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I'm given the parameterization for a manifold $\{(t^2+t, 2t-1), t\in R\}$, is this the correct method to find the manifold as the graph of a function?

Set $x=t^2+t, y=2t-1$, find $t(y)$ and then get $x(y)$? Does the function specifically need to be $y(x)$ for its graph to represent the manifold?

Also, while in this case it's clear that $F(x,y)=x(y)-y=0$ is the zero locus of a function that represents the manifold, but how do I find the zero locus in cases where the manifold cannot be represented as a graph of a function?

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    I don't see any reason why it would be wrong to express $x$ as a function of $y$. They just a couple of variables. Either can be a function of the other. Expressing $y$ as a function of $x$ is just a choice of convenience. As for zero locus, that is a concept of functions, not of manifolds, so there is no "zero locus" to find if there is no function around. – 2017-02-16
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    I meant expressing the manifold as the set of all points in the zero locus of a function – 2017-02-16
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    But if there is no function, there is no such thing as a "zero locus". If the manifold is not the zero locus of some function, then there is no such thing as the "zero locus" of the manifold. If it is the zero locus of some function, it is still the zero locus *of the function*, not of the manifold. Indeed, since the manifold *is* the zero locus, as far as the manifold is concerned, there is nothing to find - this "zero locus" is everything. It is like asking where in the universe is the universe. – 2017-02-16

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