Continuity with parameters

Idea. The function $F$ is continuous inside each interval, because every formula is a polynomial. The only danger is at the junctions, where the pieces must connect without a jump.

At $x=1$. The left value is $f(1)=2$, and the right limit is $a+b$. We need $a+b=2$.

At $x=3$. The left limit is $3a+b$ and $f(3)=3^2-4=5$, which equals the right limit. We need $3a+b=5$.

Solving. Subtracting the two equations gives $2a=3$, so $a=\frac32$ and $b=2-\frac32=\frac12$.

Remark. With these values the graph is made of pieces glued end to end. The function is continuous but not differentiable at $1$ and $3$: the slopes are $1$, $\frac32$ and $6$.