matHs probLem
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12 Feb 2008, 19:11
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It's a parable 3rd degrees/3rd class (dunno how its called but it's f(x)=ax³+bx²+cx+d oki ?
Well it's about that pic:
U've got 2 points there: P1(1|2) ; P2(-1|6)
How can u calculate the parable's equation ?
pLz help =/
Edit: That's just a part of the whole parable. The rest has been removed. U ought to calculate the whole parable u know .
Edit²: The equation is f(x) = x³ – 3x + 4 btw. I sitll dont know how to do it...
Edit³: Alright alright, I finally got the solution after 2 hours of mental activity.
First of all u've got 2 equations through those 2 points:
#1 ~ a + b + c + d = 2
#2 ~ -a +b - c + d = 6
The two points, viewable on the graph, are turning points.
That means that f'(1) = 0 & f'(-1) = 0 because the tangents on those two turning points got the gradients 0.
Now u've got the 3rd and 4th equations.
#3 ~ 3a + 2b + c = 0
#4 ~ 3a - 2b + c = 0
» f(x)=x³ - 3x + 4
tHx for ur help guys.
hoPe u've learned sth aswell :P
bb
Well it's about that pic:
U've got 2 points there: P1(1|2) ; P2(-1|6)
How can u calculate the parable's equation ?
pLz help =/
Edit: That's just a part of the whole parable. The rest has been removed. U ought to calculate the whole parable u know .
Edit²: The equation is f(x) = x³ – 3x + 4 btw. I sitll dont know how to do it...
Edit³: Alright alright, I finally got the solution after 2 hours of mental activity.
First of all u've got 2 equations through those 2 points:
#1 ~ a + b + c + d = 2
#2 ~ -a +b - c + d = 6
The two points, viewable on the graph, are turning points.
That means that f'(1) = 0 & f'(-1) = 0 because the tangents on those two turning points got the gradients 0.
Now u've got the 3rd and 4th equations.
#3 ~ 3a + 2b + c = 0
#4 ~ 3a - 2b + c = 0
» f(x)=x³ - 3x + 4
tHx for ur help guys.
hoPe u've learned sth aswell :P
bb
Don't make me laugh about ur stupidity pLz -_-
Sry if u haven't meant it that way.
tHx for ur try anyway.
U can't just insert the values by random.
Sry but I'm a bit hackled up atm.
Try to insert the x-values and show me what ur solution is and I will show u what's wrong.
f'(x) = 3ax² + 2bx + c
Insert the values you have and you got a system of 4 equations, very easy to solve.
f'(x) has to be 0 and not any y-value .
f'(x) is 0 for x=1 and x=-1...
But the value for f'(1) would have been 2, but it's 0.
f'(x) = 3ax² + 2bx + c
Insert the values you have and you got a system of 4 equations, very easy to solve.
READ.
I could easily get it out if I would have 4 points but I just have 2. I could even get it out if I would have 3 (f'(x)) but I just have 2.
And I could get it out with 2 points if I would have known that its point symetric, but it's not ffs =/
What do u mean with f'(-1) = f'(1) = 0 ?
then the derivative is f'(x) = 3ax^2 + 2bx + c
Since the points on the drawing appears as local maximum and local minimum, then you can assume that f'(-1) = 0 and f(1) = 0
together with f(-1) = 6 and f(1) = 2 that gives you 4 equations.
for the first one, that's
3a(-1)^2 + 2b(-1)^2 + c = 0
do the same with the rest and use gauss or substition to get the values for a, b, c and d.
I wanna know if ur way is correct before I'll try to bang it into my brain.
The solution is f(x) = x³ – 3x + 4 btw
>> B = [0;0;6;2];
>> A\B
ans =
1.0000
0
-3.0000
4.0000
thus a = 1, b = 0, c = -3 and d = 4
ez pz
3 -2 1 0 | 0
3 2 1 0 | 0
-1 1 -1 1 | 6
1 1 1 1 | 2
If you read the first line, it means 3a - 2b + 1c + 0d = 0
if you subtract the first and second equation, you get the value of b.
then you can add the third and fourth equation to get d.
I won't help you more than this.
I just didn't understand ur notation.
>> A = [3 -2 1 0;3 2 1 0;-1 1 -1 1;1 1 1 1];
>> B = [0;0;6;2];
That's new for me =)
tHx for ur help.
if i stil lremember from year a go you do that;
you get M first; 6-1 = 5
...................2-1 = 1
= 5
then;
Y-6=5(x+1)
y-6=5x+5/+6
y=5x+11 -> which is the answer :P
the equation is right, not sure if it's for this parabola ;/
tHx for ur help anyway =)
The excersise is to get the whole thing out with those 2 points u know.
but ive gotta eat, i might try and solve it later
Ich wollt wirklich nur wissen auf welcher Schule, also auf welche Schule du gehst.
und das als türke surprise :o ^^
kk dann passts ja
Bin grad WirtschaftGymnasium 11. =)