10.4. On the coordinate plane, the graphs of a linear and a quadratic function are plotted as shown in the figure (one of the intersection points is at the vertex of the parabola), and the line $y=k x+b$ passes through the point ( $-1 ; 2020$ ), while the coefficients $a$ and $c$ of the parabola $y=a(x-c)^{2}$ are integers. How many different values can the coefficient $k$ take? List all possible options and explain why there are no others.

![](https://cdn.mathpix.com/cropped/2024_05_06_ee9dccd3d85fbb3b219eg-2.jpg?height=457&width=440&top_left_y=794&top_left_x=1525)
Answer: two. $k=-404 ;-1010$.

Solution: Note that $k<0$. Since the line passes through the point (-1; $2020)$, we have $2020=-k+b$, from which $b=2020+k$. Then the equation of the line can be written as $y=kx+2020+k$.

Let's find the points of intersection of the line with the coordinate axes.

Point $A: y_{A}=k \cdot 0+2020+k=2020+k$. Then $A=(0 ; 2020+k)$.

Point $B: 0=k \cdot x_{B}+2020+k, x_{B}=-\frac{2020+k}{k}$. Then $B=\left(-\frac{2020+k}{k} ; 0\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ee9dccd3d85fbb3b219eg-2.jpg?height=446&width=428&top_left_y=1336&top_left_x=1522)

Since point $B$ is the vertex of the parabola, then $c=-\frac{2020+k}{k}=-1-\frac{2020}{k}$. Then the parabola is described by the equation $y=a\left(x+1+\frac{2020}{k}\right)^{2}$.

Since the parabola also passes through point $A$, then

$2020+k=a\left(\frac{2020+k}{k}\right)^{2}$, from which $a=\frac{k^{2}}{2020+k}$. Then, finally, the equation of the parabola is $y=\frac{k^{2}}{2020+k} \left(x+1+\frac{2020}{k}\right)^{2}$.

Since the coefficient is an integer, $k$ must be a divisor of the number 2020. Considering $k<0$, the possible values are $k=-1,-2,-4,-5,-10,-20,-101,-202,-404,-505,-1010,-2020$.

The coefficient $a$ is also an integer, which means that $k^{2}$ must be divisible by $2020+k$, i.e., $k^{2} \geq 2020+k$, from which $k<-44$, and $k \neq 2020$, i.e., we need to check $-101,-202,-404,-505,-1010$.

$101^{2}$ is not divisible by $1919=19 \cdot 101$.

$202^{2}=4 \cdot 101^{2}$ is not divisible by $1818=2 \cdot 9 \cdot 101$.

$404^{2}=16 \cdot 101^{2}$ is divisible by $1616=16 \cdot 101$.

$505^{2}=5^{2} \cdot 101^{2}$ is not divisible by $1515=3 \cdot 5 \cdot 101$.
$1010^{2}$ is divisible by 1010.

## Criteria:

7 points - correct answer and complete correct solution;

expression for $b$ in terms of $k-1$ point;

expression for $c$ in terms of $k-2$ points;

expression for $a$ in terms of $k-2$ points;

the last three criteria can be summed

only answer - 0 points.