At what points will be tangents to the curve

Solution:

finding the slope of the tangent by differentiating the curve

$\frac{d y}{d x}=6 x^{2}-30 x+36$

According to the question, tangent is parallel to the $x$-axis, which implies $m=0$

$6 x^{2}-30 x+36=0$

$x^{2}-5 x+6=0$

$x=3$ or $x=2$

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

$y=2(3)^{3}-15(3)^{2}+36(3)-21$

$y=6$

or

$y=2(2)^{3}-15(2)^{2}+36(2)-21$

$y=7$

equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$

$y-6=0(x-3)$

$y=6$

or

$y-7=0(x-2)$

$y=7$