2. Find the geometric locus of points $\left(x_{z} ; y_{z}\right)$, which are the vertices of parabolas $y=a x^{2}+z x+$ $c$, where $a$ and $c$ are fixed positive numbers, and $z$ takes all real values.
Answer: parabola $y_{z}=-a x_{z}^{2}+c$.

Solution. The vertex of the parabola $y=a x^{2}+z x+c$ has coordinates $x_{z}=-\frac{z}{2 a}, y_{z}=-a\left(\frac{z}{2 a}\right)^{2}+$ $c=-a x_{z}^{2}+c$. Thus, any vertex lies on the parabola $y_{z}=-a x_{z}^{2}+c$. Now we need to show that any point on this line is the vertex of the parabola $y=a x^{2}+z x+c$. If the equation of the parabola can be written in the form $y=a\left(x-x_{0}\right)^{2}+y_{0}$, then the point ( $x_{0}, ; y_{0}$ ) is the vertex of this parabola. Let's take an arbitrary point $M$ on the parabola $y_{z}=-a x_{z}^{2}+c$ with coordinates ( $x_{0}, ; y_{0}$ ), where $y_{0}=-a x_{0}^{2}+c$. But $a\left(x-x_{0}\right)^{2}+y_{0}=a\left(x-x_{0}\right)^{2}-a x_{0}^{2}+c=a x^{2}-2 a x x_{0}+c$, which corresponds to the expression $a x^{2}+z x+c$ when $z=-2 a x_{0}$. Therefore, for any point $M$ on the parabola $y_{z}=$ $-a x_{z}^{2}+c$, we have found a value of $z$ for which the point $M$ is the vertex of the parabola $y=a x^{2}+z x+c$.

Comment. The coordinates of the vertex of the parabola given in the problem statement are recorded - 1 point. The equation of the line of vertices is found - 3 points. It is shown that any point on this line is a vertex - another 3 points. Points are summed.