What is the tangent of f(x)=(x^2)/(x-8) ?

The tangent of f(x)=(x^2)/(x-8) is y=(a_{0}^2-16a_{0})/((a_{0)-8)^2}x+(8a_{0}^2)/((a_{0)-8)^2}

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tangent of f(x)=(x^2)/(x-8)

tangent of

f (x) = \frac{x ^{2}}{x - 8}

y = \frac{a _{0}^{2} - 16 a _{0}}{( a _{0} - 8 ) ^{2}} x + \frac{8 a _{0}^{2}}{( a _{0} - 8 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0}^{2}}{a _{0} - 8})

Find the slope of

f (x) = \frac{x ^{2}}{x - 8} : \frac{df ( x )}{d x} = \frac{x ^{2} - 16 x}{( x - 8 ) ^{2}}

Compute the slope of

f (x) = \frac{x ^{2}}{x - 8}

x = a_{0} : m = \frac{a _{0}^{2} - 16 a _{0}}{( a _{0} - 8 ) ^{2}}

Find the line with slope m=

\frac{a _{0}^{2} - 16 a _{0}}{( a _{0} - 8 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}^{2}}{a _{0} - 8}) : y = \frac{a _{0}^{2} - 16 a _{0}}{( a _{0} - 8 ) ^{2}} x + \frac{8 a _{0}^{2}}{( a _{0} - 8 ) ^{2}}

y = \frac{a _{0}^{2} - 16 a _{0}}{( a _{0} - 8 ) ^{2}} x + \frac{8 a _{0}^{2}}{( a _{0} - 8 ) ^{2}}

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

\frac{\partial}{\partial x} (- 2 x z)

\frac{d}{d x} (\frac{8 x - 1}{3 x + 8})

f (x) = \frac{7 x}{x + 4}, (3, 3)

\int 7 x^{11} e^{- x^{6}} d x

What is the tangent of f(x)=(x^2)/(x-8) ?
The tangent of f(x)=(x^2)/(x-8) is y=(a_{0}^2-16a_{0})/((a_{0)-8)^2}x+(8a_{0}^2)/((a_{0)-8)^2}