To find out whether a function is increasing or decreasing at one point, find the derivative of the function and **determine if the derivative is positive or negative at that point**.

First, differentiate #f(x)#

#f'(x) = d/dx (-x^3-7x^2-x+2)/(x-2)#

#= d/dx (-x^3-7x^2-x+2) * 1/(x-2)#

#= d/dx (-x^3-7x^2-x+2) * (x-2)^-1#

#= 1/(x-2) * d/dx (-x^3-7x^2-x+2) + (-x^3-7x^2-x+2) * d/dx (x-2)^-1#

#=1/(x-2) * (-3x^2-14x-1) + (-x^3-7x^2-x+2) * -(x-2)^-2 * d/dx (x-2)#

#=(-3x^2-14x-1)/(x-2) + (-x^3-7x^2-x+2) * -(x-2)^-2#

Simplifying,

#=(-3x^2-14x-1)/(x-2) - (-x^3-7x^2-x+2)/(x-2)^2#

#=((-3x^2-14x-1)(x-2))/(x-2)^2 - (-x^3-7x^2-x+2)/(x-2)^2#

#=(-3 x^3 - 8 x^2 + 27 x + 2 - (-x^3-7x^2-x+2))/(x-2)^2#

#=(-3 x^3 - 8 x^2 + 27 x + 2 +x^3 + 7x^2 + x - 2)/(x-2)^2#

#=(-2 x^3 - x^2 + 28 x)/(x-2)^2#

So to find out if the function is increasing or decreasing at #x=3#, find if #f'(3)# is positive or negative

#f'(3)=(-2 (3)^3 - (3) ^2 + 28 (3))/((3)-2)^2=21#

Since #f'(3)>0#, #f(x)# has a positive slope at x=3, and thus **#f(x)# is increasing**

graph{(y-(-x^3-7x^2-x+2)/(x-2))=0 [1.423, 4.492, -91.816, -90.28]}