The function f is defined by $f(x)=ax^3+bx^2+cx+d$, where $a,b,c$ and $d$ are real numbers.

1. State the complex conjugate root.

Since $f(x)$ has real coefficients ($a, b, c, d \in \mathbb{R}$), any complex roots must occur in conjugate pairs. Thus, another root is $1+2i$.

2. Form the quadratic factor corresponding to the complex conjugate roots.

The quadratic factor is $(x - (1-2i))(x - (1+2i))$.
Expanding this, we get:
$x^2 - x(1+2i) - x(1-2i) + (1-2i)(1+2i)$
$= x^2 - 2x + (1^2 - (2i)^2)$
$= x^2 - 2x + (1 - (-4))$
$= x^2 - 2x + 5$

3. Multiply the linear and quadratic factors to form the full polynomial.

Since $-1$ is a root, $(x+1)$ is a factor.
We can write $f(x)$ using the leading coefficient $a$:
$f(x) = a(x+1)(x^2 - 2x + 5)$
$= a(x(x^2 - 2x + 5) + 1(x^2 - 2x + 5))$
$= a(x^3 - 2x^2 + 5x + x^2 - 2x + 5)$
$= a(x^3 - x^2 + 3x + 5)$
$= ax^3 - ax^2 + 3ax + 5a$

4. Compare coefficients with the original polynomial equation to find $b$, $c$, and $d$.

Comparing $ax^3 - ax^2 + 3ax + 5a$ with $ax^3 + bx^2 + cx + d$, we obtain:
$b = -a$
$c = 3a$
$d = 5a$

1. Rearrange the equation to separate the terms with $x$ from the constant $d$.

$2x^3 - x^2 - 4x + d = 0 \implies 2x^3 - x^2 - 4x = -d$
Let $g(x) = 2x^3 - x^2 - 4x$. The real roots of $f(x) = 0$ correspond to the $x$-coordinates of the intersections between the curve $y = g(x)$ and the horizontal line $y = -d$.

2. Differentiate $g(x)$ and equate to zero to find the $x$-coordinates of the stationary points.

$g'(x) = 6x^2 - 2x - 4$
Setting $g'(x) = 0$:
$6x^2 - 2x - 4 = 0$
$3x^2 - x - 2 = 0$
$(3x + 2)(x - 1) = 0$
$x = -\frac{2}{3} \text{ or } x = 1$

3. Evaluate the $y$-coordinates of $g(x)$ at the stationary points.

When $x = -\frac{2}{3}$:
$g\left(-\frac{2}{3}\right) = 2\left(-\frac{2}{3}\right)^3 - \left(-\frac{2}{3}\right)^2 - 4\left(-\frac{2}{3}\right) = -\frac{16}{27} - \frac{12}{27} + \frac{72}{27} = \frac{44}{27}$
When $x = 1$:
$g(1) = 2(1)^3 - (1)^2 - 4(1) = -3$
The curve $y = g(x)$ has a local maximum at $\left(-\frac{2}{3}, \frac{44}{27}\right)$ and a local minimum at $(1, -3)$.

4. Deduce the inequalities for $d$ such that there is only one real root.

For the equation to have exactly one real root, the line $y = -d$ must intersect the curve $y = g(x)$ exactly once. This happens if the horizontal line is strictly above the local maximum or strictly below the local minimum.

Line above local maximum:
$-d > \frac{44}{27} \implies d < -\frac{44}{27}$
Line below local minimum:
$-d < -3 \implies d > 3$

1. Identify the vertical asymptote from the root of the original function.

The original graph $y = f(x)$ meets the $x$-axis at $(-k, 0)$, so $f(-k) = 0$. Thus, the reciprocal graph $y = \frac{1}{f(x)}$ has a vertical asymptote at $x = -k$.

2. Identify the horizontal asymptote from the end behaviour of the cubic function.

As $f(x)$ is a cubic polynomial ($f(x) = ax^3+bx^2+cx+d$), $f(x) \to \pm\infty$ as $x \to \pm\infty$. Therefore, $\frac{1}{f(x)} \to 0$, giving a horizontal asymptote at $y = 0$ (the $x$-axis).

3. Determine the coordinates of the $y$-intercept.

The curve $y = f(x)$ passes through the $y$-axis at $(0, k)$, meaning $f(0) = k$. Thus, the $y$-intercept of $y = \frac{1}{f(x)}$ is $(0, \frac{1}{k})$.

4. Determine the shape of the curves based on the signs and turning points of $f(x)$ to sketch the final graph.

For $x < -k$, $f(x) < 0 \implies y < 0$. The curve approaches $y=0$ from below as $x \to -\infty$ and goes to $-\infty$ as $x \to -k^-$.

For $x > -k$, $f(x) > 0 \implies y > 0$. The curve comes down from $\infty$ as $x \to -k^+$. Since $f(x)$ has a local maximum and then a local minimum in this region, $\frac{1}{f(x)}$ will reflect this by having a local minimum followed by a local maximum, remaining positive before asymptotically approaching $y=0$ as $x \to \infty$.