2009 AMC 12A Problems/Problem 9

Problem

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

$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3$

Solution

Solution 1

As $f(x)=ax^2 + bx + c$, we have $f(1)=a\cdot 1^2 + b\cdot 1 + c = a+b+c$.

To compute $f(1)$, set $x=-2$ in the first formula. We get $f(1) = f(-2+3) = 3(-2)^2 + 7(-2) + 4 = 12 - 14 + 4 = \boxed{2}$.

Solution 2

Combining the two formulas, we know that $f(x+3) = a(x+3)^2 + b(x+3) + c$.

We can rearrange the right hand side to $ax^2 + (6a+b)x + (9a+3b+c)$.

Comparing coefficients we have $a=3$, $6a+b=7$, and $9a+3b+c = 4$. From the second equation we get $b=-11$, and then from the third we get $c=10$. Hence $a+b+c = 3-11+10 = \boxed{2}$.

See Also

2009 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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