Q.
The maximum vertical distance d between the parabola y=−2x2+4x+3 and the line y=x−2 throughout the bounded region in the figure, is
160
103
Complex Numbers and Quadratic Equations
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Solution:
The vertical distance is given by d=−2x2+4x+3−(x−2)=−2x2+3x+5, which is a parabola opening down.
Its maximum value is the y-coordinate of the vertex which has x-coordinate equal to 2a−b=2(−2)−3=43. The y-coordinate, then, is −2(43)2+3(43)+5=8−9+818+840=849