<PRE><B>Question 423208</B>
need an explanation on how to find the x intercepts, and the vertex of this equation y is less than or equal to -x^2+3x+2.
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a = -1 ; b = 3 ; c = 2
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Vertex occurs at x = -b/(2a) = -3/(-2) = 3/2
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Vertex y-value: f(3/2) = -(3/2)^2+3(3/2)+2 = -9/4+(9/2)+2 = 17/4
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Vertex:(3/2 , 17/4)
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x-intercepts:
solve -x^2+3x+2 = 0
x = [-3 +- sqrt(9-4*-1*2)]/(-2) 
= [-3 +- sqrt(17)]/(-2)
Those are the x-intercepts
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Cheers,
Stan H.
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