49.2. It is known that the parabola $y=x^{2}+p x+q$ touches the line $y=2 x-3$ at the point $M(2 ; 1)$. Find $p$ and $q$.
49.2. If this parabola passes through the point $M(2 ; 1)$, then $1=2^{2}+2 p+q$, i.e., $2 p+q=-3$. Moreover, since the parabola is tangent to the line $y=2 x-3$ at the point $\mathrm{M}(2 ; 1)$, the derivative of the function $y=x^{2}+p x+q$ at $x=2$ will be equal to the slope of the line, i.e., 2. Thus: $y^{1}=2 x+p, 2=4+p, p=-2$. Using the previously obtained equation $2 p+q=-3$ with $p=-2$, we find that $q=1$. Therefore, we have the parabola $y=x^{2}-2 x+1=$ $=(x-1)^{2}$.