Asked by derekkkk
The points (3, 8) and (6, 8) are part of the solution set of a quadratic inequality, while (5, 9) and (7, 7) are not. The point that must also be part of the solution set is
A. (4, 7)
B. (4, 9)
C. (8, 6)
D. (6, 10)
A. (4, 7)
B. (4, 9)
C. (8, 6)
D. (6, 10)
Answers
Answered by
oobleck
(3,8) and (6,8) are part of the solution set, the vertex is at x = 9/2, so
y = a(x - 9/2)^2 + b
so we know that
a(3 - 9/2)^2 + b = 8
a(6 - 9/2)^2 + b = 8
or, 9a+4b = 32
The problem with that is that there are many values of a which fit those two points. A parabola is determined by 3 points. So you could pick an a such that none or all of the choices lie in the solution set.
y = a(x - 9/2)^2 + b
so we know that
a(3 - 9/2)^2 + b = 8
a(6 - 9/2)^2 + b = 8
or, 9a+4b = 32
The problem with that is that there are many values of a which fit those two points. A parabola is determined by 3 points. So you could pick an a such that none or all of the choices lie in the solution set.
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