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> | ||
− | + | == Solution 1 == | |
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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>. | ||
− | + | == 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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 03:53, 21 July 2022
Contents
[hide]Problem
Suppose that and . What is ?
Solution 1
As , we have .
To compute , set in the first formula. We get .
Solution 2
Combining the two formulas, we know that .
We can rearrange the right hand side to .
Comparing coefficients we have , , and . From the second equation we get , and then from the third we get . Hence .
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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