Answer
$y=-\displaystyle \frac{3}{\pi}x+2$
Graph:
Work Step by Step
First, find the t for which $x=0,y=2,$
$t^{2}+t=2$
$t^{2}+t-2=0$
$(t+2)(t-1)=0$
For both t=1 and t=-2,$\Rightarrow x=0.$
Now, the slope of the tangent at the given point.
$\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2t+1}{\pi\cdot\cos\pi t}$
For $t=1,$
$\displaystyle \frac{dy}{dx}=\frac{3}{-\pi}=-\frac{3}{\pi}$
For t=$-2.\displaystyle \quad\frac{dy}{dx}=\frac{-3}{+\pi}=-\frac{3}{\pi}$
With m=$-\displaystyle \frac{3}{\pi}$, P(0,2), an equation of the tangent is
$y-y_{1}=m(x-x_{1})$
$y-2=-\displaystyle \frac{3}{\pi}x$
$y=-\displaystyle \frac{3}{\pi}x+2$
To graph the curve, build a table of coordinates (x,y)
$x=\sin\pi t,\quad y=t^{2}+t$
Plot the points and join with a smooth curve.
To graph the line, we have (0,2): move 1 unit to the right, go down about 0.955 units (slope=$-\displaystyle \frac{3}{\pi}$), and plot the second point on the tangent, (1,1.045)
