To find out if the function is increasing or decreasing at #x=0#, take the derivative, and plug in #x=0# and see if it's positive or negative.

#f'(x)=d/dx((x^3+7x^2-5x-4)/(x-2))#

Quotient rule:

#color(white)(f'(x))=(d/dx(x^3+7x^2-5x-4)*(x-2)-(x^3+7x^2-5x-4)*d/dx(x-2))/(x-2)^2#

#color(white)(f'(x))=((3x^2+14x-5)(x-2)-(x^3+7x^2-5x-4)*1)/(x-2)^2#

#color(white)(f'(x))=((3x^2+14x-5)(x-2)-x^3-7x^2+5x+4)/(x-2)^2#

#color(white)(f'(x))=(3x^3-6x^2+14x^2-28x-5x+10-x^3-7x^2+5x+4)/(x-2)^2#

#color(white)(f'(x))=(2x^3+x^2-28x+14)/(x-2)^2#

Now, let's plug #x=0#. If the result is positive, then the function is increasing at #x=0#. If it's negative, then it's decreasing.

#color(white)=>(2x^3+x^2-28x+14)/(x-2)^2#

#=>(2(0)^3+(0)^2-28(0)+14)/(0-2)^2#

#=(0+0-0+14)/(-2)^2#

#=(14)/4#

#=7/2#

Since this value is positive, the original function #f(x)# is increasing at #x=0#.

We can verify this by looking at the graph of #f(x)#:

graph{(((x^3+7x^2-5x-4)/(x-2))) [-3, 3, -10, 6]}