Applications on derivative

Question

Show that the tangents to the curve $y=2x^{3}-2$ at $x=2$ and at $x=-2$ are parallel. I try to solve this but not getting how to start . Thanks in advance

Answer 1

the derivative of the curve, $\frac{dy}{dx} = 6x^2$. For $x=2$, the slope of the curve is equal to $24$. For $x=-2$ the slope of the curve is also equal to $24$. Hence, the tangents on the curve will have the same slope on $x=2$ and $x=-2$. Conclusion: The tangents are parallel.

Answer 2

For tangents to be parallel, their slopes should be equal.

Let $f(x)=2x^3-2$

$$f'(x)=6x^2$$ Now, slope of a tangent at a point on curve $= f'(x)$ You can put the values of $x$ as $+2$ and $-2$ to verify your answer