(a) Using the same axes, sketch the curves \(y = 6 - x - x^{2}\)

Explanation

(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 :

\(\int_{-\frac{3}{4}} ^{1} [(6 - x - x^{2}) - (3x^{2} - 2x + 3)] \mathrm {d} x\)

= \(\int_{-\frac{3}{4}} ^{1} (3 + x - 4x^{2}) \mathrm {d} x\)

= \([3x + \frac{x^{2}}{2} - \frac{4}{3}x^{3}]|_{-\frac{3}{4}} ^{1}\)

= \((3(1) + \frac{1}{2} - \frac{4}{3}) - (3(-\frac{3}{4}) + \frac{(\frac{-3}{4})^{2}}{2} - \frac{4}{3} (-\frac{3}{4})^{3}\)

= \(\frac{13}{6} - (-\frac{45}{32})\)

= \(\frac{343}{96} sq. units\)