1. The problem asks to find $h(3)$ where $h(x) = f(x) + \frac{g(x) + 2x}{4}$.\n\n2. We are given piecewise linear graphs for $f$ and $g$. We need to find $f(3)$ and $g(3)$ from the graphs.\n\n3. From the graph of $f$: The segment from $(2,1)$ to $(3,4)$ is linear. The slope is $\frac{4-1}{3-2} = 3$. At $x=3$, $f(3) = 4$.\n\n4. From the graph of $g$: The segment from $(2,4)$ to $(3,2)$ is linear. The slope is $\frac{2-4}{3-2} = -2$. At $x=3$, $g(3) = 2$.\n\n5. Substitute into $h(3)$: $$h(3) = f(3) + \frac{g(3) + 2 \cdot 3}{4} = 4 + \frac{2 + 6}{4} = 4 + \frac{8}{4} = 4 + 2 = 6.$$\n\n6. The value $h(3) = 6$ is not among the given options, so let's re-check the problem statement carefully. The problem states $h(x) = f(x) + \frac{g(x) + 2x}{4}$.\n\n7. Re-examining the graph for $f(3)$: The segment from $(3,4)$ to $(5,4)$ is horizontal, so $f(3) = 4$ is correct. For $g(3)$: The segment from $(2,4)$ to $(3,2)$ is linear with slope $-2$, so $g(3) = 2$ is correct.\n\n8. Therefore, $h(3) = 4 + \frac{2 + 6}{4} = 4 + 2 = 6$.\n\n9. Since none of the options match 6, the problem might have a typo or the options correspond to a different expression.\n\n10. If we consider $h(x) = \frac{f(x) + g(x) + 2x}{4}$ instead, then: $$h(3) = \frac{4 + 2 + 6}{4} = \frac{12}{4} = 3.$$ Still no match.\n\n11. Alternatively, if $h(x) = \frac{f(x) + g(x)}{4} + \frac{2x}{4} = \frac{f(x) + g(x)}{4} + \frac{x}{2}$, then: $$h(3) = \frac{4 + 2}{4} + \frac{3}{2} = \frac{6}{4} + 1.5 = 1.5 + 1.5 = 3.$$ No match again.\n\n12. Given the confusion, the best consistent answer from the original formula and graphs is $h(3) = 6$.