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 4 - 2017 - Paper 1

To find the constant $c$, we first need the derivative to find the minimum conditions. Setting $f'(0) = 0$:

Substituting $c$ into the function:

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

We compute the first derivative and set $f'(0) = 0$. After analysis, we find that:

$c = -\frac{5}{2}$

Using the product rule for differentiation:

$f'(x) = (g(x)h(x))' = g'(x)h(x) + g(x)h'(x)$

Letting $g(x) = (2x - 5)^2$ and $h(x) = (x + 3)$, we compute:

$g'(x) = 12x - 20 \quad \text{and} \quad h'(x) = 1$

So:

$f'(x) = (12x - 20)(x + 3) + (2x - 5)^2(1)$

After expanding and simplifying, we obtain:

$f'(x) = 12x^2 - 16x - 35$

Given that points A and B are distinct points lying on the curve, we first need to solve for the gradient. If the gradient at point A is equal to the gradient at point B, we set:

$f'(x_A) = f'(x_B)$

If point A has coordinate 3, we calculate:

$f'(3) = 12(3)^2 - 16(3) - 35 = 12(9) - 48 - 35 = 108 - 48 - 35 = 25$

Now, solving for $f'(x) = 25$ leads to:

$12x^2 - 16x - 35 = 25\Rightarrow 12x^2 - 16x - 60 = 0$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

we find the $x$ coordinates for B, resolving finally to:

The calculations give two solutions, so solving the quadratic provides distinct solutions, leading us to identify specifically the $x$ coordinate of point B.