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