Part of the graph of $ f(x) \equal{} x^3 \plus{} bx^2 \plus{} cx \plus{} d$ is shown. What is $ b$?
[asy]import graph;
unitsize(1.5cm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
dotfactor=3;

real y(real x)
{
return (x-1)*(x+1)*(x-2);
}

path bounds=(-1.5,-1)--(1.5,-1)--(1.5,2.5)--(-1.5,2.5)--cycle;

pair[] points={(-1,0),(0,2),(1,0)};
draw(bounds,white);
draw(graph(y,-1.5,1.5));
drawline((0,0),(1,0));
drawline((0,0),(0,1));
dot(points);
label("$(-1,0)$",(-1,0),SE);
label("$(1,0)$",(1,0),SW);
label("$(0,2)$",(0,2),NE);

clip(bounds);[/asy]$ \textbf{(A)}\minus{}\!4 \qquad \textbf{(B)}\minus{}\!2 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 4$
1. Given the function \( f(x) = x^3 + bx^2 + cx + d \) and the points on the graph \((-1,0)\), \((1,0)\), and \((0,2)\), we can infer that \( f(x) \) has roots at \( x = -1 \) and \( x = 1 \). Therefore, \( f(x) \) can be factored as:
   \[
   f(x) = m(x - 1)(x + 1)(x - r)
   \]
   where \( r \) is the third root and \( m \) is a constant.

2. Expanding the factored form:
   \[
   f(x) = m(x - 1)(x + 1)(x - r) = m(x^2 - 1)(x - r) = m(x^3 - rx^2 - x + r)
   \]
   Comparing this with the general form \( f(x) = x^3 + bx^2 + cx + d \), we get:
   \[
   x^3 + bx^2 + cx + d = m(x^3 - rx^2 - x + r)
   \]
   Therefore, the coefficients must match:
   \[
   b = -mr, \quad c = -m, \quad d = mr
   \]

3. Using the point \((0,2)\) which lies on the graph, we substitute \( x = 0 \) and \( f(0) = 2 \):
   \[
   f(0) = m(0 - 1)(0 + 1)(0 - r) = m(-1)(1)(-r) = mr
   \]
   Given \( f(0) = 2 \), we have:
   \[
   mr = 2
   \]

4. From the coefficient comparison, we know \( b = -mr \). Substituting \( mr = 2 \):
   \[
   b = -2
   \]

Conclusion:
\[
\boxed{-2}
\]