Show that the area of the triangle in the figure above is given by: A(x) = -x2/2 + 4x - 15/2

From the figure, the base $BC = 3-x$ and the height $AB = x+2$.

The area of a triangle is given by $A = \frac{1}{2} \times \text{base} \times \text{height}$.

The problem statement gives $A(x) = -\frac{x^2}{2} + 4x - \frac{15}{2}$. There seems to be a discrepancy between the derived area and the given area function. Let's re-check the calculation.

Re-evaluating the problem statement for 23.1: "Show that the area of the triangle in the figure above is given by: $A(x) = -\frac{x^2}{2} + 4x - \frac{15}{2}$".
The image shows the base as $3-x$ and the height as $x+2$.
Let's assume the given area function in the question is correct and the diagram might be misleading or there's a different interpretation of the sides. However, based on the diagram, the area is $A(x) = -\frac{x^2}{2} + \frac{x}{2} + 3$.

If we are to show that the area is $A(x) = -\frac{x^2}{2} + 4x - \frac{15}{2}$, then the sides of the triangle in the diagram must be different from what is labeled.
Let's assume the question implies that the area is $A(x) = -\frac{x^2}{2} + 4x - \frac{15}{2}$ and we need to work with this function for the subsequent parts, even if it doesn't directly match the diagram's dimensions. This is a common type of question where the diagram is illustrative but the function provided is the one to use.

Given the instruction "Show that the area... is given by: $A(x) = -\frac{x^2}{2} + 4x - \frac{15}{2}$", and the diagram showing sides $x+2$ and $3-x$, these two pieces of information are contradictory. If I use the diagram, I get $A(x) = -\frac{x^2}{2} + \frac{x}{2} + 3$. If I assume the given formula is correct, then the diagram's labels are incorrect.

I will proceed by assuming the question intends for us to use the provided formula $A(x) = -\frac{x^2}{2} + 4x - \frac{15}{2}$ for parts 23.2 and 23.3, and that the "Show that" part of 23.1 is either a misstatement or refers to a different set of dimensions not explicitly shown or derivable from the current diagram. Since I cannot derive the given formula from the diagram, I will state this discrepancy and then use the given formula for the subsequent parts.

Discrepancy Note: Based on the provided diagram with base $3-x$ and height $x+2$, the area is $A(x) = \frac{1}{2}(3-x)(x+2) = -\frac{x^2}{2} + \frac{x}{2} + 3$. This does not match the formula $A(x) = -\frac{x^2}{2} + 4x - \frac{15}{2}$ given in the question for 23.1. I will proceed with the given formula $A(x) = -\frac{x^2}{2} + 4x - \frac{15}{2}$ for the following calculations, assuming it is the intended function for the problem.

Given the area function:

To confirm this is a maximum, we can check the second derivative: $A''(x) = -1$, which is less than 0, indicating a maximum.

The value of $x$ for which the area will be maximum is $\boxed{x=4}$.

Substitute $x=4$:

The maximum area of the given triangle is $\boxed{0.5 \text{ square units}}$.