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

The equation of the given curve is $y=7x^3+11$. 
$\therefore \frac{dy}{dx}=21x^2$
Thus slope of the tangent to a curve at $x=2$ is $={\left(\frac{dy}{dx}\right)}_{x=2}=(21x^2{)}_{x=2}=84$
and slope of the tangent to a curve at $x=-2$ is $={\left(\frac{dy}{dx}\right)}_{x=-2}=(21x^2{)}_{x=-2}=84$
It is observed that the slopes of the tangents at both the are equal.
Hence, the two tangents are parallel.