Differentiability when gluing two pieces

Continuity at $1$. The left limit is $1-1+3=3=f(1)$ and the right limit is $\alpha+\beta$. So $\alpha+\beta=3$. (1)

One-sided derivatives at $1$.
$$f'_-(1)=\lim_{x\to1^-}\frac{x^2-x+3-3}{x-1}=\lim_{x\to1^-}\frac{x(x-1)}{x-1}=1,$$
$$f'_+(1)=\lim_{x\to1^+}\frac{\alpha x+\beta-3}{x-1}=\lim_{x\to1^+}\frac{\alpha(x-1)}{x-1}=\alpha,$$
where we used $\beta-3=-\alpha$ from (1). So $\alpha=1$, and then $\beta=2$.

Answer. $\alpha=1$ and $\beta=2$.

Remark. Geometrically, the line $y=x+2$ is the tangent to the parabola at the point $(1,3)$: we have glued the parabola to its own tangent.