Answer
$$\frac{{{d^2}y}}{{d{x^2}}} < 0{\text{ we can conclude that the graph is concave downward}}$$
Work Step by Step
$$\eqalign{
& x = {t^2} + 5t + 4,{\text{ }}y = 4t,{\text{ }}t = 0 \cr
& {\text{By theorem 10}}{\text{.7}}{\text{ the slope is}} \cr
& \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}},{\text{ }}dx/dt \ne 0 \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{d}{{dt}}\left[ {4t} \right]}}{{\frac{d}{{dt}}\left[ {{t^2} + 5t + 4} \right]}} \cr
& \frac{{dy}}{{dx}} = \frac{4}{{2t + 5}} \cr
& {\text{Evaluate at }}t = 0 \cr
& m = \frac{{dy}}{{dx}} = \frac{4}{{2\left( 0 \right) + 5}} \cr
& m = \frac{4}{5} \cr
& {\text{Calculate the second derivative }}\frac{{{d^2}y}}{{d{x^2}}} \cr
& \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left[ {\frac{{dy}}{{dx}}} \right]}}{{dx/dt}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left[ {\frac{4}{{2t + 5}}} \right]}}{{2t + 5}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \frac{{2\left( 4 \right)}}{{{{\left( {2t + 5} \right)}^2}}}}}{{2t + 5}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{8}{{{{\left( {2t + 5} \right)}^3}}} \cr
& {\text{Evaluate at }}t = 0 \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = 0}} = - \frac{8}{{{{\left( {2\left( 0 \right) + 5} \right)}^3}}} \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = 0}} = - \frac{8}{{125}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} < 0{\text{ we can conclude that the graph is concave downward}} \cr} $$
