The second derivative test tells you the concavity of a graph but what's the point if you can tell the concavity by the leading coefficient? For example, in $-x^3$, I know it's concave down. The second derivative would tell me that as well. But, please explain why it's needed.
 A: You can't tell the concavity of a graph from the leading coefficient. First of all, only polynomials have a leading coefficient, and even for such functions, this does not tell you about its concavity. For example, $f(x) = x^3 + 3x^2$ has a positive leading coefficient, but it has second derivative $6x + 6$, so it is concave down for $x < - 1$ and concave up for $x > -1$.
Added Later: Simpler still, the function $f(x) = -x^3$ which you claim is concave down is not. It has second derivative $-6x$, so it is concave up for $x < 0$ and concave down for $x > 0$.
A: The second derivative test is about determining if a critical point is a local extremum, not about telling the concavity.
A: concave down is determined by negative  coefficient, or has decreasing(not increasing slope)  in your case  if $y=-x^3$ 
then
$dy/dx=-3*x^2$
A differentiable function f is concave on an interval if its derivative function f ′ is monotonically decreasing on that interval: a concave function has a decreasing slope. ("Decreasing" here means non-increasing, rather than strictly decreasing, and thus allows zero slopes.)
For a twice-differentiable function f, if the second derivative, f ′′(x), is positive (or, if the acceleration is positive), then the graph is convex; if f ′′(x) is negative, then the graph is concave. Points where concavity changes are inflection points.
If a convex (i.e., concave upward) function has a "bottom", any point at the bottom is a minimal extremum. If a concave (i.e., concave downward) function has an "apex", any point at the apex is a maximal extremum.
