Hans et al, I think I've found something that works. I know ahead of time that there will only be two crossings of the clipping path boundary (I make sure of the by the parameters I choose): beginfig(0); % initialize a, h, and h for f(x)=ax^2+bx+c=a(x-h)^2+k numeric a, h, k; a=-1; h=-3; k=2; % initialize scale numeric u; 10u=3in; % draw grid for k=-5u step 1u until 5u: draw (-5u,k)--(5u,k) withcolor 0.85white; draw (k,-5u)--(k,5u) withcolor 0.85white; endfor; % draw axes drawarrow (-5u,0)--(5u,0); drawarrow (0,-5u)--(0,5u); % label axes label.rt(btex $x$ etex, (5.2u,0)); label.top(btex $y$ etex, (0,5.2.u)); label.bot(btex $5$ etex, (5u,0)); label.lft(btex $5$ etex, (0,5u)); % define function f(x)=a|x-b|+c vardef f(expr x)= a*(x-h)*(x-h)+k enddef; % draw line with given point and slope path F; F:=(-6,f(-6)); for x=-6 step .1 until 6: F:=F--(x,f(x)); endfor; % the line F:=F scaled 1u; % clipping path path cpath; cpath:=(-5,-5)--(5,-5)--(5,5)--(-5,5)--cycle; cpath:=cpath scaled 1u; % clip the path F pair A, B; F:=F cutbefore cpath; F:=reverse F; F:=F cutbefore cpath; drawdblarrow F; endfig; end.