Answer
$$\eqalign{
& {\text{At }}t = 1,{\text{ the slope is }} - \frac{1}{{{e^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} > 0{\text{ we can conclude that the graph is concave upward}} \cr} $$
Work Step by Step
$$\eqalign{
& x = {e^t},{\text{ }}y = {e^{ - t}},{\text{ }}t = 1 \cr
& {\text{By theorem 10}}{\text{.7}}{\text{ the slope is}} \cr
& \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}},{\text{ }}dx/dt \ne 0 \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{d}{{dt}}\left[ {{e^{ - t}}} \right]}}{{\frac{d}{{dt}}\left[ {{e^t}} \right]}} \cr
& \frac{{dy}}{{dx}} = \frac{{ - {e^{ - t}}}}{{{e^t}}} \cr
& \frac{{dy}}{{dx}} = - {e^{ - 2t}} \cr
& {\text{Evaluate at }}t = 1 \cr
& \frac{{dy}}{{dx}} = - {e^{ - 2\left( 1 \right)}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{{e^2}}} \cr
& {\text{Calculate the second derivative }}\frac{{{d^2}y}}{{d{x^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left[ {\frac{{dy}}{{dx}}} \right]}}{{dx/dt}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left[ { - {e^{ - 2t}}} \right]}}{{{e^t}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{2{e^{ - 2t}}}}{{{e^t}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = 2{e^{ - 3t}} \cr
& {\text{Evaluate at }}t = 1 \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = 1}} = 2{e^{ - 3\left( 1 \right)}} = 2{e^{ - 3}} \cr
& \cr
& {\text{At }}t = 1,{\text{ the slope is }} - \frac{1}{{{e^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} > 0{\text{ we can conclude that the graph is concave upward}} \cr} $$
