Question about the sketch of a parametric curve

Question

The problem is so sketch the curve parametrized by $x = \sqrt{t}$ and $y = t-5$. Solving, I get that $t = x^2$ and so $y = x^2 - 5$.

In the solutions of my textbook, the graph is only drawn for $x \geq 0$ (the half of the parabola that lies to the right of the origin). Can anyone explain why this is? Why must $x$ be greater than or equal to $0$?

Answer 1

$x$ cannot be negative because $x=\sqrt t$, and $\sqrt t$ is never negative. When you try to solve $x=\sqrt t$ by “squaring both sides,” you get an equation with more solutions.

Answer 2

Because $t$ cannot be less than $0$, lest you take the square root of a negative.

Answer 3

As a general rule when you solve equations with radicals, you have to put the C.E. (or existance condition), so: $t\geq0$ and because the square root is always positive $x\geq0$. Now you can square: $x^2=t$.