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

Revision as of 16:14, 7 December 2015

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 $p(x) - q(x) = 0$ . 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. Thus, the answer is $\boxed{C}$ .

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==Solution==
-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>.
+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) = 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. Thus, the answer is <math>\boxed{C}</math>.
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Difference between revisions of "1999 AHSME Problems/Problem 12"

Revision as of 16:14, 7 December 2015

Problem

Solution

See Also