Answer
$y=3x-3$ and $y=3x-7$
Work Step by Step
$y=x^3-3x^2+3x-3$ and $3x-y=15$
Let $g:3x-y=15$.
Find the slope of $g$:
$3x-y=15$
$y=3x-15$
$m_g=3$
Find $\frac{dy}{dx}$:
$\frac{dy}{dx}=\frac{d}{dx}(x^3-3x^2+3x-3)=3x^2-6x+3$
Find the values of $x$ such that the tangent lines to the curve are parallel to $g$:
$\frac{dy}{dx}=m_g$
$3x^2-6x+3=3$
$3x^2-6x=0$
$x(3x-6)=0$
$x=0\vee x=2$
Find the tangent line to the curve at $(0,y(0))$:
$y-y(0)=m_g(x-0)$
$y-(0^3-3\cdot 0^2+3\cdot 0-3)=3x$
$y-(-3)=3x$
$y=3x-3$
Find the tangent line to the curve at $(2,y(2))$:
$y-y(2)=m_g(x-2)$
$y-(2^3-3\cdot 2^2+3\cdot 2-3)=3(x-2)$
$y-(-1))=3x-6$
$y=3x-7$
Thus, the tangent lines are $y=3x-3$ and $y=3x-7$.
