If the minimum of x²+4x+K is -10, find the value of K. Express 4x²-3x-1 in the form a(x+b)²+c and 8+2x-x² in the form a-(x+b)². Find the maximum and the value of x where it occurs.

For $x^2 + 4x + K$, we have $a=1$, $b=4$, and $c=K$.

We are given that the minimum value is $-10$.

Multiply both sides by 4:

Add 16 to both sides:

Divide by 4:

The value of $K$ is $\boxed{-6}$.

: Express $4x^2 - 3x - 1$ in the form $a(x+b)^2 + c$ where $a$, $b$ and $c$ are constant. Hence find the coordinate of the turning point.

Take half of the coefficient of $x$ ($\frac{-3}{4}$), square it, and add and subtract it.
Half of $\frac{-3}{4}$ is $\frac{-3}{8}$. Squaring it gives $\left(\frac{-3}{8}\right)^2 = \frac{9}{64}$.

The expression in the form $a(x+b)^2 + c$ is $\boxed{4\left(x - \frac{3}{8}\right)^2 - \frac{25}{16}}$.

For a quadratic in the form $a(x+b)^2 + c$, the turning point is at $(-b, c)$.
From our expression, $b = -\frac{3}{8}$ and $c = -\frac{25}{16}$.
So, $-b = -(-\frac{3}{8}) = \frac{3}{8}$.
The coordinate of the turning point is $\boxed{\left(\frac{3}{8}, -\frac{25}{16}\right)}$.

: Express $8 + 2x - x^2$ in the form $a - (x+b)^2$ where $a$ and $b$ are constant. Hence find the maximum and the exact value of $x$ where it occurs.

Take half of the coefficient of $x$ (which is $-2$), square it, and add and subtract it.
Half of $-2$ is $-1$. Squaring it gives $(-1)^2 = 1$.

Rearrange to the form $a - (x+b)^2$:

The expression in the form $a - (x+b)^2$ is $\boxed{9 - (x - 1)^2}$.

The term $(x-1)^2$ is always greater than or equal to 0. To maximize $9 - (x-1)^2$, we need to minimize $(x-1)^2$.
The minimum value of $(x-1)^2$ is 0, which occurs when $x-1=0$, so $x=1$.
When $(x-1)^2 = 0$, the expression becomes $9 - 0 = 9$.
The maximum value is $\boxed{9}$.
The exact value of $x$ where it occurs is $\boxed{1}$.