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quelqu'un pourrait-il m'aider?(prob de mathématique)

Messagepar -Prada » Mer Oct 07, 2009 7:47 pm

Bonjour, si quelqu'un comprend comment trouver la règle de cette fonction racine carrée aidez-moi S.V.P!

je connais seulement 2 points de la courbe, soit : (0,0),(36,48 ) :shock:

c'est une fonction sous la forme f(x)=a√(x-h)+k

Le sommet est réellement (0,0) mais il faut absolument que je le prouve... Je n'ai pas le paramètre '' a'' non plus, j'ai vraiment juste 2 points et là je suis réellement au désespoir...
Peu importe de quelle manière je m'y prend, j'obtiens toujours une inéquation à 3 inconnus et je ne peux pas aller plus loin puisque j'ai seulement 2 points dans ma courbe.

Est-ce que par miracle il y aurait quelqu'un de bon en maths parmis les membres ?..
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Messagepar alex12 » Mer Oct 07, 2009 7:49 pm

Ouahhh...
C'est du quel niveau ça ? :lol:
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Messagepar -Prada » Mer Oct 07, 2009 7:53 pm

Sec 5, maths sciences naturelles :roll:
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Messagepar astrolus » Mer Oct 07, 2009 8:07 pm

Bonjour,

C'est racine de (x-h) (et + k en dehors de la racine) ou racine de ((x-h)+k) (tout dans la racine) ?
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Messagepar Mushee » Mer Oct 07, 2009 8:09 pm

J'ai oublié ce que signifient a, h et k?
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Messagepar Mushee » Mer Oct 07, 2009 8:12 pm

As-tu une image de la courbe?
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Messagepar -Prada » Mer Oct 07, 2009 8:13 pm

heum c'est la racine carré de (x-h), le K est en dehors.

Mushee, h et k sont les coordonnées du sommet de la courbe. A c'est le taux de variation

oui j'ai une image, je vais prendre une photo...
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Messagepar astrolus » Mer Oct 07, 2009 8:13 pm

la courbe passe par (0,0) donc f(0) = 0
racine(-h.k) = 0 donc h.k = 0 donc soit h = 0 soit k = 0.

Ensuite f(36) = 48
racine((36-h)+k) = 48

Si h = 0
racine(36.k) = 48
k = 64

si k = 0
racine(36 - h) = 48
h = -12

Il y a deux solutions :
h = 0 et k = 64
ou
h = -12 et k = 0
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Messagepar alex12 » Mer Oct 07, 2009 8:14 pm

Ouahhhhh Content d'être en CST ! :lol: :lol:
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Messagepar astrolus » Mer Oct 07, 2009 8:14 pm

ooops j'ai écrit avant d'avoir ta réponse.

Si k est en-dehors de la racine, alors ce que j'ai écrit ne s'applique pas.

Désolée !

PS : en plus j'avais pas vu le "a " !!!!
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Messagepar -Prada » Mer Oct 07, 2009 8:17 pm

Lol :lol: l'image s'en vient là.. je connecte ma calculatrice sur l'ordi et je vais l'avoir.
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Messagepar -Prada » Mer Oct 07, 2009 8:25 pm

Image

voilà :)
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Messagepar Mushee » Mer Oct 07, 2009 8:27 pm

Ca serait pas simplement f(x)= 8√x ?
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Messagepar -Prada » Mer Oct 07, 2009 8:28 pm

oui oui c'est ça, le problème c'est qu'il faut que je prouve que le sommet est bien (0,0) même si je le sais... :?
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Messagepar Mushee » Mer Oct 07, 2009 8:34 pm

Pour prouver que (x,y) = (0,0), il suffit pas de prendre l'équation, remplacer y par 0 et résoudre, puis remplacer x par 0 puis résoudre.. et si les 2 solutions te donnent 0, ca prouve que ton sommet et bel et bien (0,0) ?
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