Answer
$x=t,\ \ \ y=1-t$, $\ \ $ $z =2t$
Work Step by Step
Work done in Geogebra CAS (free online).
From the parametric equations, for the point (0,1,0), t is 0..
CAS entry: k:=0
Next, define r(t) and r'(t)
CAS entry: $r(t):=(t,\exp(-t),2t-t^{2})$
CAS entry: $r1(t)=$Derivative(r(t))
To plot the point (0,1,0):
CAS entry: $A=r(k)$
To find the tangent vector at (0,1,0):
CAS entry: $u:=r1(k)$
We now use the point-vector facility of geogebra to draw the vector u from point A. Then, the tangent line.
The tangent line contains $(0, 1,0)$ and is parallel to the vector $\langle 1,-1,2\rangle$.
Its parametric equations are (see sec 12-5):
$x=t,\ \ \ y=1-t$, $\ \ $ $z =2t$
