jmc

A point $(x,y)$ is on the graph if and only if $y=(x+2{)}^4-100$, so we seek to determine all pairs of negative integers $(x,y)$ which satisfy this equation. We can obtain pairs by plugging in $-1,-2,-3,$ and so on for $x$: $$\begin{array}{rl}x=-1& \Rightarrow y=1^4-100=-99\\ x=-2& \Rightarrow y=0^4-100=-100\\ x=-3& \Rightarrow y=(-1{)}^4-100=-99\\ x=-4& \Rightarrow y=(-2{)}^4-100=-84\\ x=-5& \Rightarrow y=(-3{)}^4-100=-19\end{array}$$Starting at $x=-6$, the $y$-coordinates obtained in this way are positive. To be sure that there are no more solutions, we can solve the equation $$(x+2{)}^4-100<0,$$which yields $-2-\sqrt[4]{100}<x<-2+\sqrt[4]{100}$ (in decimals, this is roughly $-5.16<x<1.16$). Thus, the graph of $y=(x+2{)}^4-100$ passes through $\boxed{5}$ points with negative integer coordinates.