Determine the point of tangency #(x_1, y_1)# then the slope #m#
From the given equation #f(x) = (4x^3)/(3-5x)^5#
#y_1=f(2)=(4(2)^3)/(3-5(2))^5=32/(-7)^5=-32/16807#
#color(red)((x_1, y_1)=(2, -32/16807))#
Obtain the slope #m#
#f(x) = (4x^3)/(3-5x)^5#
#f' (x)=((3-5x)^5d/dx(4x^3)-4x^3*d/dx(3-5x)^5)/((3-5x)^5)^2#
#f' (x)=((3-5x)^5*12x^2-4x^3*5(3-5x)^4*(-5))/((3-5x)^5)^2#
Use #x=2# to find the slope #m=f' (2)#
#f' (x)=((3-5x)^5*12x^2-4x^3*5(3-5x)^4*(-5))/((3-5x)^5)^2#
#f' (2)=((3-5(2))^5*12(2)^2-4(2)^3*5(3-5(2))^4*(-5))/((3-5(2))^5)^2#
#f' (2)=((-7)^5*48+32*25(-7)^4)/(-7)^10#
#f' (2)=((-7)*48+32*25)/(-7)^6#
#color(red)(f' (2)=(464)/117649)##color(red)" the slope"#
The Tangent line by Point-Slope Form
#(y-y_1)=m(x-x_1)#
#(y-(-32/16807))=(464/117649)(x-2)#
#y+32/16807=(464/117649)(x-2)#
#y=(464/117649)(x-2)-32/16807#
#color(red)(y=(464x)/117649-1152/117649)#
Also
#color(red)(464x-117649y=1152)#
Kindly see the graph, the tangent #464x-117649y=1152# is almost coincident with the x_axis. Take note the slope #m=0.003944~=0# at the point #(2, -0.0019)#
graph{(y - (4x^3)/(3-5x)^5)(464x-117649y-1152)=0[-3,3,-1.5,1.5]}
God bless....I hope the explanation is useful.
