\(f(1) = -1 \text{.}\)

\(-2 x + 6 ≥ 0\)
\(-2 x ≥ -6\)
\(x ≤ 3\)
Dus het randpunt is \((3 , -5) \text{.}\)

\(x > 1\) geeft \(-5 ≤ f(x) < -1 \text{.}\)

\(-3 + 5 \sqrt{2 x + 6} = 7\)
\(5 \sqrt{2 x + 6} = 10\)
\(\sqrt{2 x + 6} = 2\)
\(2 x + 6 = 4\)
\(2 x = -2\)
\(x = -1 \text{.}\)

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

\(f(x) ≤ 7\) geeft \(-3 ≤ x ≤ -1 \text{.}\)

\(f(x) = 6\)
\(-5 ⋅ {}^{4}\!\log(-2 x + 1) + 6 = 6\)
\(-5 ⋅ {}^{4}\!\log(-2 x + 1) = 0\)
\({}^{4}\!\log(-2 x + 1) = 0\)
\(-2 x + 1 = 4^{0} = 1\)
\(-2 x = 0\)
\(x = 0\)

Bereking van het domein geeft
\(-2 x + 1 > 0\)
\(-2 x > -1\)
\(x < \frac{1}{2}\)
Dus de verticale asymptoot is de lijn \(x = \frac{1}{2} \text{.}\)

\(f(x) > 6\) geeft \(0 < x < \frac{1}{2} \text{.}\)