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 $\boxed{0\textbf{ (C)}}$ .

1962 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
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1962 AHSME Problems/Problem 19

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