find an equation for the tangent line to the given curve at the point where x=x0. y=(6x-5)(3+3x); x0=0
$\displaystyle x_0 = 0$
$\displaystyle y_0 = -15$
find the derivative of $\displaystyle y$, then find $\displaystyle y'(x_0)$, the slope of the line tangent to the curve at $\displaystyle (x_0,y_0)$.
now use the point-slope form of a linear equation to find the equation of the tangent line.
$\displaystyle y - y_0 = y'(x_0)(x - x_0)$