(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 t...

Answer Details

(a) Sketching the two curves on the same axes.

Curve 1: \(y = 6 - x - x^{2}\)

This is an inverted (\(\cap\)) parabola since the coefficient of \(x^{2}\) is negative.

• y-intercept: when \(x = 0,\ y = 6\), so it passes through \((0,\,6)\).
• x-intercepts: when \(y = 0,\ 6 - x - x^{2} = 0 \Rightarrow x^{2} + x - 6 = 0 \Rightarrow (x+3)(x-2) = 0\), giving \(x = -3\) and \(x = 2\), i.e. the points \((-3,\,0)\) and \((2,\,0)\).
• Turning point: \(\dfrac{dy}{dx} = -1 - 2x = 0 \Rightarrow x = -\tfrac12\). Then \(y = 6 - (-\tfrac12) - (-\tfrac12)^{2} = 6\tfrac14\). Maximum at \(\left(-\tfrac12,\ 6\tfrac14\right)\).

Curve 2: \(y = 3x^{2} - 2x + 3\)

This is an upright (\(\cup\)) parabola since the coefficient of \(x^{2}\) is positive.

• y-intercept: when \(x = 0,\ y = 3\), so it passes through \((0,\,3)\).
• x-intercepts: \(b^{2} - 4ac = (-2)^{2} - 4(3)(3) = 4 - 36 = -32 < 0\), so the curve does not cut the x-axis.
• Turning point: \(\dfrac{dy}{dx} = 6x - 2 = 0 \Rightarrow x = \tfrac13\). Then \(y = 3(\tfrac13)^{2} - 2(\tfrac13) + 3 = 2\tfrac23\). Minimum at \(\left(\tfrac13,\ 2\tfrac23\right)\).

Plotting both on the same axes, the \(\cap\) parabola sits above the \(\cup\) parabola between their two crossing points, enclosing the finite shaded region:

(b) x-coordinates of the points of intersection.

At the points of intersection the two y-values are equal:

\[6 - x - x^{2} = 3x^{2} - 2x + 3\]\[0 = 3x^{2} - 2x + 3 - 6 + x + x^{2}\]\[0 = 4x^{2} - x - 3\]

Factorising: \(4x^{2} - 4x + 3x - 3 = 4x(x-1) + 3(x-1) = (4x + 3)(x - 1) = 0\).

\[x = -\frac{3}{4} \quad\text{or}\quad x = 1\]

So the curves intersect where \(x = -\tfrac34\) and \(x = 1\).

(c) Area of the finite region bounded by the two curves.

Between \(x = -\tfrac34\) and \(x = 1\), the curve \(y = 6 - x - x^{2}\) lies above \(y = 3x^{2} - 2x + 3\), so the enclosed area is

\[A = \int_{-3/4}^{1}\big[(6 - x - x^{2}) - (3x^{2} - 2x + 3)\big]\,dx = \int_{-3/4}^{1}\big(3 + x - 4x^{2}\big)\,dx\]\[A = \left[\,3x + \frac{x^{2}}{2} - \frac{4x^{3}}{3}\,\right]_{-3/4}^{1}\]

At \(x = 1:\quad 3(1) + \dfrac{1}{2} - \dfrac{4}{3} = \dfrac{18 + 3 - 8}{6} = \dfrac{13}{6}\).

At \(x = -\tfrac34:\quad 3\left(-\tfrac34\right) + \dfrac{(-3/4)^{2}}{2} - \dfrac{4}{3}\left(-\tfrac34\right)^{3} = -\dfrac{9}{4} + \dfrac{9}{32} + \dfrac{4}{3}\cdot\dfrac{27}{64} = -\dfrac{9}{4} + \dfrac{9}{32} + \dfrac{9}{16} = -\dfrac{45}{32}\).

\[A = \frac{13}{6} - \left(-\frac{45}{32}\right) = \frac{208}{96} + \frac{135}{96} = \frac{343}{96} \approx 3.57\ \text{square units}\]

The area of the finite region bounded by the two curves is \(\dfrac{343}{96} \approx 3.57\) square units.