document.write( "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
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document.write( "C equals 125, what is the value of α?
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document.write( "A) -3
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document.write( "B) -5
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document.write( "C) 3
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document.write( "D) 5 \n" );
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Algebra.Com's Answer #734664 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "I will use k instead of α for the constant we are looking for.... \n" ); document.write( "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. \n" ); document.write( "So the vertex is C(k,k^2). \n" ); document.write( "The x-intercepts are A(0,0) and B(2k,0). \n" ); document.write( "Triangle ABC then has base 2k and altitude k; its area is (1/2) base times height = k^2. \n" ); document.write( "So k^2 = 125, making k=5*sqrt(5)... which is not one of the answer choices. \n" ); document.write( "The problem as you show it is flawed. \n" ); document.write( "Perhaps the area was supposed to be 25 instead of 125, making D the answer....? \n" ); document.write( " \n" ); document.write( " |