What is the tangent of f(x)=((x))/(x-4) ?

The tangent of f(x)=((x))/(x-4) is y=-4/((a_{0)-4)^2}x+(a_{0}^2)/((a_{0)-4)^2}

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tangent of f(x)=((x))/(x-4)

tangent of

f (x) = \frac{( x )}{x - 4}

y = - \frac{4}{( a _{0} - 4 ) ^{2}} x + \frac{a _{0}^{2}}{( a _{0} - 4 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0}}{a _{0} - 4})

Find the slope of

f (x) = \frac{x}{x - 4} : \frac{df ( x )}{d x} = - \frac{4}{( x - 4 ) ^{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 = - \frac{4}{( a _{0} - 4 ) ^{2}}

Find the line with slope m=

- \frac{4}{( a _{0} - 4 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}}{a _{0} - 4}) : y = - \frac{4}{( a _{0} - 4 ) ^{2}} x + \frac{a _{0}^{2}}{( a _{0} - 4 ) ^{2}}

y = - \frac{4}{( a _{0} - 4 ) ^{2}} x + \frac{a _{0}^{2}}{( a _{0} - 4 ) ^{2}}

\frac{\partial}{\partial x} (sin (x^{2} + 5))

(8 - r^{2}) d r = r^{3} sin (θ) d θ

\int_{0}^{1} \frac{x - 6}{x ^{2} - 6 x + 8} d x

3 x \frac{d y}{d x} = x^{2} - 4

x \to 4 lim (2 x^{2} - 2 x)

What is the tangent of f(x)=((x))/(x-4) ?
The tangent of f(x)=((x))/(x-4) is y=-4/((a_{0)-4)^2}x+(a_{0}^2)/((a_{0)-4)^2}