Answer
Points$: (-1.816, 5.089), (-0.184, 2.911)$
Work Step by Step
Given curve, $y=x^3+3x^2+x+3$. 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 $x$,
$$\frac{dy}{dx}=3x^2+6x+1$$.Now, the slope needs to be $0$ or $dy/dx=0$.
$$\frac{dy}{dx}=3x^2+6x+1=0$$
$$\therefore x=-1-\sqrt{\frac{2}{3}}, -1+\sqrt{\frac{2}{3}}.$$Or, $x=-1.816, -0.184.$
Thus, the curve $y$ has horizontal tangents at $2$ points$: (-1.816, 5.089), (-0.184, 2.911)$.
