Answer
$ f_{(x)} = 2\cos{(\frac{x}{2})} + 4$
Work Step by Step
$f_{(x)}' = -\sin{(\frac{x}{2})} \longrightarrow f_{(x)} = \int [-\sin{(\frac{x}{2})}]\ dx = -\int \sin{(\frac{x}{2})}\ dx$
Let $u = \dfrac{x}{2} \longrightarrow du = \dfrac{dx}{2} \longrightarrow dx = 2du$
$-\int \sin{(\frac{x}{2})}\ dx = -\int \sin{(u)}\ 2\ du = -2\int \sin{(u)}\ du = -2(-\cos{(u)}) + C$
$\longrightarrow 2\cos{(u)} + C = 2\cos{(\frac{x}{2})} + C \longrightarrow f_{(x)} = 2\cos{(\frac{x}{2})} + C$
Since the graph passes through the point $(0, 6)$, then
$f_{(0)} = 6 \longrightarrow 2\cos{(\frac{0}{2})} + C = 6 \longrightarrow 2\cos{0} + C = 6$
$\longrightarrow 2(1) + C = 6 \longrightarrow 2 + C = 6 \longrightarrow C = 4$
Hence $ f_{(x)} = 2\cos{(\frac{x}{2})} + 4$
