1. **State the problem:** We need to express the area $A$ of the figure as a function of one variable $x$. The figure consists of a rectangle and two right triangles on top. 2. **Identify dimensions:** - Rectangle has height $x+1$ and base $(x-5)+(x+3) = 2x - 2$. - Green triangle has base $x-5$ and height $x$. - Blue triangle has base $x+3$ and height $x$. 3. **Formula for area:** - Area of rectangle: $A_{rect} = \text{base} \times \text{height} = (2x - 2)(x + 1)$. - Area of green triangle: $A_{green} = \frac{1}{2} (x-5)(x)$. - Area of blue triangle: $A_{blue} = \frac{1}{2} (x+3)(x)$. 4. **Total area:** $$ A(x) = A_{rect} + A_{green} + A_{blue} = (2x - 2)(x + 1) + \frac{1}{2} x (x-5) + \frac{1}{2} x (x+3) $$ 5. **Expand and simplify:** $$ (2x - 2)(x + 1) = 2x^2 + 2x - 2x - 2 = 2x^2 - 2 $$ $$ \frac{1}{2} x (x-5) = \frac{1}{2} (x^2 - 5x) = \frac{1}{2} x^2 - \frac{5}{2} x $$ $$ \frac{1}{2} x (x+3) = \frac{1}{2} (x^2 + 3x) = \frac{1}{2} x^2 + \frac{3}{2} x $$ 6. **Sum all parts:** $$ A(x) = 2x^2 - 2 + \left(\frac{1}{2} x^2 - \frac{5}{2} x\right) + \left(\frac{1}{2} x^2 + \frac{3}{2} x\right) $$ 7. **Combine like terms:** $$ A(x) = 2x^2 - 2 + \frac{1}{2} x^2 + \frac{1}{2} x^2 - \frac{5}{2} x + \frac{3}{2} x $$ $$ = 2x^2 - 2 + x^2 - x $$ $$ = 3x^2 - x - 2 $$ **Final answer:** $$ \boxed{A(x) = 3x^2 - x - 2} $$