1.
ClDrw
FnOff
For(A, .05, .25, .05)
    DrawF 1/((1-x)^2 + 2*A*x)
End

2.
By setting the derivative of y equal to zero, we find that x = 1 - a is a critical point.

3.
When a is very close to zero, 1 - a is very close to 1. Since y(1) = 1/(2a), there is a critical point at approximately (1, 1/(2a)).
Note that since a is in the denominator of the y-coordinate, as a gets closer and closer to zero, the y-coordinate gets larger and larger. The bump in the graph gets taller and taller.
When a = 0, we can no longer have a critical point at (1, 1/(2a)). When a = 0, our function becomes y = 1/((1 - x)^2), which is not defined at x = 1. In effect, the bump in the graph gets infinitely tall.

4.
When a > 1, the critical point x = 1 - a is less than zero and hence not in the domain of our function. Therefore, if a > 1, no member of this family of curves has a critical point and is not very interesting. The bump in the graph gets smaller and smaller as a gets closer and closer to 1, and when a exceeds 1, the bump goes away.

5.
The parameter a determines the height of the bump in the curves, i.e., the maximum value of the function, when a is between 0 and 1. The smaller a is, the higher the maximum value. When a = 0, this function is not bounded from above; when a > 1, this function has no critical points. The parameter a also determines the x-coordinate of the maximum, when a is between 0 and 1. As a gets larger, the critical point shifts to the left.