Sketch the graph of $(x-2)^2 + 4(y+2)^2 = 16$, stating the coordinates of all the vertices. On the same diagram, sketch the graph of $y = \frac{x^2 - 2x + 6}{x + 1}$, stating the equations of any asymptotes, the coordinates of turning points and the points of intersection with the axes. Find the range of values of $m$, where $m > 0$, such that $(x-2)^2 + 4m^2 \left( \frac{x^2 - 2x + 6}{x + 1} + 2 \right)^2 = 16$ has no real solutions.

1. Rewrite the given equation into the standard form of an ellipse.

$(x-2)^2 + 4(y+2)^2 = 16$
Divide by 16 throughout:
$\frac{(x-2)^2}{16} + \frac{(y+2)^2}{4} = 1$

2. Identify the centre, major, and minor axes lengths to determine the vertices.

Comparing with $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$:
Centre $(h, k) = (2, -2)$
Horizontal semi-axis $a = \sqrt{16} = 4$
Vertical semi-axis $b = \sqrt{4} = 2$

Horizontal vertices are at $(2 \pm 4, -2)$:
$\Rightarrow (-2, -2)$ and $(6, -2)$

Vertical vertices are at $(2, -2 \pm 2)$:
$\Rightarrow (2, 0)$ and $(2, -4)$

3. Sketch the ellipse indicating the calculated vertices.

Draw a smooth oval shape centered at $(2, -2)$. Ensure the curve passes through the points $(-2, -2)$, $(6, -2)$, $(2, 0)$, and $(2, -4)$. The graph should touch the $x$-axis at $(2, 0)$ and not cross it.

1. Rewrite the rational function using algebraic division to identify the asymptotes.

y = rac{x^2 - 2x + 6}{x + 1} = rac{x(x + 1) - 3(x + 1) + 9}{x + 1} = x - 3 + rac{9}{x + 1}
The vertical asymptote is $x = -1$.
As $x \to \pm \infty$, $\frac{9}{x + 1} \to 0$, so the oblique asymptote is $y = x - 3$.

2. Find the coordinates of the turning points using differentiation.

$\frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{9}{(x + 1)^2}$
Set $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ for turning points:
$1 - \frac{9}{(x + 1)^2} = 0 \implies (x + 1)^2 = 9 \implies x + 1 = \pm 3$
$x = 2 \quad \text{or} \quad x = -4$
When $x = 2$, $y = 2 - 3 + \frac{9}{3} = 2$.
When $x = -4$, $y = -4 - 3 + \frac{9}{-3} = -10$.
The turning points are $(2, 2)$ and $(-4, -10)$.

3. Determine the points of intersection with the axes.

For the $y$-intercept, set $x = 0$:
$y = \frac{0 - 0 + 6}{0 + 1} = 6$
For the $x$-intercepts, set $y = 0$:
$x^2 - 2x + 6 = 0$
Discriminant $\Delta = (-2)^2 - 4(1)(6) = -20 < 0$.
Since the discriminant is negative, there are no real roots, meaning the curve does not intersect the $x$-axis.
The point of intersection with the axes is $(0, 6)$.

4. Sketch the graph of the rational function on the same diagram.

Draw the two branches of the rational curve on the same axes used in part (i). Ensure the branches approach the asymptotes $x = -1$ and $y = x - 3$, passing through $(0, 6)$, with a local minimum at $(2, 2)$ and a local maximum at $(-4, -10)$.

1. Relate the given equation to algebraic bounds.

Let $y = \frac{x^2 - 2x + 6}{x + 1}$. The given equation is:
$(x-2)^2 + 4m^2(y+2)^2 = 16$
For any real solution $x$, we know $(x-2)^2 \ge 0$, which implies:
$4m^2(y+2)^2 \le 16$
$m^2(y+2)^2 \le 4$
Since $m > 0$, taking the square root gives:
$m|y+2| \le 2 \implies |y+2| \le \frac{2}{m}$

2. Use the range of $y$ from part (ii) to find the condition for no real solutions.

From part (ii), the turning points of $y$ are $(2, 2)$ and $(-4, -10)$. The range of values for $y$ is $y \ge 2$ or $y \le -10$.
Evaluating $|y+2|$ for these ranges:
For $y \ge 2$, $y + 2 \ge 4 \implies |y+2| \ge 4$.
For $y \le -10$, $y + 2 \le -8 \implies |y+2| \ge 8$.
Thus, the minimum value of $|y+2|$ for any valid $x$ is $4$.
For the equation to have no real solutions, the necessary values of $|y+2|$ must lie strictly outside the possible region $|y+2| \le \frac{2}{m}$.
$4 > \frac{2}{m}$
Since $m > 0$,
$4m > 2 \implies m > \frac{1}{2}$