Difference between revisions of "2003 AMC 12B Problems/Problem 20"
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− | From the graph, we have <math>f(0)=2</math> so <math>d=2</math>. Also from the graph, we have <math>f(1)=a+b+c+2=0</math>. But we also have from the graph <math>f(-1)=-a+b-c+2=0</math>. Summing <math>f(1)+f(2)</math> we get <math>2b | + | From the graph, we have <math>f(0)=2</math> so <math>d=2</math>. Also from the graph, we have <math>f(1)=a+b+c+2=0</math>. But we also have from the graph <math>f(-1)=-a+b-c+2=0</math>. Summing <math>f(1)+f(2)</math> we get <math>2b+4=0</math> so <math>b = -2 \Rightarrow \mathrm{(B)}</math>. |
Solution by franzliszt | Solution by franzliszt |
Latest revision as of 17:01, 23 November 2020
Contents
Problem
Part of the graph of is shown. What is ?
Solution
Solution 1
Since
It follows that . Also, , so .
Solution 2
Two of the roots of are , and we let the third one be . Then Notice that , so .
Solution 3
Notice that if , then vanishes at and so implies by coefficient, .
Solution 4
The roots of this equation are , letting be the root not shown in the graph. By Vieta, we know that and . Therefore, . Setting the two equations for equal to each other, . We know that the y-intercept of the polynomial is , so . Plugging in for , .
Therefore,
Solution 5
From the graph, we have so . Also from the graph, we have . But we also have from the graph . Summing we get so .
Solution by franzliszt
See also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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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