Show that the tangents to the curve $y = 7{x^3} + 11$ at the points where $x = 2$ and $x = - 2$ are parallel.

We have,
$y = 7{x^3} + 11$ …(i)

Differentiating (i) w.r.t. $x$, we get
$\cfrac{{dy}}{{dx}} = 7 \cdot (3{x^2}) + 0 = 21{x^2}$

Therefore the slope of tangent at $x = 2$ is ${\left( {\cfrac{{dy}}{{dx}}} \right)_{x = 2}} = 21{(2)^2} = 84$
and slope of tangent at $x = - 2$ is ${\left( {\cfrac{{dy}}{{dx}}} \right)_{x = - 2}} = 21{( - 2)^2} = 84$

Hence the slope of tangents at $x = 2$ and $x = - 2$ are equal. Therefore, these tangents are parallel.