1962 AMC 12 Problem 19

Attempt Problem 19 of the 1962 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1962 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

If the parabola y=ax2+bx+cy=ax^2+bx+c passes through the points (−1,12),(-1,12), (0,5),(0,5), and (2,−3),(2,-3), the value of a+b+ca+b+c is:
−4-4
−2-2
00
11
22
Answer: C
Concepts:system of equationsquadraticsubstitution
Small hint:
The requested sum is the value of the parabola at x=1x=1
Big hint:
Use the three given points to solve for a,a, b,b, and cc
Solution:
The point (0,5)(0,5) gives c=5.c=5. The other two points give a−b=7,4a+2b=−8. a-b=7,\qquad 4a+2b=-8. Solving yields a=1a=1 and b=−6.b=-6. Hence a+b+c=1−6+5=0.a+b+c=1-6+5=0.
Thus, the correct answer is C.
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