\(f(-2)=-7\text{.}\)

\(-6x-3≥0\)
\(-6x≥3\)
\(x≤-\frac{1}{2}\)
Dus het randpunt is \((-\frac{1}{2}, 2)\text{.}\)

\(x>-2\) geeft \(-7<f(x)≤2\text{.}\)

\(-2+3\sqrt{5x-5}=13\)
\(3\sqrt{5x-5}=15\)
\(\sqrt{5x-5}=5\)
\(5x-5=25\)
\(5x=30\)
\(x=6\text{.}\)

\(5x-5≥0\)
\(5x≥5\)
\(x≥1\)
Dus het randpunt is \((1, -2)\text{.}\)

\(f(x)≤13\) geeft \(1≤x≤6\text{.}\)

\(f(x)=6\)
\(-5⋅{}^{4}\!\log(-3x+4)+6=6\)
\(-5⋅{}^{4}\!\log(-3x+4)=0\)
\({}^{4}\!\log(-3x+4)=0\)
\(-3x+4=4^0=1\)
\(-3x=-3\)
\(x=1\)

Bereking van het domein geeft
\(-3x+4>0\)
\(-3x>-4\)
\(x<1\frac{1}{3}\)
Dus de verticale asymptoot is de lijn \(x=1\frac{1}{3}\text{.}\)

\(f(x)≥6\) geeft \(1≤x<1\frac{1}{3}\text{.}\)