1951 AMC 12 Problem 41

Question

The formula expressing the relationship between

x

and

y

in the table is:

$x$	$2$	$3$	$4$	$5$	$6$
$y$	$0$	$2$	$6$	$12$	$20$

y=2x-4

y=x^2-3x+2

y=x^3-3x^2+2x

y=x^2-4x

y=x^2-4

The first differences

2,4,6,8

suggest a quadratic rule

The values are

(x-1)(x-2)

The

y

-values factor naturally as

\begin{gathered} 0=1\cdot0,\quad 2=2\cdot1,\quad 6=3\cdot2,\\ 12=4\cdot3,\quad 20=5\cdot4. \end{gathered}

Thus

y=(x-1)(x-2)

and hence

y=x^2-3x+2.

Thus, the correct answer is B.

Problem 41 in other years