\(f(4)=-6\text{.}\)

\(2x-4≥0\)
\(2x≥4\)
\(x≥2\)
Dus het randpunt is \((2, 4)\text{.}\)

\(x<4\) geeft \(-6<f(x)≤4\text{.}\)

\(5-4\sqrt{-3x+3}=-7\)
\(-4\sqrt{-3x+3}=-12\)
\(\sqrt{-3x+3}=3\)
\(-3x+3=9\)
\(-3x=6\)
\(x=-2\text{.}\)

\(-3x+3≥0\)
\(-3x≥-3\)
\(x≤1\)
Dus het randpunt is \((1, 5)\text{.}\)

\(f(x)>-7\) geeft \(-2<x≤1\text{.}\)

\(f(x)=9\)
\(4⋅{}^{3}\!\log(-x+4)+5=9\)
\(4⋅{}^{3}\!\log(-x+4)=4\)
\({}^{3}\!\log(-x+4)=1\)
\(-x+4=3^1=3\)
\(-x=-1\)
\(x=1\)

Bereking van het domein geeft
\(-x+4>0\)
\(-x>-4\)
\(x<4\)
Dus de verticale asymptoot is de lijn \(x=4\text{.}\)

\(f(x)<9\) geeft \(1<x<4\text{.}\)