Answer
a) $y=4x $; $y=-3x+18$
b) See graph
Work Step by Step
a) First, we have to differentiate the function $G(x) = 4x^{2}-x^{3}$, and we will get tangent line equation:
$G^{'}(x) = 8x - 3x^{2}$
To get the gradient of the tangent line, we have to substitute each point with the points.
For point $(2,8)$
$G^{'}(2) = 8(2) - 3(2)^{2} = 4$
For point $(3,9)$
$G^{'}(3) = 8(3) - 3(3)^{2} = -3$
Equation of the tangent line at point $(2,8)$:
$y-8=4(x-2)$
$y=4x$
Equation of the tangent line at point $(3,9)$:
$y-9=-3(x-3)$
$y= -3x+18$
b) Graph the function and the tangents in the two points.