Answer
$$\frac{{{d^2}y}}{{d{x^2}}} < 0{\text{ we can conclude that the graph is concave downward}}$$
Work Step by Step
$$\eqalign{
& x = 2 + \sec \theta ,{\text{ }}y = 1 + 2\tan \theta ,{\text{ }}\theta = \frac{\pi }{6} \cr
& {\text{By theorem 10}}{\text{.7}}{\text{ the slope is}} \cr
& \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}},{\text{ }}dx/dt \ne 0,{\text{ for this exercise consider }}t = \theta \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{d}{{d\theta }}\left[ {1 + 2\tan \theta } \right]}}{{\frac{d}{{d\theta }}\left[ {2 + \sec \theta } \right]}} \cr
& \frac{{dy}}{{dx}} = \frac{{2{{\sec }^2}\theta }}{{\sec \theta \tan \theta }} \cr
& \frac{{dy}}{{dx}} = \frac{{2\sec \theta }}{{\tan \theta }} \cr
& {\text{Evaluate at }}\theta = \frac{\pi }{6} \cr
& {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \pi /6}} = \frac{{2\sec \left( {\pi /6} \right)}}{{\tan \left( {\pi /6} \right)}} \cr
& {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \pi /6}} = \frac{{\frac{{4\sqrt 3 }}{3}}}{{\frac{{\sqrt 3 }}{3}}} = 4 \cr
& {\text{Calculate the second derivative }}\frac{{{d^2}y}}{{d{x^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{d\theta }}\left[ {\frac{{dy}}{{dx}}} \right]}}{{dx/d\theta }} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{d\theta }}\left[ {\frac{{2\sec \theta }}{{\tan \theta }}} \right]}}{{\sec \theta \tan \theta }} = \frac{{\frac{d}{{d\theta }}\left[ {2\csc \theta } \right]}}{{\sec \theta \tan \theta }} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 2\csc \theta \cot \theta }}{{\sec \theta \tan \theta }} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - 2\left( {\frac{1}{{\sin \theta }}} \right)\left( {\frac{{\cos \theta }}{{\sin \theta }}} \right)\left( {\frac{{\cos \theta }}{1}} \right)\left( {\frac{{\cos \theta }}{{\sin \theta }}} \right) \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - 2{\cot ^3}\theta \cr
& {\text{Evaluate at }}\theta = \frac{\pi }{6} \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\theta = \frac{\pi }{6}}} = - 2{\cot ^3}\left( {\frac{\pi }{6}} \right) \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\theta = \frac{\pi }{6}}} = - 6\sqrt 3 \cr
& \frac{{{d^2}y}}{{d{x^2}}} < 0{\text{ we can conclude that the graph is concave downward}} \cr} $$
