1951 AHSME Problems/Problem 41

Problem

The formula expressing the relationship between $x$ and $y$ in the table is: \[\begin{tabular}{|c|c|c|c|c|c|}\hline x & 2 & 3 & 4 & 5 & 6\\ \hline y & 0 & 2 & 6 & 12 & 20\\ \hline\end{tabular}\]

$\textbf{(A)}\ y = 2x-4\qquad\textbf{(B)}\ y = x^{2}-3x+2\qquad\textbf{(C)}\ y = x^{3}-3x^{2}+2x$ $\textbf{(D)}\ y = x^{2}-4x\qquad\textbf{(E)}\ y = x^{2}-4$

Solution

Just plug the $x,y$ pair $(6,20)$ into each of the 5 answer choices:

(A): $2(6)-4=8\ne20$

(B): $6^2-3(6)+2=20$

(C): $6^3-3(6^2)+2(6)=120\ne20$

(D): $6^2-4(6)=12\ne20$

(E): $6^2-4=32\ne20$

The only one that works is $\boxed{\textbf{(B)}}$.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 40
Followed by
Problem 42
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