10. 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 $y = f(x) - k$, $k \in \mathbb{R}$, passes through the origin, find the value of the constant $k$ - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 1

For $y = f(x + c)$ to have a minimum point at the origin, the value should not vary while $x$ is set to $0$. We need to evaluate $f(0)$:

$c = -5$

This is because moving $x$ to $0$ shifts the minima of $f(x)$ to the origin.

To find the derivative $f'(x)$, we need to apply the product rule here:

Let $u = (2x - 5)^{2}$ and $v = (x + 3)$. Then:

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

First, we find $u'$:

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

And for $v'$:

$v' = 1.$

Now substituting back:

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

Expanding this leads us to:

$4(2x^{2} + 6x - 5x - 15) + (4x^{2} - 20x + 25)$

This simplifies to:

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

Knowing that the coordinates of point A are $3$, we find the gradient:

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

To find point B, we need to solve for $x$ using $f'(x) = 25$ with:

$12x^{2} - 16x - 35 = 25.$

Rearranging gives:

$12x^{2} - 16x - 60 = 0.$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{16 \pm \sqrt{(-16)^{2} - 4 \cdot 12 \cdot (-60)}}{2 \cdot 12} = \frac{16 \pm \sqrt{256 + 2880}}{24} = \frac{16 \pm \sqrt{3136}}{24} = \frac{16 \pm 56}{24}$

This gives:

$x = 3\text{ (point A)}$ and $x = -\frac{5}{3} \text{ (point B)}.$