Let $\displaystyle {x}{\left({t}\right)}={t}^{{3}}-{4}\cdot{t}^{{2}}+{1}$

$\displaystyle {\quad\text{and}\quad}{y}{\left({t}\right)}={t}^{{3}}-{5}\cdot{t}^{{2}}+{2}\cdot{t}+{8}$

Box 1-4 answers should be integers.

Box 5 answer should be an integer or fraction.

Box 6 answer should be a decimal (2 places).

At $\displaystyle {t}={2}$

$\displaystyle {x}{\left({t}\right)}=$

$\displaystyle {y}{\left({t}\right)}=$

$\displaystyle \frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}=$

$\displaystyle \frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}=$

$\displaystyle \frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ = tangent slope =

$\displaystyle \text{speed =}$