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

$\displaystyle {\quad\text{and}\quad}{y}{\left({t}\right)}={t}^{{2}}-{6}\cdot{t}+{3}$

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 $\displaystyle {t}={0}$

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

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

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

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

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

$\displaystyle \text{speed =}$