Question:
Show that the tangents to the curve $y=7 x^{3}+11$ at the points where $x=2$ and $x=-2$ are parallel.
Solution:
The equation of the given curve is $y=7 x^{3}+11$.
$\therefore \frac{d y}{d x}=21 x^{2}$
The slope of the tangent to a curve at $\left(x_{0}, y_{0}\right)$ is $\left.\frac{d y}{d x}\right]_{\left(x_{0}, y_{0}\right)}$
Therefore, the slope of the tangent at the point where x = 2 is given by,
$\left.\frac{d y}{d x}\right]_{x=-2}=21(2)^{2}=84$
It is observed that the slopes of the tangents at the points where x = 2 and x = −2 are equal.
Hence, the two tangents are parallel.