Answer
$$\eqalign{
& f\left( x \right) = \sqrt {x + 4} \cr
& {\text{Linear approximation at }}\left( {0,2} \right) \to y = 2 + \frac{1}{4}x \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \sqrt {x + 4} ,{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\sqrt {x + 4} } \right] \cr
& f'\left( x \right) = \frac{1}{{2\sqrt {x + 4} }} \cr
& \cr
& {\text{The tangent line at the point }}\left( {c,f\left( c \right)} \right){\text{ is}} \cr
& y = f\left( c \right) + f'\left( c \right)\left( {x - c} \right) \cr
& {\text{We have the point }}\left( {0,2} \right) \cr
& y = 2 + f'\left( 0 \right)\left( {x - 0} \right) \cr
& y = 2 + \frac{1}{{2\sqrt {0 + 4} }}\left( {x - 0} \right) \cr
& y = 2 + \frac{1}{4}\left( {x - 0} \right) \cr
& y = 2 + \frac{1}{4}x \cr
& \cr
& {\text{Graph}} \cr} $$
