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