What is the tangent of y=(x-1)/(x-2)+1,(3,3) ?

The tangent of y=(x-1)/(x-2)+1,(3,3) is y=-x+6

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tangent of y=(x-1)/(x-2)+1,(3,3)

tangent of

y = \frac{x - 1}{x - 2} + 1, (3, 3)

y = - x + 6

Solution steps

Find the slope of

y = \frac{x - 1}{x - 2} + 1 : \frac{d y}{d x} = - \frac{1}{( x - 2 ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 1

Find the line with slope m=

- 1

and passing through

(3, 3) : y = - x + 6

y = - x + 6

ln (x^{2} + y) - x y^{2} = 10

\frac{cos ( x )}{1 + a \cdot sin ( x )}

\int 14 e^{- 14 x} d x

\int x^{3} - \frac{1}{2} x^{2} + 4 d x

f (x) = e^{x} (x^{4} + 3), (0, 3)

What is the tangent of y=(x-1)/(x-2)+1,(3,3) ?
The tangent of y=(x-1)/(x-2)+1,(3,3) is y=-x+6