Question

find the slope of the tangent to the graph of y = sqrt(x+2) at x= -1
PLEASE SHOW ALL STEPS THANK YOU! :)

Answer 1

\[\Large y(x) = \sqrt{x+2}\]



\[\Large y^{\prime}(x) = \frac{d}{dx}(\sqrt{x+2})\]



\[\Large y^{\prime}(x) = \frac{1}{2\sqrt{x+2}}\]



\[\Large y^{\prime}(-1) = \frac{1}{2\sqrt{-1+2}}\]



\[\Large y^{\prime}(-1) = \frac{1}{2\sqrt{1}}\]



\[\Large y^{\prime}(-1) = \frac{1}{2(1)}\]



\[\Large y^{\prime}(-1) = \frac{1}{2}\]



So the slope of the tangent line at x = -1 is \[\Large \frac{1}{2}\]