Question 1119197: Q57 For some constant α, the graph of the quadratic function f(x) = – x^2+ 2α x is a parabola with x-intercepts at A and B and vertex C. If the area of the triangle whose vertices are A, B, and
C equals 125, what is the value of α?
A) -3
B) -5
C) 3
D) 5
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
I will use k instead of α for the constant we are looking for....
The vertex of the parabola is when x = -b/(2a) = -2k/-2 = k; the value of the function at x=k is -k^2+2k^2 = k^2.
So the vertex is C(k,k^2).
The x-intercepts are A(0,0) and B(2k,0).
Triangle ABC then has base 2k and altitude k; its area is (1/2) base times height = k^2.
So k^2 = 125, making k=5*sqrt(5)... which is not one of the answer choices.
The problem as you show it is flawed.
Perhaps the area was supposed to be 25 instead of 125, making D the answer....?
|
|
|
