Answer
$y=3x$ and $y=3x+4$
Work Step by Step
$f(x)=x^{3}+2$
$f'(x)=Mtangent=3x^{2}$
Parallel: Mtangent=Mline
$3x-y-4=0$
$y=3x-4$
$Mline=3$
$3x^{2}=3$
$x^{2}=1$
$x=+-1$
going back to the curve:
$f(1)=3, f(-1)=1$
$(1,3) and (-1,1)$
Equations:
$y-3=3(x-1)$
$y-3=3x-3$
$y=3x$
$y-1=3(x+1)$
$y-1=3x+3$
$y=3x+4$
