Sketch synthesis: domain, intercepts, asymptotes, monotonicity, and concavity

Setting $n(x)=0$ gives $(x+1)^{1/3}=0$, so $x=-1$; thus the $x$-intercept is $(-1,0)$. Evaluating at $x=0$ gives $n(0) = 1^{1/3} = 1$, so the $y$-intercept is $(0,1)$. Since $n'(x) = \frac{1}{3}(x+1)^{-2/3} = \frac{1}{3(x+1)^{2/3}}$ is positive for all $x \neq -1$ (a square root of a real number under the cube gives a nonnegative denominator), $n(x)$ is increasing on the entire real line, with a vertical tangent (not asymptote) at $x=-1$ since $n(-1)$ is defined and finite.

For concavity, $n''(x) = -\frac{2}{9}(x+1)^{-5/3} = -\frac{2}{9(x+1)^{5/3}}$. For $x < -1$, $(x+1)^{5/3} < 0$, making $n''(x) > 0$ (concave up); for $x > -1$, $(x+1)^{5/3} > 0$, making $n''(x) < 0$ (concave down). Since concavity switches sign across $x=-1$ and $n(-1)=0$ is defined, the point $(-1, 0)$ is an inflection point — matching the same location as the $x$-intercept.