Let \(f(x) = x^3 - 12x + 16\).

**Domain of definition, range and intersections with axes:** The domain of definition of \(f\) is all real numbers, and we will determine the function's range later.
We can factor the function as \(f(x) = (x-2)^2(x+4)\), so it intersects the \(x\)-axis at the points \((-4{,}0)\) and \((2{,}0)\), and intersects the \(y\)-axis at the points \((0, 16)\).

**Monotonicity:** We have \(f'(x) = 3x^2 - 12 = 3(x-2)(x+2)\). So we have critical points at \(x = \pm 2\). From the factorization we can easily tell that \(f'(x) > 0\) on the intervals \((-\infty, -2)\) and \((2, \infty)\), so the function is increasing on these intervals. Conversely \(f'(x) < 0\) on the interval \((-2, 2)\), where the function is decreasing. So we know that at \(x = -2\) there is a local maximum, namely \(f(-2) = 32\), and at \(x = 2\) there is a local minimum, namely \(f(2) = 0\).

**Convexity:** Now we compute \(f''(x) = 6x\). So we have \(f''(x) > 0\) for \(x \in (0, \infty)\) and \(f''(x) < 0\) for \(x \in (-\infty, 0)\). So the function is concave on the interval \((-\infty, 0)\) and convex on \((0, \infty)\). At \(x = 0\) we have an inflection point.

**Limits at boundary points:** We have \(\lim_{x \to \infty} f(x)= \infty\) and \(\lim_{x \to -\infty} f(x)= -\infty\) (as you can surely easily compute). This helps us sketch the graph around "infinities". And now we also know, since \(f\) is continuous, that the range of \(f\) is \(\mathbb R\).

**Asymptotes:** We have \(\lim_{x \to \infty} f(x)/x = \lim_{x \to -\infty} f(x)/x =\infty\), as you can surely easily compute. From here it follows that the function does not have asymptotes either at \(\infty\) or at \(-\infty\).