The equation y = x (x + 2) - ^ 2 ( x + 1 ) represents a straight...

y = \sin ( x + 2 )sin x - \sin^{2} \left(x + 1\right)$<br />$= \dfrac{1}{2}\left[2\sin \left(x + 2\right)sin x - 2\sin^{2} \left(x + 1\right)\right]$</p><p>\left[∵ 2sin Asin B = \cos \left(A - B\right) - \cos \left(A + B\right) \text{ and } \cos 2A = 1 - 2\sin^{2} A\right]$
$= \dfrac{1}{2}\left[\cos \left(x + 2 - x\right) - \cos \left(2x + 2\right) - \left(1 - \cos 2\left(x + 1\right)\right)\right]$
$= \dfrac{1}{2}\left[\cos 2 - \cos \left(2x + 2\right) - 1 + \cos \left(2x + 2\right)\right]$
$= \dfrac{1}{2}\left[1 - \cos 2\right] = - \dfrac{1}{2}2\sin^{2} 1$
$= - \sin^{2} 1 ($ constant and negative $)$

Hence, the straight line lies in the third and fourth quadrants only.