Find the slope tangent line to the curve of \(f(x) =( \sqrt{}x+2) -1\) at X=2. Find the equation of tangent line to the function f at x=2 and graph both the function and the tangent line.
\(m= lim_{x\rightarrow 2} = f(x)-f(2) / x-2 \)
\(\sqrt{}x+2 -1 - (\sqrt{}x+2 -1) / x-2\)
I believe that the slope will be 1/4 however I'm not sure exactly how to get the answer.
f(x) = sqrt(x+2) -1
f' = 1/ (2(sqrtx+2)) at x = 2 this will give you the slope AT x=2 1 /(2(sqrt2+2)) = 1/4