## Calculus: Early Transcendentals 8th Edition

$G'(a) = 8a - 3a^2$ 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)
$G'(x) = \lim\limits_{h \to 0}\frac{[4(x+h)^2 - (x+h)^3] - [4x^2 - x^3]}{h}$ $G'(x) = \lim\limits_{h \to 0}\frac{[4(x^2 +2xh +h^2) - (h^3 + 3xh^2 +3x^2h +x^3) - 4x^2 + x^3]}{h}$ $G'(x) = \lim\limits_{h \to 0}\frac{4x^2 +8xh +4h^2 - h^3 - 3xh^2 - 3x^2h - x^3 - 4x^2 + x^3}{h}$ $G'(x) = \lim\limits_{h \to 0}\frac{8xh +4h^2 - h^3 - 3xh^2 - 3x^2h}{h}$ $G'(x) = \lim\limits_{h \to 0}\frac{h(8x +4h - h^2 - 3xh - 3x^2)}{h}$ $G'(x) = \lim\limits_{h \to 0} 8x + 4h - 3xh - 3x^2 - h^2$ 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 $y -y_{1} = m(x-x_{1})$ $y - 8 = 4(x-2)$ $y = 4x - 8 +8$ $y = 4x$ $G'(3) = 8(3) - 3(3)^2$ $G'(3) = 24 - 27 = -3$ $-3 = \frac{y-9}{x-3}$ $y-9 = -3(x-3)$ $y = -3x + 18$