1962 AHSME Problems/Problem 19

Problem

If the parabola $y = ax^2 + bx + c$ passes through the points $( - 1, 12)$, $(0, 5)$, and $(2, - 3)$, the value of $a + b + c$ is:

$\textbf{(A)}\ -4\qquad\textbf{(B)}\ -2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$

Solution

Substituting in the $(x, y)$ pairs gives the following system of equations: \[a-b+c=12\] \[c=5\] \[4a+2b+c=-3\] We know $c=5$, so plugging this in reduces the system to two variables: \[a-b=7\] \[4a+2b=-8\] Dividing the second equation by 2 gives $2a+b=-4$, which can be added to the first equation to get $3a=3$, or $a=1$. So the solution set is $(1, -6, 5)$, and the sum is $\boxed{0\textbf{ (C)}}$.

See Also

1962 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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