The curves $C_1$ and $C_2$ have equations $y = \frac{2x^2 + 9}{x^2 - 4}$ and $\frac{y^2}{25} - \frac{x^2}{9} = 1$ respectively.

1. Find the asymptotes and axis intercepts of the curve $C_1$.

For $C_1: y = \frac{2x^2 + 9}{x^2 - 4}$

• Vertical asymptotes: Setting the denominator to zero gives $x^2 - 4 = 0 \implies x = -2$ and $x = 2$.
• Horizontal asymptote: As $x \to \pm\infty$, $y \to \frac{2x^2}{x^2} = 2 \implies y = 2$.
• $y$-intercept: When $x = 0$, $y = \frac{9}{-4} = -2.25$. The coordinate is $(0, -2.25)$.
• $x$-intercept: Setting $y = 0$ yields $2x^2 + 9 = 0$, which has no real solutions.

2. Find the asymptotes and axis intercepts of the curve $C_2$.

For $C_2: \frac{y^2}{25} - \frac{x^2}{9} = 1$

• This equation represents a standard hyperbola opening along the $y$-axis.
• Asymptotes: Replacing $1$ with $0$ gives $\frac{y^2}{25} - \frac{x^2}{9} = 0 \implies y = \pm \frac{5}{3}x$.
• $y$-intercepts: When $x = 0$, $y^2 = 25 \implies y = \pm 5$. The coordinates are $(0, 5)$ and $(0, -5)$.
• $x$-intercepts: Setting $y = 0$ yields $-\frac{x^2}{9} = 1$, which has no real solutions.

3. Determine the coordinates of the points of intersection between $C_1$ and $C_2$.

To find the intersections, we can use a graphing calculator to solve the system of equations:
$y = \frac{2x^2 + 9}{x^2 - 4} \quad \text{and} \quad \frac{y^2}{25} - \frac{x^2}{9} = 1$
Graphing both curves and locating their points of intersection yields the coordinates:
$(1.31, -5.46)$, $(-1.31, -5.46)$, $(2.75, 6.78)$ and $(-2.75, 6.78)$ (all given to 3 significant figures).

4. Sketch both curves on the same diagram.

(A valid sketch must be drawn on a Cartesian plane including both curves.)

• Draw the vertical lines $x=2$, $x=-2$ and horizontal line $y=2$ as dotted lines for the asymptotes of $C_1$.
• Draw the diagonal dotted lines $y = \frac{5}{3}x$ and $y = -\frac{5}{3}x$ for the asymptotes of $C_2$.
• Sketch $C_1$: The middle branch is a 'U'-shape opening downwards with a maximum at $(0, -2.25)$, bounded by $x = \pm 2$. The outer branches lie above $y=2$ and asymptotically approach $x = \pm 2$ and $y = 2$.
• Sketch $C_2$: A hyperbola with the upper branch vertex at $(0, 5)$ and the lower branch vertex at $(0, -5)$, bounded tightly by its diagonal asymptotes.
• Ensure the calculated intersection points are correctly positioned relatively on the sketch and labeled.

1. Relate the given inequality to the graphs of $C_1$ and $C_2$.

From the equation of $C_2$, $\frac{y^2}{25} - \frac{x^2}{9} = 1 \implies \frac{y^2}{25} = 1 + \frac{1}{9}x^2 \implies y = \pm 5\sqrt{1 + \frac{1}{9}x^2}$.
Thus, $y = 5\sqrt{1 + \frac{1}{9}x^2}$ represents the upper branch of $C_2$ (where $y \ge 0$).
The inequality $5\sqrt{1 + \frac{1}{9}x^2} \ge \frac{2x^2 + 9}{x^2 - 4}$ requires us to find the range of $x$ for which the upper branch of $C_2$ lies above or intersects $C_1$.

2. Determine the solution using the intersection points and asymptotes from part (i).

From the sketch in part (i), the $x$-coordinates of the intersections between the upper branch of $C_2$ and $C_1$ are found (using a graphing calculator) to be $x \approx -2.75$ and $x \approx 2.75$.
The vertical asymptotes of $C_1$ are at $x = -2$ and $x = 2$.
Observing the relative positions of the two curves, the upper branch of $C_2$ is above or meets $C_1$ when $x \le -2.75$, in the entire interval between the vertical asymptotes $-2 < x < 2$, and when $x \ge 2.75$.