y = x2 + 2x - 6
⇒ y + 7 = (x + 1)2
∴ vertex (-1, -7)
For Intersection point of line and parabola,
We have 3x = x2 + 2x - 6
⇒ x2 - x - 6 = 0
⇒ x = 3 or x = 2
∴ Required area = \(\int\limits_{-2}^3\)[3x - (x2 + 2x - 6)]dx
= \(\int\limits_{-2}^3\)(-x2 + x + 6)dx
= \([-\frac{x^3}3+\frac{x^2}2+6x]_{-2}^3\)
= -27/3 + 9/2 + 18 - 8/3 - 2 + 12
= -35/3 + 28 + 9/2
= -35/3 + 65/2
= \(\frac{-70+195}6\)
= \(\frac{125}6\) squre units
