(a) Using the same axes, sketch the curves \(y = 6 - x - x^{2}\) and \(y = 3x^{2} - 2x + 3\).
(b) Find the x- coordinates of the points of intersection of the two curves in (a).
(c) Calculatethe area of the finite region bounded by the two curves in (a).
(a) \(y = 6 - x - x^{2}\)
When x = 0, y = 6.
When y = 0, \(x^{2} + x - 6 = 0\)
\(x^{2} + 3x - 2x - 6 = 0 \implies x(x + 3) - 2(x + 3) = 0\)
\(x = -3 ; x = 2\)
Graph passes through (0, 6), (-3, 0) (2, 0).
Turning point : \(\frac{\mathrm d y}{\mathrm d x} = - 1 - 2x\)
\(-1 - 2x = 0 \implies x = \frac{-1}{2}\)
When \(x = \frac{-1}{2}\), \(y = 6 - (-\frac{1}{2}) - (\frac{-1}{2})^{2} = 6\frac{1}{4}\).
Turning point : \((-\frac{1}{2}, 6\frac{1}{4})\).
\(y = 3x^{2} - 2x + 3\)
When x = 0, y = 3.
\(b^{2} - 4ac = (-2)^{2} - 4(3)(3) = 4 - 36 < 0\)
Graph does not cut x - axis.
\(\frac{\mathrm d y}{\mathrm d x} = 6x - 2\)
At turning point, \(6x - 2 = 0 \implies x = \frac{2}{6} = \frac{1}{3}\)
When \(x = \frac{1}{3}, y = 3(\frac{1}{3})^{2} - 2(\frac{1}{3}) + 3 = 2\frac{2}{3}\)
Turning point : \((\frac{1}{3}, 2\frac{2}{3})\).
(b) For the x- coordinates, we use the equation.
\(3x^{2} - 2x + 3 = 6 - x - x^{2}\)
\(4x^{2} - x - 3 = 0 \implies 4x^{2} - 4x + 3x - 3 = 0\)
\(4x(x - 1) + 3(x - 1) = 0 \implies (4x + 3)(x - 1) = 0\)
\(x = -\frac{3}{4} ; 1\)
(c) Area of shaded portion :
\(\int_{-\frac{3}{4}} ^{1} [(6 - x - x^{2}) - (3x^{2} - 2x + 3)] \mathrm {d} x\)
= \(\int_{-\frac{3}{4}} ^{1} (3 + x - 4x^{2}) \mathrm {d} x\)
= \([3x + \frac{x^{2}}{2} - \frac{4}{3}x^{3}]|_{-\frac{3}{4}} ^{1}\)
= \((3(1) + \frac{1}{2} - \frac{4}{3}) - (3(-\frac{3}{4}) + \frac{(\frac{-3}{4})^{2}}{2} - \frac{4}{3} (-\frac{3}{4})^{3}\)
= \(\frac{13}{6} - (-\frac{45}{32})\)
= \(\frac{343}{96} sq. units\)
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