[example: A First Vanishing Set]
In $\mathbb A^2_k$, write a point as $(a,b)$ with $a,b\in k$. Evaluating the polynomial $y-x^2$ at this point means substituting $x=a$ and $y=b$, so
\begin{align*}
(y-x^2)(a,b)=b-a^2.
\end{align*}
By the definition of a vanishing set, $(a,b)\in V(y-x^2)$ exactly when $(y-x^2)(a,b)=0$. Therefore
\begin{align*}
(a,b)\in V(y-x^2)\quad\Longleftrightarrow\quad b-a^2=0\quad\Longleftrightarrow\quad b=a^2.
\end{align*}
Thus every point of $V(y-x^2)$ has the form $(a,a^2)$ with $a\in k$. Conversely, for every $t\in k$,
\begin{align*}
(y-x^2)(t,t^2)=t^2-t^2=0,
\end{align*}
so $(t,t^2)\in V(y-x^2)$. Hence
\begin{align*}
V(y-x^2)=\{(t,t^2):t\in k\}.
\end{align*}
For the polynomial $xy$, substitution gives
\begin{align*}
(xy)(a,b)=ab.
\end{align*}
Again by the definition of a vanishing set,
\begin{align*}
(a,b)\in V(xy)\quad\Longleftrightarrow\quad ab=0.
\end{align*}
Since $k$ is a field, it has no zero divisors, so $ab=0$ holds exactly when $a=0$ or $b=0$. If $a=0$, then $(a,b)=(0,b)$ lies on the $y$-axis; if $b=0$, then $(a,b)=(a,0)$ lies on the $x$-axis. Therefore
\begin{align*}
V(xy)=\{(0,b):b\in k\}\cup\{(a,0):a\in k\}.
\end{align*}
So $V(y-x^2)$ is the parabola parametrized by $t\mapsto(t,t^2)$, while $V(xy)$ is the union of the two coordinate axes.
[/example]