Show that the tangents to the curve y


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


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.

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