The diagram shows part of the curve $y = \frac{16}{(x+1)^2} - 1$, cutting the $x$-axis at $M$. The tangent at the point $L$ on the curve cuts the $x$-axis at $N$.
The gradient of this tangent is $-4$.
Calculate the area of the shaded region $LMN$. Continuation of working space for Question 15.

1. Find the coordinates of $M$, where the curve cuts the $x$-axis.

At $M$, $y = 0$:
$\frac{16}{(x+1)^2} - 1 = 0$
$(x+1)^2 = 16$
Since $x > 0$ from the diagram,
$x + 1 = 4 \implies x = 3$
So, $M$ is $(3, 0)$.

2. Find the $x$-coordinate of $L$ using the given gradient of the tangent.

$y = 16(x+1)^{-2} - 1$
$\frac{dy}{dx} = -32(x+1)^{-3}$
Given the gradient is $-4$:
$-32(x+1)^{-3} = -4$
$(x+1)^3 = 8$
$x + 1 = 2 \implies x = 1$

3. Find the $y$-coordinate of $L$ and the equation of the tangent to find $N$.

When $x = 1$, $y = \frac{16}{(1+1)^2} - 1 = 4 - 1 = 3$.
So, $L$ is $(1, 3)$.

Equation of the tangent at $L$:
$y - 3 = -4(x - 1)$
$y = -4x + 7$
At $N$, the tangent cuts the $x$-axis ($y = 0$):
$-4x + 7 = 0 \implies x = 1.75$
So, $N$ is $(1.75, 0)$.

4. Calculate the area under the curve from $x = 1$ to $x = 3$.

$\text{Area under curve} = \int_{1}^{3} \left( \frac{16}{(x+1)^2} - 1 \right) dx$
$= \left[ -\frac{16}{x+1} - x \right]_{1}^{3}$
$= \left( -\frac{16}{4} - 3 \right) - \left( -\frac{16}{2} - 1 \right)$
$= (-4 - 3) - (-8 - 1)$
$= -7 - (-9) = 2$

5. Calculate the area of the shaded region $LMN$.

The shaded area is the area under the curve minus the area of the triangle formed by the tangent, the $x$-axis, and the vertical line $x = 1$.
$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}$
$= \frac{1}{2} \times (1.75 - 1) \times 3$
$= \frac{1}{2} \times 0.75 \times 3 = 1.125$

$\text{Shaded Area} = 2 - 1.125 = 0.875$