monroe17

Suppose that f(x)=((5x)/((2-2x)^5) Find an equation for the tangent line to the graph of at x=2.

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monroe17

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y= -5/16
x= 2
y'=((5(4x+1))/((32(x-1)^6)))
y'(2)=45/32
45/32=slope
(y-(-5/16))=45/32(x-2)
(y+5/16)=45/32(x-2)
(y+5/16)=45/32x-19/32
y=45/32x-29/32
this is what I came up with and it's wrong... Can someone help me?

eyust707

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This step is incorrect:
(y+5/16)=45/32(x-2)
(y+5/16)=45/32x-19/32
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eyust707

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eyust707

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monroe17

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45/32x-19/32... that's what i keep getting

eyust707

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|dw:1349146416842:dw|

eyust707

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monroe17

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haha ohh.. stupid me

eyust707

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=) great job otherwise!!! you did all the hard stuff perfectly! +fan

monroe17

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yay! thank you!