Determine the equation (s) of tangent (s) line to the curve

finding the slope of the tangent by differentiating the curve

$\frac{d y}{d x}=12 x^{2}-3$

$\mathrm{m}($ tangent $)=12 \mathrm{x}^{2}-3$

the slope of given line is $-\frac{1}{9}$, so the slope of line perpendicular to it is 9

$12 x^{2}-3=9$

$x=1$ or $-1$

since this point lies on the curve, we can find y by substituting $x$

$y=6$ or 4

therefore, the equation of the tangent is given by equation of tangent is given by

$y-y_{1}=m(\operatorname{tangent})\left(x-x_{1}\right)$

$y-6=9(x-1)$

or

$y-4=9(x+1)$