Asked by Anonymous
the graph of y= (ax+b)/(x-1)(x-4) has a horizontal tangent at the point (2,-1. Find a ANd B.
Please help I have ne idea where to start
Please help I have ne idea where to start
Answers
Answered by
Steve
Assuming you meant
y= (ax+b)/((x-1)(x-4))
y' = (ax^2+2bx-(4a+5b))/((x-1)(x-4))^2
Now, y'=0 when ax^2+2bx-(4a+5b) = 0
when x=2, that means
4a+4b-4a-5b=0
b=0
So, y = ax/((x-1)(x-4))
at x=2, y=-1, so
2a/-2 = -1
a = 1
and so
y = x/((x-1)(x-4))
y= (ax+b)/((x-1)(x-4))
y' = (ax^2+2bx-(4a+5b))/((x-1)(x-4))^2
Now, y'=0 when ax^2+2bx-(4a+5b) = 0
when x=2, that means
4a+4b-4a-5b=0
b=0
So, y = ax/((x-1)(x-4))
at x=2, y=-1, so
2a/-2 = -1
a = 1
and so
y = x/((x-1)(x-4))
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