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