## Calculus (3rd Edition)

Slope of the tangent line= $c'(t)=\frac{(t^{3}-6t)'}{(3t^{2}-2t)'}=\frac{3t^{2}-6}{6t-2}=3$ $\implies 3t^{2}-6=3(6t-2)$ $\implies 3t^{2}-6=18t-6$ $\implies 3t^{2}=18t\implies t^{2}=6t$ $\implies t= 0, 6$ When $t=0$, the point is $(x,y)= (3(0)^{2}-2(0), (0)^{3}-6(0))=(0,0)$ When $t=6$, the point is $(3(6)^{2}-2(6), (6)^{3}-6(6))=(96,180)$