= Solution
In the affine chart $z=1$, the equation is $y^2=x^2(x+1)$, irreducible because $x+1$ is not a square in $k(x)$. The <partial derivatives> show the unique singular point is $P=[0:0:1]$; the point at infinity is smooth.
A line $y=tx$ through $P$ meets the curve again at
$$
x=t^2-1,\qquad y=t(t^2-1).
$$
Thus $t\mapsto[t^2-1:t(t^2-1):1]$ is birational, with inverse $t=y/x$ away from $P$. The incidence closure has one point over every smooth $Q$ and two points over $P$, corresponding to $t=\pm1$. This is the <normalization of a nodal curve> and separates the two branches.
Dominated by projective space does not imply being projective space. A smooth quadric surface $Y\simeq\mathbb P^1\times\mathbb P^1$ is rational and hence dominated by $\mathbb P^2$, but is not any $\mathbb P^k$; for example its Picard rank is $2$, not $1$.
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