## Calculus 8th Edition

$y=-8x+12$
$y=4x-3x^2, (2,-4)$ $y'=4-6x$ Plug in x-value. $y'=4-6(2)$ $y'=-8$ This is the slope of the tangent line to the curve at the given point. The question asks for an equation of the tangent line. We can use the point slope form. $y-y_{1}=m(x-x_{1})$ $y+4=-8(x-2)$ We can then rewrite this in standard form. $y=-8x+12$