(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).
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} = 6frac{1}{4}).
Turning point : ((-frac{1}{2}, 6frac{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 = 2frac{2}{3})
Turning point : ((frac{1}{3}, 2frac{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)