jmc

The graph has a horizontal asymptote $y=0,$ a hole at $x=1$, and a vertical asymptote at $x=2$. Since $q(x)$ is a quadratic, and we have a horizontal asymptote at $y=0,$ $p(x)$ must be linear (have degree 1).

Since we have a hole at $x=1$, there must be a factor of $x-1$ in both $p(x)$ and $q(x)$. Lastly, since there is a vertical asymptote at $x=2$, the denominator $q(x)$ must have a factor of $x-2$. Since $q(x)$ is quadratic, we know that $q(x)=b(x-1)(x-2)$ for some $b.$ It follows that $p(x)=a(x-1),$ for some constant $a.$ Since $p(2)=2$, we have $a(2-1)=2$ and $a=2.$ Since $q(-1)=18,$ we have $b(-1-1)(-1-2)=18$ and hence $b=3.$

So $p(x)=2(x-1)=2x-2$ and $q(x)=3(x-1)(x-2)=3x^2-9x+6,$ so $p(x)+q(x)=\boxed{3x^2-7x+4}.$