Answer
$G'(a) = 8a - 3a^2 $
Curve (represented in red): $y = 4x^2 - x^3$
Tangent line at $(3,9)$ $y = -3x + 18$ (represented in blue)
Tangent line at $(2,8)$ $y = 4x$ (represented in blue)
Work Step by Step
$G'(x) = \lim\limits_{h \to 0}\frac{[4(x+h)^2 - (x+h)^3] - [4x^2 - x^3]}{h}$
$G'(x) = \lim\limits_{h \to 0}\frac{[4(x^2 +2xh +h^2) - (h^3 + 3xh^2 +3x^2h +x^3) - 4x^2 + x^3]}{h}$
$G'(x) = \lim\limits_{h \to 0}\frac{4x^2 +8xh +4h^2 - h^3 - 3xh^2 - 3x^2h - x^3 - 4x^2 + x^3}{h}$
$G'(x) = \lim\limits_{h \to 0}\frac{8xh +4h^2 - h^3 - 3xh^2 - 3x^2h}{h}$
$G'(x) = \lim\limits_{h \to 0}\frac{h(8x +4h - h^2 - 3xh - 3x^2)}{h}$
$G'(x) = \lim\limits_{h \to 0} 8x + 4h - 3xh - 3x^2 - h^2$
Substitute $h$ for $0$
$G'(x) = \lim\limits_{h \to 0} 8x + 4h - 3xh - 3x^2 - h^2$
$G'(x) = 8x + 4(0) - 3x(0) - 3x^2 - 0^2$
$G'(x) = 8x - 3x^2 $
$G'(2) = 8(2) - 3(2)^2 $
$G'(2) = 16 - 3(4) $
$G'(2) = 16 - 12 = 4$
Find the tangent line
$y -y_{1} = m(x-x_{1})$
$y - 8 = 4(x-2)$
$y = 4x - 8 +8$
$y = 4x$
$G'(3) = 8(3) - 3(3)^2 $
$G'(3) = 24 - 27 = -3$
$-3 = \frac{y-9}{x-3}$
$y-9 = -3(x-3)$
$y = -3x + 18$
