Difference between revisions of "2009 AMC 12A Problems/Problem 9"
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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>. | ||
Latest revision as of 04:53, 21 July 2022
Contents
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
