Given a quadratic satisfy the conditions of the limit

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SUMMARY

The discussion centers on solving the quadratic equation ax^2 + bx + c under specific limit conditions: f(0) = 0, lim (x->-1) f(x) = 3, and lim (x->2) f(x) = 6. The values derived are a = 2, b = -1, and c = 0, fulfilling all conditions. The solution process involves substituting values into the limits and expressing variables in terms of one another to find the coefficients accurately.

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Plutonium88
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Homework Statement


Determine values of a, b, c in the formula ax^2+bx^2 +c that satisfy the conditions:
f(0)=0 // Limx->-1 F(x)=3 // limx->2 f(x)=6


The Attempt at a Solution



1.
F(0)=0 therefore x=0

so f(0)=a(0)^2+b(0)+c
so f(0)=c = 0
so c=0

2.
Lim f(x) = 3, x->-1

so f(x)=ax^2+bx+c
3 = a(-1)^2+(-b) + 0
so a= 3+b or b= a-3

3.
f(x)=6 x->2

6 = ax^2 + bx + c
6 = 4a + 2b
sub in a.
6=4(3+b) +2b
-6 = 6b
b=-1

now do the same but sub in b. to solve for a
6=4a+2b
6=4a + 2(a-3)
6=4a + 2a -6
12=6a
a=2

So therefore values are a=2, b=-1 and c=0 to satisfy those conditons.

I believe this is the correct way to go about it, but i just wanted some one to check for me if possible, and also i was curious is it because you are expressing a in terms of be and b in terms of a when taking the limit, and substituting that value into the next limit restriction 6=4a+2b that it makes it follow all of the previous conditions... (ie: c=0, b=a-3 a=3+b)

I was just curious if some one could give me some more understanding if this is correct.
 
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Your solution is correct, but after getting b=-1, you get a=3+b=2 directly.


ehild
 
ehild said:
Your solution is correct, but after getting b=-1, you get a=3+b=2 directly.


ehild

thank you sir.
 

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