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Mathematics 22 Online
OpenStudy (anonymous):

Find an equation of the tangent line to the curve at the point (7, 2). http://www.webassign.net/cgi-bin/symimage.cgi?expr=y%20%3D%20%28x%20-%205%29%2F%28x%20-%206%29

OpenStudy (anonymous):

take the derivative using the quotient rule, plug in 7 to get the slope.

OpenStudy (anonymous):

Differentiate it: Quotient rule gives: (x-6)-(x-5) --------- (x-6)^2 Plugging in 7 you get: -1? So then you have point slope form: y-y_0=m(x-x_0) y-2=-(x-7) y=-x+9

OpenStudy (anonymous):

i get \[y'-\frac{-1}{(x-6)^2}\]

OpenStudy (anonymous):

malevolence has it, but i would have simplified the numerator first

OpenStudy (anonymous):

:P I thought about it but I'm lazy xP

OpenStudy (anonymous):

Thank you guys!

OpenStudy (anonymous):

No problem :P

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