Computing the genus of a plane curve

Let $$b(x)=x^4 + 3x^3 + 3x^2 + 2x + 1$$, and let $$a(x)\in \mathbb Z[x]$$ be a separable polynomial. Let $$C$$ be the plane curve defined by $$(y^2+(x+x^2+x^3)a(x))^2-a(x)^2b(x)=0$$. I would need to show that if the degree of $$a(x)$$ is large enough, then $$C$$ has a finite number of $$\mathbb Q$$-rational points. My guess is that when the degree of $$a(x)$$ is large, then $$C$$ has geometric genus $$\geq 2$$, so that the claim would follow from Faltings' theorem. However, $$C$$ might have singularities, so I wouldn't know how to compute, or at least bound from below, its geometric genus. Is there any general principle/result that I can apply? (Notice that unfortunately the curve $$y^2=b(x)$$ has infinitely many rational points)

• $C$ is a cover of $y^2=b(x)$. What does Riemann-Hurwitz tell you? – Felipe Voloch Apr 11 at 19:55
• ...that as soon as the cover is ramified in at least one point, C must have genus at least 2. Thanks a lot for the suggestion! – user36371 Apr 12 at 8:42