# Equation of birationally equivalent curve?

I'm reading through Shafarevich volume 1 and am a bit confused about some of his examples. So I get how if we use the line $y=tx$ then we can show certain curves are birational to a line. One of his later examples takes a function $f$ with terms of degree $n$, $n-1$, and $n-2$, i.e.,$f=u_{n-2}+u_{n-1}+u_n$, where the $u_i$ are homogeneous of degree $i$. We again set $y=tx$, divide out the $x^{n-2}$ at which point we have $f=a(t)x^2+b(t)x+c(t)$. He then uses $s=2ax+b$ to complete the square at which point he declares that $f$ is birational to $s^2=p(t) = b^2-4ac$ (hyperelliptic curve).

What's confusing is that in the earlier example of $y^2=x^2+x^3$ he did the same thing, found $x =t^2-1$ (analogous to the $s$ above) which isn't the equation of a line (which the curve is birational to). How is it that our function $s^2(t)$ above was the birational curve but this function $x(t)$ isn't?

$$y=x^3+x,\;\;\; \mathrm{and} \;\;\;y^2=x^3+x.$$ The first curve is rational, because clearly for each value of $x$ there is a unique value of $y$, so there is a one to one map from a line to the curve, $x\to (x, x^3+x)$. On the other hand in the second example there is no such an obvious map, because if you know the $x$ coordinate of a point on the curve you can not say what is the value of $y$ -- the quadratic equation has two solution. And indeed, the curve in the second example is an elliptic curve, it is not rational.