find an equation of the tangent line to the curve at the given point. graph the curve and the tangent line y=x/x-1 at (2,2) )

Answer 1

y = x/(x-1) = 1 + 1/(x-1)
y' = -1/(x-1)^2
y'(2) = -1

so, the tangent line is

y-2 = -1(x-2)

see the graphs at

http://www.wolframalpha.com/input/?i=ploy+y%3Dx%2F%28x-1%29%2C+y%3D-%28x-2%29%2B2

Answer 2

thanks. what rule did you use to find it?

Answer 3

find an equation of the tangent line to the curve at the given point. graph the curve and the tangent line x-3^1/2 at (1,2)

it is suppose to be square root of x-3

Answer 4

find y' at the given point (h,k).
That is the slope of the tangent line.

And then just plug in that slope (call it m) into the point-slope form of the line:

y-k = m(x-h)

Don't forget your Algebra I now that your'e doing calculus. So, what do you get for

y=√(x-3) at (1,2)?