# Calculus

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

• Calculus -

f'(x) = ( a(x^2 - 5x + 4) - (2x-5)(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)

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