Answer
$$(\frac{16}{27}, \frac{29} 9) \space and \space(-2,-4)$$
Work Step by Step
1. Calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$
$\frac{dy}{dt} = \frac{d(1 + 4t - t^2)}{dt} = 4 - 2t$
$\frac{dx}{dt} = \frac{d(2t^3)}{dt} = 6t^2$
2. Find at which points the tangent line has slope 1:
Slope of the tangent line = $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4-2t}{6t^2}$:
$\frac{4-2t}{6t^2} = 1$
$4 - 2t = 6t^2$
$6t^2 + 2t - 4 = 0$
$t_1 = \frac{-b + \sqrt {b^2 -4ac}}{2a} = \frac{-2 + \sqrt {2^2 -4(6)(-4)}}{2(6)} = \frac{-2 + \sqrt {4 + 96}}{12} = \frac{-2 + 10}{12} = \frac 2 3$
$t_2 = \frac{-2 - \sqrt {4 + 96}}{12} = \frac{-2 - 10}{12} = -1$
Therefore, this slope occurs when t = $2/3$ and when $t = -1$
3. Identify the points for these values of $t$:
$x = 2(\frac 2 3)^3 = \frac{16}{27}$
$y = 1 + 4(\frac 2 3) - (\frac 2 3)^2 = \frac 99 + \frac {24} 9 - \frac 4 9 = \frac {29} 9$
$x = 2(-1)^3 = -2$
$y = 1 + 4(-1) - (-1)^2 = 1 - 4 - 1 = -4$
These points are: $(\frac{16}{27}, \frac{29} 9)$ and $(-2,-4)$
