Let f(x) = -x2 + 4x + 5 and g(x) = 2x + 2. The points A and C are on the x-axis and the line y=2x+2 respectively. The point B is the turning point of f. Calculate the maximum length of EH for f > g, w

The function is given by $f(x) = -x^2 + 4x + 5$.
For a parabola in the form $ax^2 + bx + c$, the x-coordinate of the turning point is given by $x = -\frac{b}{2a}$.
Here, $a = -1$ and $b = 4$.

$$x_B = -\frac{4}{2(-1)} = -\frac{4}{-2} = 2$$

Substitute $x_B = 2$ into $f(x)$ to find the y-coordinate:

$$y_B = f(2) = -(2)^2 + 4(2) + 5 = -4 + 8 + 5 = 9$$

The coordinates of B are $(2, 9)$.

$$\boxed{\text{B}(2, 9)}$$

The point C is given as $(3, 8)$.
To find point A, we need to find the x-intercepts of $f(x)$ by setting $f(x) = 0$:

$$-x^2 + 4x + 5 = 0$$

Multiply by -1:

$$x^2 - 4x - 5 = 0$$

Factor the quadratic equation:

$$(x - 5)(x + 1) = 0$$

The x-intercepts are $x = 5$ and $x = -1$. From the diagram, A is the negative x-intercept, so A is $(-1, 0)$.
Now, calculate the slope $m$ of the line passing through A$(-1, 0)$ and C$(3, 8)$:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 0}{3 - (-1)} = \frac{8}{4} = 2$$

Using the point-slope form $y - y_1 = m(x - x_1)$ with point A$(-1, 0)$:

$$y - 0 = 2(x - (-1))$$ $$y = 2(x + 1)$$ $$y = 2x + 2$$

Thus, the equation of the line is $g(x) = 2x + 2$.

The length of EH is the vertical distance between $f(x)$ and $g(x)$, which is $D(x) = f(x) - g(x)$.

$$D(x) = (-x^2 + 4x + 5) - (2x + 2)$$ $$D(x) = -x^2 + 4x + 5 - 2x - 2$$ $$D(x) = -x^2 + 2x + 3$$

This is a downward-opening parabola, so its maximum value occurs at its turning point. The x-coordinate of the turning point is $x = -\frac{b}{2a}$.
Here, $a = -1$ and $b = 2$.

$$x = -\frac{2}{2(-1)} = -\frac{2}{-2} = 1$$

Substitute $x = 1$ into $D(x)$ to find the maximum length:

$$D(1) = -(1)^2 + 2(1) + 3 = -1 + 2 + 3 = 4$$

The maximum length of EH is 4 units.

$$\boxed{4 \text{ units}}$$

The function $k(x)$ is given as $k(x) = f(x+m) = -x^2 - 2mx - m^2 + 4x + 4m + 5$.
We can rewrite $k(x)$ by grouping terms:

$$k(x) = -x^2 + (4 - 2m)x + (-m^2 + 4m + 5)$$

For $g(x)$ to be a tangent to $k(x)$, there must be exactly one point of intersection between the two functions. Set $k(x) = g(x)$:

$$-x^2 + (4 - 2m)x + (-m^2 + 4m + 5) = 2x + 2$$

Rearrange the equation to form a quadratic equation $Ax^2 + Bx + C = 0$:

$$-x^2 + (4 - 2m - 2)x + (-m^2 + 4m + 5 - 2) = 0$$ $$-x^2 + (2 - 2m)x + (-m^2 + 4m + 3) = 0$$

Multiply by -1 to make the leading coefficient positive:

$$x^2 - (2 - 2m)x - (-m^2 + 4m + 3) = 0$$ $$x^2 + (2m - 2)x + (m^2 - 4m - 3) = 0$$

For a tangent, the discriminant ($\Delta = B^2 - 4AC$) of this quadratic equation must be zero.
Here, $A = 1$, $B = (2m - 2)$, and $C = (m^2 - 4m - 3)$.

$$(2m - 2)^2 - 4(1)(m^2 - 4m - 3) = 0$$

Expand the terms:

$$(4m^2 - 8m + 4) - (4m^2 - 16m - 12) = 0$$ $$4m^2 - 8m + 4 - 4m^2 + 16m + 12 = 0$$

Combine like terms:

$$(-8m + 16m) + (4 + 12) = 0$$ $$8m + 16 = 0$$ $$8m = -16$$ $$m = -2$$

The value of $m$ is -2.

$$\boxed{m = -2}$$