The equation of the tangent line is of the form:
#y=color(orange)(a)x+color(violet)(b)#
where #a# is the slope of this straight line.
To find the slope of this tangent line to #f(x)# at point #x=5# we should differentiate #f(x)#
#f(x)# is a quotient function of the form #(u(x))/(v(x))#
where #u(x)=x-3# and #v(x)=(x-4)^2#
#color(blue)(f'(x)=(u'(x)v(x)-v'(x)u(x))/(v(x))^2)#
#u'(x)=x'-3'#
#color(red)(u'(x)=1)#
#v(x)# is a composite function so we have to apply chain rule
let #g(x)=x^2# and #h(x)=x-4#
#v(x)=g(h(x))#
#color(red)(v'(x)=g'(h(x))*h'(x))#
#g'(x)=2x# then
#g'(h(x))=2(h(x))= 2(x-4)#
#h'(x)=1#
#color(red)(v'(x)=g'(h(x))*h'(x))#
#color(red)(v'(x)=2(x-4)#
#color(blue)(f'(x)=(u'(x)v(x)-v'(x)u(x))/(v(x))^2)#
#f'(x)=(1*(x-4)^2-2(x-4)(x-3))/((x-4)^2)^2#
#f'(x)=((x-4)^2-2(x-4)(x-3))/(x-4)^2#
#f'(x)=((x-4)(x-4-2(x-3)))/(x-4)^4#
#f'(x)=((x-4)(x-4-2x+6))/(x-4)^4#
#f'(x)=((x-4)(-x+2))/(x-4)^4#
simplifying the common factor #x-4# between numerator and denominator
#color(blue)(f'(x)=(-x+2)/(x-4)^3)#
Because the tangent line passes through the point #x=5# so we can find the value of slope #a# by substituting #x=5# in # f'(x)#
#color(orange)(a=f'(5))#
#a=(-5+2)/(5-4)^3#
#a=-3/1^3#
#color(orange)(a=-3)#
Given the abscissa of point of tangency #color(brown)(x=5)# lets
lets find its ordinate #y=f(5)#
#color(brown)(y=f(5))=(5-3)/(5-4)^4#
#y=2/1#
#color(brown)(y=2)#
Having the coordinates of tangency point #color(brown)((5;2))# and the slope #color(orange)(a=-3)# lets find #color(violet)(b)#
lets substitute all the known values in the equation of the tangent line to find value #color(violet)(b)#
#color(brown)(y)=color(orange)(a)color(brown)(x)+color(violet)(b)#
#2=-3(5)+color(violet)(b)#
#2=-15+color(violet(b)#
#17=color(violet)(b)#
therefore, the equation of the tangent line at point #color(brown)((5;2))# is:
#y=-3x+17#
