Question

Show that the tangents to the curve $y=2x^3-3$ at the points where $x=2$ and $x=-2$.

Answer 1

Equation of tangent to any curve $y=f\left(x\right)$ at $(x_1,y_1)$ is
$(y-y_1)=\frac{dy}{dx}{|}_{(x_1,y_1)}(x-x_1)$.
For $y=2x^3-3$ when $x=2$ and $x=-2$ , $y=13$ and $y=-19$ respectively.
Equation of tangents to $y=2x^3-3$ at $(2,13)$ and $(-2,-19)$ are respectively,
$(y-13)=\left(24\right)(x-2)$ and $(y+19)=\left(24\right)(x+2)$.
Clearly, they are parallel.