Answer
Points$: (-2, 21), (1, -6)$
Work Step by Step
Given curve, $y=2x^3+3x^2-12x+1$. The slope of the curve at a point is equal to the slope of the tangent at that point. Since horizontal lines (tangent) have slope $0$, we need to find how many points on the curve $y$ have slope $0$. For slope, we will differentiate $y$ with respect to $x$,
$$\frac{dy}{dx}=6x^2+6x-12$$.Now, the slope needs to be $0$ or $dy/dx=0$.
$$\frac{dy}{dx}=6x^2+6x-12=0\implies x^2+x-2=0$$
$$\therefore x=-2, 1.$$
Thus, the curve $y$ has horizontal tangents at $2$ points$: (-2, 21), (1, -6)$.
