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.

Here are the solutions to the problems.

: If the minimum of $x^2 + 4x + K$ is $-10$, find the value of $K$.

Step 1: Identify the coefficients of the quadratic equation $ax^2 + bx + c$. For $x^2 + 4x + K$, we have $a=1$, $b=4$, and $c=K$.

Step 2: Use the formula for the minimum value of a quadratic, which is $\frac{4ac - b^2}{4a}$. We are given that the minimum value is $-10$. $-10 = \frac{4(1)(K) - (4)^2}{4(1)}$

Step 3: Solve the equation for $K$. $-10 = \frac{4K - 16}{4}$ Multiply both sides by 4: $-40 = 4K - 16$ Add 16 to both sides: $-40 + 16 = 4K$ $-24 = 4K$ Divide by 4: $K = \frac{-24}{4}$ $K = -6$ 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.

Step 1: Factor out the coefficient of $x^2$ from the terms involving $x$. $y = 4x^2 - 3x - 1$ $y = 4\left(x^2 - \frac{3}{4}x\right) - 1$

Step 2: Complete the square inside the parenthesis. 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 $\left(\frac{-3}{8}\right)^2 = \frac{9}{64}$. $y = 4\left(x^2 - \frac{3}{4}x + \frac{9}{64} - \frac{9}{64}\right) - 1$

Step 3: Group the perfect square trinomial and distribute the 4. $y = 4\left(\left(x - \frac{3}{8}\right)^2 - \frac{9}{64}\right) - 1$ $y = 4\left(x - \frac{3}{8}\right)^2 - 4\left(\frac{9}{64}\right) - 1$ $y = 4\left(x - \frac{3}{8}\right)^2 - \frac{9}{16} - 1$

Step 4: Combine the constant terms. $y = 4\left(x - \frac{3}{8}\right)^2 - \frac{9}{16} - \frac{16}{16}$ $y = 4\left(x - \frac{3}{8}\right)^2 - \frac{25}{16}$ The expression in the form $a(x+b)^2 + c$ is $\boxed{4\left(x - \frac{3}{8}\right)^2 - \frac{25}{16}}$.

Step 5: Find the coordinate of the turning point. 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.

Step 1: Rearrange the quadratic into standard form and factor out $-1$. $y = -x^2 + 2x + 8$ $y = -(x^2 - 2x - 8)$

Step 2: Complete the square inside the parenthesis. 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 $(-1)^2 = 1$. $y = -(x^2 - 2x + 1 - 1 - 8)$

Step 3: Group the perfect square trinomial. $y = -((x - 1)^2 - 9)$

Step 4: Distribute the negative sign. $y = -(x - 1)^2 + 9$ Rearrange to the form $a - (x+b)^2$: $y = 9 - (x - 1)^2$ The expression in the form $a - (x+b)^2$ is $\boxed{9 - (x - 1)^2}$.

Step 5: Find the maximum value and the value of $x$ where it occurs. 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 $\boxed{1}$.