The matrix m = 1 & 5 2 & 7 and matrix n = 2 & 0 6 & 1 find:

$$
nm = \begin{pmatrix} 2 & 0 \\ 6 & 1 \end{pmatrix} \begin{pmatrix} 1 & 5 \\ 2 & 7 \end{pmatrix}
$$

$$
nm = \begin{pmatrix} (2 \times 1) + (0 \times 2) & (2 \times 5) + (0 \times 7) \\ (6 \times 1) + (1 \times 2) & (6 \times 5) + (1 \times 7) \end{pmatrix}
$$

$$
nm = \begin{pmatrix} 2 + 0 & 10 + 0 \\ 6 + 2 & 30 + 7 \end{pmatrix}
$$
$$
nm = \begin{pmatrix} 2 & 10 \\ 8 & 37 \end{pmatrix}
$$
The product $nm$ is $\boxed{\begin{pmatrix} 2 & 10 \\ 8 & 37 \end{pmatrix}}$.

The relationship is $y = kx^2$.

**a)** Find the constant $k$.

$$
96 = k(4^2)
$$
$$
96 = 16k
$$

$$
k = \frac{96}{16}
$$
$$
k = 6
$$
The constant $k$ is $\boxed{6}$.

**b)** Find $y$ where $x=5$.

$$
y = 6(5^2)
$$
$$
y = 6(25)
$$
$$
y = 150
$$
The value of $y$ is $\boxed{150}$.

**c)** Find $x$ where $y=24$.

$$
24 = 6x^2
$$

$$
x^2 = \frac{24}{6}
$$
$$
x^2 = 4
$$

$$
x = \pm\sqrt{4}
$$
$$
x = \pm 2
$$
The values of $x$ are $\boxed{2 \text{ or } -2}$.

$$
x^2 = \frac{5}{2}
$$

$$
x = \pm\sqrt{\frac{5}{2}}
$$

$$
x = \pm\frac{\sqrt{5}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}
$$
$$
x = \pm\frac{\sqrt{10}}{2}
$$
The solutions for $x$ are $\boxed{\pm\frac{\sqrt{10}}{2}}$.

The relationship is $P = k \frac{m}{n^2}$.

**a)** Find the constant $k$.

$$
6 = k \frac{27}{2^2}
$$
$$
6 = k \frac{27}{4}
$$

$$
k = \frac{6 \times 4}{27}
$$
$$
k = \frac{24}{27}
$$

$$
k = \frac{8}{9}
$$
The constant $k$ is $\boxed{\frac{8}{9}}$.

**b)** Find $P$ when $n=-6$ and $m=-2$.

$$
P = \frac{8}{9} \frac{-2}{(-6)^2}
$$
$$
P = \frac{8}{9} \frac{-2}{36}
$$

$$
P = \frac{8}{9} \times \left(-\frac{1}{18}\right)
$$
$$
P = -\frac{8}{162}
$$
$$
P = -\frac{4}{81}
$$
The value of $P$ is $\boxed{-\frac{4}{81}}$.

**c)** Find $n$ when $m=8$ and $P=1$.

$$
1 = \frac{8}{9} \frac{8}{n^2}
$$
$$
1 = \frac{64}{9n^2}
$$

$$
9n^2 = 64
$$
$$
n^2 = \frac{64}{9}
$$

$$
n = \pm\sqrt{\frac{64}{9}}
$$
$$
n = \pm\frac{8}{3}
$$
The values of $n$ are $\boxed{\frac{8}{3} \text{ or } -\frac{8}{3}}$.

**a)** Find the coordinates of A and B.

$$
(x-4)(x+2) = 0
$$

$$
x-4=0 \Rightarrow x=4
$$
$$
x+2=0 \Rightarrow x=-2
$$

The coordinates of A are $\boxed{(-2, 0)}$ and B are $\boxed{(4, 0)}$.

**b)** Find the minimum point on the graph.

$$
y = (1)^2 - 2(1) - 8
$$
$$
y = 1 - 2 - 8
$$
$$
y = -9
$$
The minimum point on the graph is $\boxed{(1, -9)}$.