Let $\displaystyle {x}={4}{t}^{{2}},{y}=-{2}{t}^{{3}}+{18}{t}^{{2}}$.

Determine $\displaystyle \frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}\right.}}$ as a function of $\displaystyle {t}$, then find the slope of the parametric curve at $\displaystyle {t}={6}$.

    $\displaystyle \frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}\right.}}$ =   Preview Question 6 Part 1 of 5  

    $\displaystyle \frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}\right.}}{\left({6}\right)}=$Preview Question 6 Part 2 of 5  

Determine $\displaystyle \frac{{{d}^{{2}}{y}}}{{\left.{d}{x}\right.}^{{2}}}$ as a function of $\displaystyle {t}$, then find the concavity of the parametric curve at $\displaystyle {t}={6}$.

    $\displaystyle \frac{{{d}^{{2}}{y}}}{{\left.{d}{x}\right.}^{{2}}}$ =   Preview Question 6 Part 3 of 5  

    $\displaystyle \frac{{{d}^{{2}}{y}}}{{\left.{d}{x}\right.}^{{2}}}{\left({6}\right)}=$Preview Question 6 Part 4 of 5  

At $\displaystyle {t}={6}$, the parametric curve has