Express $-2x^{2} + 3x + 2$ in the form $a(x + b)^{2} + c$ where $a$, $b$ and $c$ are constants. Use your answer from part (a) to explain why the curves with equations $y = -2x^{2} + 3x + 2$ and $(x - 9)^{2} + (y - 15)^{2} = 25$ will not intersect.

1. Factor out the coefficient of $x^2$ from the $x$ terms

$-2x^{2} + 3x + 2 = -2\left(x^{2} - \frac{3}{2}x\right) + 2$

2. Complete the square for the expression inside the bracket

$= -2\left[ \left(x - \frac{3}{4}\right)^{2} - \left(\frac{3}{4}\right)^{2} \right] + 2$
$= -2\left[ \left(x - \frac{3}{4}\right)^{2} - \frac{9}{16} \right] + 2$

3. Expand and simplify to reach the final form

$= -2\left(x - \frac{3}{4}\right)^{2} + \frac{9}{8} + 2$
$= -2\left(x - \frac{3}{4}\right)^{2} + \frac{25}{8}$

1. Determine the maximum $y$-value of the parabola using the result from part (a)

From part (a), $y = -2x^2 + 3x + 2 = -2\left(x - \frac{3}{4}\right)^2 + \frac{25}{8}$.
Since $-2\left(x - \frac{3}{4}\right)^2 \leqslant 0$ for all real $x$, the maximum value of $y$ for the parabola is $\frac{25}{8} = 3.125$.

2. Identify the center and radius of the circle

The equation of the circle is $(x - 9)^2 + (y - 15)^2 = 25$.
Comparing with $(x - a)^2 + (y - b)^2 = r^2$, the center is $(9, 15)$ and the radius is $r = \sqrt{25} = 5$.

3. Determine the minimum $y$-value of the circle and conclude

The minimum $y$-value of the circle is the $y$-coordinate of the center minus the radius:
Minimum $y = 15 - 5 = 10$.
Since the maximum $y$-value of the parabola ($3.125$) is strictly less than the minimum $y$-value of the circle ($10$), the highest point of the parabola lies below the lowest point of the circle. Hence, the two curves will not intersect.