Figure 2 shows a sketch of part of the curve $y = f(x)$, $x \in \mathbb{R}$, where $$f(x) = (2x - 5)^{2}(x + 3)$$ (a) Given that (i) the curve with equation $y = f(x) - k$, $x \in \mathbb{R}$, passes through the origin, find the value of the constant $k$ - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 3

To find the value of the constant $c$, we need to ensure that $f(c) = 0$ at the minimum point, meaning $c$ needs to be such that:

$2c - 5 = 0 \implies c = \frac{5}{2}.$

We also need to check that this indeed gives a minimum by inspecting the second derivative or the behavior of the first derivative around this point.

To find $f'(x)$, we can use the product rule on the function:

$f(x) = (2x - 5)^{2}(x + 3).$

Letting $u = (2x - 5)^{2}$ and $v = (x + 3)$, we apply the product rule:

$f'(x) = u'v + uv',$

where:

$u' = 2(2x - 5)(2) = 4(2x - 5)\ ext{ and } v' = 1.$

Thus,

$f'(x) = 4(2x - 5)(x + 3) + (2x - 5)^{2} \ ext{ (after expansion and simplification)}$

This can be shown to equal $12x^{2} - 16x - 35$ after completing the algebraic steps.

Since the gradients at points A and B are equal, we know:

$f'(a) = f'(b) \text{ (where a and b are the respective x-coordinates)}.$

After substituting values for $f'(x)$, we can equate:

$12b^{2} - 16b - 35 = 0.$

Using the quadratic formula to solve for $b$,

$b = \frac{-(-16) \pm \sqrt{(-16)^{2} - 4(12)(-35)}}{2(12)}.$

Determining the roots gives us the x-coordinate values for point B.