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Difference between revisions of "2009 AMC 12A Problems/Problem 9"

(New page: == Problem == Suppose that <math>f(x+3)=3x^2 + 7x + 4</math> and <math>f(x)=ax^2 + bx + c</math>. What is <math>a+b+c</math>? <math>\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 0 \qquad \textbf...)
 
 
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<math>\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3</math>
 
<math>\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3</math>
  
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== Solution ==
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== Solution 1 ==
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=== Solution 1 ===
 
  
 
As <math>f(x)=ax^2 + bx + c</math>, we have <math>f(1)=a\cdot 1^2 + b\cdot 1 + c = a+b+c</math>.  
 
As <math>f(x)=ax^2 + bx + c</math>, we have <math>f(1)=a\cdot 1^2 + b\cdot 1 + c = a+b+c</math>.  
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To compute <math>f(1)</math>, set <math>x=-2</math> in the first formula. We get <math>f(1) = f(-2+3) = 3(-2)^2 + 7(-2) + 4 = 12 - 14 + 4 = \boxed{2}</math>.
 
To compute <math>f(1)</math>, set <math>x=-2</math> in the first formula. We get <math>f(1) = f(-2+3) = 3(-2)^2 + 7(-2) + 4 = 12 - 14 + 4 = \boxed{2}</math>.
  
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=== Solution 2 ===
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== Solution 2 ==
  
 
Combining the two formulas, we know that <math>f(x+3) = a(x+3)^2 + b(x+3) + c</math>.
 
Combining the two formulas, we know that <math>f(x+3) = a(x+3)^2 + b(x+3) + c</math>.
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{{AMC12 box|year=2009|ab=A|num-b=8|num-a=10}}
 
{{AMC12 box|year=2009|ab=A|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 03:53, 21 July 2022

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 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 • Answer Key • 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
All AMC 12 Problems and Solutions

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