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 1 - 2017 - Paper 1

To find $c$, we examine the expression for $f(x)$, which is:

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

For the curve to have a minimum at the origin, we need to have:

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

To find the derivative of $f(x)$, we apply the product rule:

Let:

$u = (2x-5)^{2}, \quad v = (x+3).$

Then:

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

Calculating $u'$:

$u' = 2(2x-5)(2) = 4(2x-5)$

And $v'$:

$v' = 1.$

Thus:

$f'(x) = 4(2x-5)(x+3) + (2x-5)^{2}(1).$

Expanding the first term:

$= (4(2x-5)(x)+12(2x-5)) + (2x-5)^{2}.$

Simplifying this expression will yield:

$= 12x^{2} - 16x - 35.$

Let the coordinates of point A be $(a, f(a))$ and point B be $(b, f(b))$. Given that the gradients at these points are equal:

$f'(a) = f'(b).$

Using the derivative found previously:

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

This simplifies to:

$12a^{2} - 16a = 12b^{2} - 16b.$

Assuming that both points are distinct, we can solve for the values of $b$ given the coordinates of point A. The specific value can be determined with further information about point A.