Answer
The equation of the tangent is $y=3x+3$.
The graph of the curve and the tangent is below.
Work Step by Step
Find the value of $t$ corresponding to $(x,y)=(0,3)$:
$(x,y)=(0,3)$
$(t^2-t,t^2+t+1)=(0,3)$
$t^2-t=0$ and $t^2+t+1=3$
$t(t-1)=0$ and $t^2+t-2=0$
$t(t-1)=0$ and $(t+2)(t-1)=0$
$t=0,1$ and $t=-2,1$
$t=1$
Find the slope of the tangent to the curve at $t=1$:
$m=\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2t+1}{2t-1}=\frac{2\cdot 1+1}{2\cdot 1-1}=\frac{3}{1}=3$
Find the equation of the tangent to the curve at $(0,3)$:
$y-3=m(x-0)$
$y-3=3x$
$y=3x+3$
Here is the graph of the curve and the tangent.
