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 6 - 2022 - Paper 1

The line $y = k$ intersects the curve $C$ at one point when the horizontal line is tangential to a point on the curve. This occurs when the value of $k$ is equal to the maximum value of the function since the function decreases from there. The maximum value is at the point $(2, 8)$, so $k$ must be less than or equal to this value. Additionally, the minimum value of the function occurs at the point $(6, 0)$, meaning $k$ must also be greater than or equal to this value if intersecting in the interval opposite to that maximum value. Therefore, the set of values of $k$ is as follows:

ext{Set of values: } ext{All } k ext{ such that } k < 8 ext{ or } k > 0: ext{ } (- ext{∞}, 0) igcup (8, ext{∞})

The equation of a cubic function that passes through the points $(0, 0)$, $(2, 8)$, and $(6, 0)$ can be expressed in the form:

$f(x) = a(x)(x - 2)(x - 6)$

Using the point $(2, 8)$ to find the coefficient $a$:

ightarrow 0 = 8 ext{ (holds true, but does not contribute)}$$ Next, we can find $a$ by equating and simplifying: Using another point or calculating based on repeated shapes in cubic formulas would yield: The general formula can also be assumed: Thus, upon inserting the values assessed through derivatives and graphical inspections, we determine: $$f(x) = rac{1}{4}(x)(x - 2)(x - 6)$$