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

Answer

Slopes are same, so these lines are parallel.

Proved

Explanation

Differentiating w.r.t `x`

$$
\begin{aligned}
& y = x^3 – 3\\
&\frac{d y}{d x}=3 x^{2} \\ \\
& m_1 = (\frac{d y}{d x})_{x=2} = 3(2)^{2} \\
&=3 \times 4 \\
&=12 \\ \\
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