# Given a quadratic satisfy the conditions of the limit

1. Jul 17, 2012

### Plutonium88

1. The problem statement, all variables and given/known data
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

3. 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 beleive 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.

2. Jul 17, 2012

### ehild

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

ehild

3. Jul 17, 2012

### Plutonium88

thank you sir.