Answer
$a=\frac{3}{4}$, $b=\frac{9}{4}$, see explanations.
Work Step by Step
a. For the function
$f(x)=\begin{cases} ax,\ x\lt2 \\ ax^2-bx+3,\ x\geq2 \end{cases}$
to be differentiable, we need $\lim_{x\to2^-}f(x)=f(2)$ and $\lim_{x\to2^-}f'(x)=f'(2)$.
Thus we have
$\begin{cases} 2a=4a-2b+3 \\ a=2(2)a-b \end{cases}$ or $\begin{cases} 2a-2b+3=0 \\ 3a-b=0 \end{cases}$.
Plug $b=3a$ from the second equation into the first equation; we get $2a-2(3a)+3=0$, which gives $a=\frac{3}{4}$ and $b=\frac{9}{4}$
b. Graph the function
$f(x)=\begin{cases} \frac{3}{4}x,\ x\lt2 \\ \frac{3}{4}x^2-\frac{9}{4}x+3,\ x\geq2 \end{cases}$
as shown in the figure. It can be seen that the line is connected to the parabola at $x=2$ smoothly because the function is differentiable at all $x$ values, including this point.
