Answer
$y=3x+3$
Graph:
Work Step by Step
$x=t^{2}-t,y=t^{2}+t+1;(0,3).\quad $
First, find the t for which $x=0,y=3,$
$x=t^{2}-t$
$0=t(t-1)\quad \Rightarrow t=0$ or $t=1.$
When$ t=0,\quad y=0+0+1,$
When $t=1, \quad y=1+1+1=3$
so, t=1.
Now, the slope of the tangent at the given point.
$\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2t+1}{2t-1}$
For $t=1,$
$\displaystyle \frac{dy}{dx}=\frac{2+1}{2-1}=3$
With m=3, P(0,3), an equation of the tangent is
$y-y_{1}=m(x-x_{1})$
$y-3=3x$
$y=3x+3$
To graph the curve, build a table of coordinates (x,y)
$x=t^{2}-t,\quad y=t^{2}+t+1$
Plot the points and join with a smooth curve.
To graph the line, we have (0,3), move 1 unit to the right, go up 3 units (slope=3),
and plot the second point on the tangent, (1,6)
