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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 $0$?

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

Duncan Ramage
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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: $x^2=t$.

Matteo
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$x$ 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.

Steve Kass
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  • can you explain more why $\sqrt{t}$ is never negative? Why do we ignore the negative roots? – mXdX Aug 21 '19 at 03:15
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    If $t>0$, there are two solutions to $x^2= t$: One is $x=\sqrt t$ and the other is $x=-\sqrt t$. For example, there are two $x$’s for which $x^2=4$. One is $x=\sqrt4=2$ and the other is $x=-\sqrt 4=-2$. It's a common misunderstanding to think $\sqrt4$ is $\pm2$. It’s not. It’s $2$. The function $f(t)=\sqrt t$ is a function of $t$ and has at most one value for a given $t$. By definition, it's the non-negative solution to $x=t^2$. We don't just ignore negative roots in general, but “squaring both sides” can yield extra solutions, and we have to weed them out. Here, they’re “negative roots.” – Steve Kass Aug 21 '19 at 17:52