Keep a quadratic strictly negative

The leading coefficient must satisfy $4k<0$, so $k<0$. The degenerate case $k=0$ gives $4x+1$, which is not negative for every real $x$.

The discriminant must be negative: $[4(2k+1)]^2-4(4k)(4k+1)=16(3k+1)<0$. Thus $k<-1/3$, which already implies $k<0$. At $k=-1/3$ the graph touches zero, so the endpoint is excluded.