Asked 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
            
            
        Answers
                    Answered by
            Reiny
            
    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)
    
= 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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