Consider the projective plane $\mathbb{P}^2_{\overline{\mathbb{C}(t)}}$ over the algebraic closure of the function field $\mathbb{C}(t)$.

Take the point $p_0 = [0:1:0]\in \mathbb{P}^2_{\overline{\mathbb{C}(t)}}$ and eight more general points $p_1,\dots,p_8\in \mathbb{P}^2_{\overline{\mathbb{C}(t)}}$ which are not defined over $\mathbb{C}(t)$. Assume that there is a cubic curve $C \subset \mathbb{P}^2_{\overline{\mathbb{C}(t)}}$, defined over $\mathbb{C}(t)$, passing through $p_0,p_1,\dots,p_8$. We can write $C$ as $$C = \{ \alpha_0(t) x^3 + \alpha_1(t)x^2y + \alpha_2(t)x^2z + \alpha_3(t)xy^2 + \alpha_4(t)xyz + \alpha_5(t)xz^2 + \alpha_6(t)y^3 + \alpha_7(t)y^2z + \alpha_8(t)yz^2 + \alpha_9(t)z^3 = 0 \}$$ where $\alpha_0,\dots,\alpha_9$ are polynomials in $t$, and extend $C$ to a fibration on $\mathbb{P}^1$ as $$C' = \{A_0(s,t) x^3 + A_1(s,t)x^2y + A_2(s,t)x^2z + A_3(s,t)xy^2 + A_4(s,t)xyz + A_5(s,t)xz^2 + A_6(s,t)y^3 + A_7(s,t)y^2z + A_8(s,t)yz^2 + A_9(s,t)z^3 = 0 \}$$ where now $A_0,\dots,A_9$ are homogeneous polynomials on $\mathbb{P}^1$.

Can we find such a curve $C'$ such that the polynomials $A_0,\dots,A_9$ appearing in the expression of $C'$ are of degree one or at least $A_9$ is of degree one?