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Difference between revisions of "1999 AHSME Problems/Problem 12"

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==Problem==
 
==Problem==
  
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What is the maximum number of points of intersection of the graphs of two different fourth degree polynomial functions <math> y \equal{} p(x)</math> and <math> y \equal{} q(x)</math>, each with leading coefficient 1?
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What is the maximum number of points of intersection of the graphs of two different fourth degree polynomial functions <math> y=p(x)</math> and <math> y=q(x)</math>, each with leading coefficient 1?
  
 
<math> \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8</math>
 
<math> \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8</math>
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==Solution==
 
==Solution==
  
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Since the two graphs are fourth degree polynomials, then they can have at most <math>4</math> intersections, giving the answer of <math>\boxed{D}</math>.
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The intersections of the two polynomials, <math>p(x)</math> and <math>q(x)</math>, are precisely the roots of the equation <math>p(x)=q(x) \rightarrow p(x) - q(x) = 0</math>. Since the leading coefficients of both polynomials are <math>1</math>, the degree of <math>p(x) - q(x) = 0</math> is at most three, and the maximum point of intersection is three, because a third degree polynomial can have at most three roots. Thus, the answer is <math>\boxed{C}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 19:50, 6 March 2016

Problem

What is the maximum number of points of intersection of the graphs of two different fourth degree polynomial functions $y=p(x)$ and $y=q(x)$, each with leading coefficient 1?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$

Solution

The intersections of the two polynomials, $p(x)$ and $q(x)$, are precisely the roots of the equation $p(x)=q(x) \rightarrow p(x) - q(x) = 0$. Since the leading coefficients of both polynomials are $1$, the degree of $p(x) - q(x) = 0$ is at most three, and the maximum point of intersection is three, because a third degree polynomial can have at most three roots. Thus, the answer is $\boxed{C}$.

See Also

1999 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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