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2009 AMC 12A Problems/Problem 9

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Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See Also

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 (Problems • Resources)
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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2009_AMC_12A_Problems/Problem_9"

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

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