax2+bx+c form |

demute__ | |Steve| using eqnarray instead of displaymath solved the problem. |

|Steve| | Don't use eqnarray. It puts ugly spaces in the equation. Use align (as I suggested above). |

TRWBW | |Steve|: the big idea here is if r|p, then p=r*s, p'=r'*s+r*s'. so r|r'*s, which has smaller degree. which is impossible. so gcd(q,q')=1 |Steve|: impossible by induction |

powerfox | I have z/ (z^2 + z +1) dz If I'm integrating it by parts I have a very strong integral... Maybe I should do something else? |

|Steve| | Don't do it by parts. u = z^2 + z + 1, du/2 = z dz. |

TRWBW | |Steve|, du=(2z+1)dz |

|Steve| | oops! *grumble, grumble* Stupid calculus. *grumble grumble* |

powerfox | Thanks. I got it :) You are saving my stydy task, folks! Thanks! |

Cale | gcd(f,f') is a good way to notice multiple roots of f. |

|Steve| | TRWBW: how did you get gcd(q,q') = 1? |