Answer
$$\eqalign{
& {\text{At }}t = 3,{\text{ the slope is }} - \frac{4}{5} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = 0,{\text{ Neither concave upward nor downward}} \cr} $$
Work Step by Step
$$\eqalign{
& x = 2 + 5t,{\text{ }}y = 1 - 4t,{\text{ }}t = 3 \cr
& {\text{By theorem 10}}{\text{.7}}{\text{ the slope is}} \cr
& \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}},{\text{ }}dx/dt \ne 0 \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{d}{{dt}}\left[ {1 - 4t} \right]}}{{\frac{d}{{dt}}\left[ {2 + 5t} \right]}} \cr
& \frac{{dy}}{{dx}} = - \frac{4}{5} \cr
& {\text{Evaluate at }}t = 3 \cr
& \frac{{dy}}{{dx}} = - \frac{4}{5} \cr
& {\text{Calculate the second derivative }}\frac{{{d^2}y}}{{d{x^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left[ {\frac{{dy}}{{dx}}} \right]}}{{dx/dt}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left[ { - \frac{4}{5}} \right]}}{5} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = 0 \cr
& {\text{Evaluate at }}t = 3 \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = 3}} = 0 \cr
& \cr
& {\text{At }}t = 3,{\text{ the slope is }} - \frac{4}{5} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = 0,{\text{ Neither concave upward nor downward}} \cr} $$
