Answer
$$\eqalign{
& {\text{At }}t = 5,{\text{ the slope is }}10 \cr
& \frac{{{d^2}y}}{{d{x^2}}} > 0{\text{ we can conclude that the graph is concave upward}} \cr} $$
Work Step by Step
$$\eqalign{
& x = t - 6,{\text{ }}y = {t^2},{\text{ }}t = 5 \cr
& {\text{By theorem 10}}{\text{.7}}{\text{ the slope is}} \cr
& \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}},{\text{ }}dx/dt \ne 0 \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{d}{{dt}}\left[ {{t^2}} \right]}}{{\frac{d}{{dt}}\left[ {t - 6} \right]}} \cr
& \frac{{dy}}{{dx}} = \frac{{2t}}{1} \cr
& {\text{Evaluate at }}t = 5 \cr
& \frac{{dy}}{{dx}} = 2\left( 5 \right) \cr
& \frac{{dy}}{{dx}} = 10 \cr
& {\text{Calculate the second derivative }}\frac{{{d^2}y}}{{d{x^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left[ {\frac{{dy}}{{dx}}} \right]}}{{dx/dt}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left[ {2t} \right]}}{1} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = 2 \cr
& {\text{Evaluate at }}t = 5 \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = 5}} = 2 \cr
& \cr
& {\text{At }}t = 5,{\text{ the slope is }}10 \cr
& \frac{{{d^2}y}}{{d{x^2}}} > 0{\text{ we can conclude that the graph is concave upward}} \cr} $$
