What is the tangent of y= 1/(x-5),(4,-1) ?

The tangent of y= 1/(x-5),(4,-1) is y=-x+3

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tangent of y= 1/(x-5),(4,-1)

tangent of

y = \frac{1}{x - 5}, (4, - 1)

y = - x + 3

Solution steps

Find the slope of

y = \frac{1}{x - 5} : \frac{d y}{d x} = - \frac{1}{( x - 5 ) ^{2}}

Compute the slope of

y = \frac{1}{x - 5}

(4, - 1) : m = - 1

Find the line with slope m=

- 1

and passing through

(4, - 1) : y = - x + 3

y = - x + 3

\frac{d}{d y} (\frac{1}{y} e^{x y})

y^{2} = (x y y^{^{'}} + 1) ln (x)

y^{^{'}} = 1 - y

t^{2} y^{^{'}} + 6 t y - y^{3} = 0

y^{^{''}} - 2 y^{^{'}} + 4 y = 0

What is the tangent of y= 1/(x-5),(4,-1) ?
The tangent of y= 1/(x-5),(4,-1) is y=-x+3