$x=2t^3$ and $y=4+16t-10t^2$
First find the slope and set it equal to one:
$m=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} =\dfrac{16-20t}{6t^2}$
Setting equal to one:
$\dfrac{16t-20t}{6t}=1$ Factor out the 4 on the top and reduce coefficients:
$\dfrac{4(4-5t)}{6t^2}=1 \to \frac{2(4-5t)}{3t^2}=1$ Move the one over:
$\dfrac{2(4-5t)}{3t^2}-1=0$ Multiply both sides by $3t^2$:
$2(4-5t)-3t^2=0$ Factor back in the two:
$-3t^2-10t+8$ Now use the quadratic formula:
$\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Which works out to:
$\dfrac{10 \pm \sqrt{100 - (-12)(8)}}{-6} \to \dfrac{10 \pm \sqrt{100-(-96)}}{-6} \to \dfrac{10 \pm \sqrt{196}}{-6}$ Simplified gives values of t:
$t=\dfrac{10+14}{-6}=-4$
$t=\dfrac{10-14}{-6}=\frac{2}{3}$
When $t=\frac{2}{3}$:
$x=2(\frac{2}{3}^3) = .2962962963$
$y=4+16(\frac{2}{3})-10(\frac{2}{3})^2=8.000000003$
I get the correct value when $t=-4$ so I just left it out. I don't see how my answer here could be wrong but the computer rejected it. Did I miss something?