#f(x)= (2x^2-3x)/(x-1)#
Step 1: Find #f'(x)# , using the quotient rule
#f'(x)= ((x-1)(4x-3)-(2x^2-3x)(1))/(x-1)^2#
#f'(x)=(4x^2 -4x-3x+3-2x^2 +3x)/(x-1)^2#
#f'(x)=(4x^2 -4xcancel(-3x)+3-2x^2 cancel(+3x))/(x-1)^2#
#color(blue)(f'(x)= (2x^2-4x+3)/(x-1)^2)#
Step 2: Find the slope, remember #m= f'(x)#
#f'(5)= (2(5)^2-4(5)+3)/(5-1)^2 => (50-20+3)/4^2 => color(blue)(33/16)#
Step 3: Find the value of #f(x)# when #x= 5#
#f(5)= (2(5)^2 -3(5))/(5-1) => (2(25)-15)/(4) => (50-15)/4 => color(blue) (35/4) #
Step 4: Using the point slope formula to find equation of the line
#=>y-y_1 = m(x-x_1)#
#=> y-(35/4) = 33/16(x-5)# #color(red)(Distribute)#
#=> y -35/4 = 33/16x -165/16#
#=> y = 33/16x -165/16+140/16# Add #color(red)(35/4 hArr 140/14)#
#=>color(red)( y= 33/16 x -25/16)#
