25: D( f ) = (−∞,1) ∪ (1,∞); f is continuous there.
y-intercept: f (0) = 0; x-intercepts: f = 0 gives x = 0.
f is not symmetric since D( f ) is not symmetric.
Limits at endpoints:

Interpretation as asymptotes:
No horizontal asymptote at −∞, but a chance for oblique.

Asymptote y = x + 1 there.
Vertical asymptote at x = 1.
No horizontal asymptote at ∞, but a chance for oblique.

Asymptote y = x + 1 there.

Now determine monotonicity using f ′.

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Answer