## Task A-2.1.

Determine all real numbers $k$ for which the vertex of the parabola with the equation

$$
y=4 x^{2}-4(k+1) x+k^{2}+4 k-1
$$

lies on the parabola whose equation is $y=4 x^{2}-2 x-8$.
## Solution.

Let's write the equation of the parabola $y=4 x^{2}-4(k+1) x+k^{2}+4 k-1$ in vertex form.

$$
\begin{aligned}
y & =4\left(x^{2}-(k+1) x+\left(\frac{k+1}{2}\right)^{2}\right)-4\left(\frac{k+1}{2}\right)^{2}+k^{2}+4 k-1= \\
& =4\left(x-\frac{k+1}{2}\right)^{2}-(k+1)^{2}+k^{2}+4 k-1= \\
& =4\left(x-\frac{k+1}{2}\right)^{2}+2 k-2
\end{aligned}
$$

Thus, the coordinates of the vertex are

$$
x_{0}=\frac{k+1}{2}, \quad y_{0}=2 k-2
$$

By substituting into the parabola $y=4 x^{2}-2 x-8$, we have

$$
\begin{gathered}
y_{0}=4 x_{0}^{2}-2 x_{0}-8 \\
2 k-2=4\left(\frac{k+1}{2}\right)^{2}-2 \frac{k+1}{2}-8 \\
2 k-2=(k+1)^{2}-(k+1)-8 \\
2 k-2=k^{2}+k-8 \\
k^{2}-k-6=0
\end{gathered}
$$

The solutions to the resulting quadratic equation are $k=3$ and $k=-2$, which are also the required values of the real number $k$.

Note: The coordinates of the vertex can be found directly by substituting into the formula for the vertex of a quadratic function. If a student immediately writes the coordinates of the vertex according to the formula, they receive 3 points for that part.