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Show that the tangents to the curve y...

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

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Consider `y=2 x^{3}-4` as the equation of curve
By differentiating both sides w.r.t.` x`
`dy / dx=6 x^{2}`

We get
`(dy / dx)_{x=2}=6(2)^{2}=24`

When `m_{1}=24`
`dy / dx=6 x^{2}`

`(dy / dx)_{x=2}=6(2)^{2}=24`

...
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