What is $a+b+c$ for \( f(x+3)=3x^2+7x+4 \) and \( f(x)=ax^2+bx+c \)?

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To find \( a + b + c \) for the function \( f(x) = ax^2 + bx + c \) given \( f(x+3) = 3x^2 + 7x + 4 \), the problem requires substituting \( x+3 \) into the quadratic form and equating coefficients. By expanding \( f(x+3) \) and matching it to \( 3x^2 + 7x + 4 \), the values of \( a \), \( b \), and \( c \) can be determined. The correct values lead to \( a = 3 \), \( b = 1 \), and \( c = -5 \). Thus, \( a + b + c = 3 + 1 - 5 = -1 \. The final answer is -1.
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Suppose that $$f(x+3)=3x^2+7x+4$$ and $$f(x)=ax^2+bx+c$$. What is $a+b+c$?
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Congratulations to the following members for their correct solutions:

1) MarkFL
2) Bacterius
3) caffeinemachine
4) Siron

Solution (from Siron):
We have
$$f(x)=ax^2+bx+c = a(x+3)^2+b(x+3)+c = a(x^2+6x+9)+bx+3b + c = ax^2+6ax+9a + bx+3b + c = ax^2+(6a+b)x+(9a+3b+c)$$

Since $f(x+3)=3x^2+7x+4 = ax^2+(6a+b)+(9a+3b+c)$ we obtain the following system of equations
$$\left \{ \begin{array}{lll} a = 3 \\ 6a+b = 7 \\ 9a+3b+c = 4 \end{array} \right. \Rightarrow \left \{ \begin{array}{lll} a = 3 \\b =-11 \\ c c = 10 \end{array} \right.$$

Hence $a+b+c = 3-11+10 = 2$
 

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