## Calculus: Early Transcendentals 8th Edition

$y=3x+3$ Graph:
$x=t^{2}-t,y=t^{2}+t+1;(0,3).\quad$ 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)