Difference between revisions of "2003 AMC 12B Problems/Problem 20"
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===Solution 5=== | ===Solution 5=== | ||
− | 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
[hide]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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.