What is the tangent of y= 9/(x-5)-1 ?

The tangent of y= 9/(x-5)-1 is y=-9/((a_{0)-5)^2}x+(18a_{0}-45)/((a_{0)-5)^2}-1

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tangent of y= 9/(x-5)-1

tangent of

y = \frac{9}{x - 5} - 1

y = - \frac{9}{( a _{0} - 5 ) ^{2}} x + \frac{18 a _{0} - 45}{( a _{0} - 5 ) ^{2}} - 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{9}{a _{0} - 5} - 1)

Find the slope of

y = \frac{9}{x - 5} - 1 : \frac{d y}{d x} = - \frac{9}{( x - 5 ) ^{2}}

Compute the slope of

y = \frac{9}{x - 5} - 1

x = a_{0} : m = - \frac{9}{( a _{0} - 5 ) ^{2}}

Find the line with slope m=

- \frac{9}{( a _{0} - 5 ) ^{2}}

and passing through

(a_{0}, \frac{9}{a _{0} - 5} - 1) : y = - \frac{9}{( a _{0} - 5 ) ^{2}} x + \frac{18 a _{0} - 45}{( a _{0} - 5 ) ^{2}} - 1

y = - \frac{9}{( a _{0} - 5 ) ^{2}} x + \frac{18 a _{0} - 45}{( a _{0} - 5 ) ^{2}} - 1

\int \frac{x}{x ^{4} + 4} d x

\int 2 x + y e^{x y} d x

\frac{d}{d x} (tan (\frac{x}{5}))

\frac{d}{d x} ((11 - 3 x)^{\frac{2}{3}} (x - 2)^{4})

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

What is the tangent of y= 9/(x-5)-1 ?
The tangent of y= 9/(x-5)-1 is y=-9/((a_{0)-5)^2}x+(18a_{0}-45)/((a_{0)-5)^2}-1