Răspuns:
f : R -> R
f(x) = x²- (2 - 2m) x + 5 + m + m²
M(x₀,y₀) ∉ Gf ∩ Ox ⇔ f(x₀) = y₀ ≠ 0 ⇔
x₀²- (2 - 2m) x₀ + 5 + m + m² ≠ 0 ⇔ Δ<0
Δ = (2 - 2m) ²-4·(5 + m + m²) = 4+4m²-8m-20-4m-4m² = -12m-16
Δ<0 ⇔ -12m-16 < 0 ⇔
-16 < 12m ⇔
-16/12 < m ⇔
-4/3 < m ⇔ m ∈ (-4/3, +∞)
