Figure 1 shows a sketch of a curve C with equation $y = f(x)$ where $f(x)$ is a cubic expression in $x$ - Edexcel - A-Level Maths Pure - Question 7 - 2022 - Paper 1

Since $y = k$ intersects $C$ at only one point, we need to find where the line is tangent to the curve. To validate this, we need to set the condition that the discriminant of the resulting equation must equal zero. Therefore, the set of values can be expressed as:

ext{{Either }} k > 8 ext{{ or }} k < 0, ext{ which yields } oxed{(k ext{ : } k > 8) igcup (k ext{ : } k < 0)}

The general form of a cubic equation is $y = ax^3 + bx^2 + cx$. We know:

1. C passes through $(0, 0)$, so $f(0) = 0$.
2. From the turning points, we derive constraints on the coefficients using $(2, 8)$ and $(6, 0)$. By substituting these points back we can find values for $a$, $b$, and $c$. After careful substitution, the equation of C can be determined to be:

f(x) = rac{1}{4}(x - 2)(x - 6)(x - 0) Or similarly in expanded form if required.