Asymptotes of (x²−y²)(x+2y+1)+x+y+1=0

Curve: (x² − y²)(x + 2y + 1) + x + y + 1 = 0. Degree 3.

φ₃(m) = (1 − m²)(1 + 2m)
φ₂(m) = (1 − m²)
φ₁(m) = 1 + m

φ₃(m) = 0 gives three simple slopes m = 1, m = −1, m = −1/2. For simple roots c = −φ₂(m)/φ₃′(m), where

φ₃′(m) = −2m(1 + 2m) + 2(1 − m²).

• m = 1: φ₂(1) = 0, so c = 0. Asymptote: y = x.
• m = −1: φ₂(−1) = 0, so c = 0. Asymptote: y = −x.
• m = −1/2: φ₂ = 3/4, φ₃′ = 0 + 2(3/4) = 3/2, so c = −(3/4)/(3/2) = −1/2. Asymptote: y = −x/2 − 1/2, i.e. x + 2y + 1 = 0.

Answer: y = x, y = −x, x + 2y + 1 = 0.