Calculus
posted by Nevin .
The graph of f(x)=(ax+b)/(x^2  5x + 4) has a horizontal tangent line at (2, 1). Find a and b

f'(x) = ( a(x^2  5x + 4)  (2x5)(ax+b) )/(x^2  5x +4)^2
= 0 for a horizontal line
when x = 2
a(2) + (2a+b) = 0
2a + 2a + b = 0
b = 0
also f(2) = 1
2a/(4  10 + 4) = 1
2a/2 = 1
a = 1
a=1 , b=0
so f(x) = x/(x^2  5x + 4)