Question:
At what points will be tangents to the curve $y=2 x^{3}-15 x^{2}+36 x-21$ be parallel to the $x$ - axis? Also, find the equations of the tangents to the curve at these points.
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
$y-7=0(x-2)$
$y=7$