Difference between revisions of "2002 AMC 12A Problems/Problem 19"
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pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6); | pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6); | ||
real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6}; | real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6}; | ||
− | real[] yticks={-6,-5,-4,-3,-2 | + | real[] yticks={-6,-5,-4,-3,-2,1,2,3,4,5,6}; |
path graph = P1--P2--P3--P4--P5; | path graph = P1--P2--P3--P4--P5; |
Revision as of 20:08, 1 June 2017
Problem
The graph of the function is shown below. How many solutions does the equation have?
Solution
First of all, note that the equation has two solutions: and .
Given an , let . Obviously, to have , we need to have , and we already know when that happens. In other words, the solutions to are precisely the solutions to ( or ).
Without actually computing the exact values, it is obvious from the graph that the equation has two and has four different solutions, giving us a total of solutions.
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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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