What is the geometric place of the vertex points of the parabolas given by the equation $y=x^{2}+t x+1$? ( $t$ is a real parameter.)
Transform the equation of the parabola: $y=x^{2}+t x+1=\left(x+\frac{t}{2}\right)^{2}-\frac{t^{2}}{4}+1$. (Completing the square.) From this form, we can read off the coordinates of the vertex of the parabola: $x=-\frac{t}{2}, y=1-\frac{t^{2}}{4}$.

Substituting $t=-2 x$ into the second equation, we get the relationship $y=-x^{2}+1$, which is the equation of a parabola.

For every point on this parabola, by back-substitution, we can determine which parameter $t$ of the parabola with the equation $y=x^{2}+t x+1$ corresponds to its vertex.

This means that the geometric locus of the vertices of the parabolas is the downward-opening parabola described by the equation $y=-x^{2}+1$, whose axis is the $y$-axis and whose vertex is the point $(0 ; 1)$.