1969 AHSME Problems/Problem 7

Problem

If the points $(1,y_1)$ and $(-1,y_2)$ lie on the graph of $y=ax^2+bx+c$ , and $y_1-y_2=-6$ , then $b$ equals:

$\text{(A) } -3\quad \text{(B) } 0\quad \text{(C) } 3\quad \text{(D) } \sqrt{ac}\quad \text{(E) } \frac{a+c}{2}$

Solution

Because the two points are on the quadratic, $y_1 = a + b + c$ and $y_2 = a - b + c$ . Because $y_1 - y_2 = -6$ , $(a+b+c)-(a-b+c)=-6$ $2b=-6$ $b=-3$ The answer is $\boxed{\textbf{(A)}}$ .

1969 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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
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1969 AHSME Problems/Problem 7

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