What is the tangent of f(x)=(x^2)/(x-3) ?

The tangent of f(x)=(x^2)/(x-3) is y=(a_{0}^2-6a_{0})/((a_{0)-3)^2}x+(3a_{0}^2)/((a_{0)-3)^2}

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tangent of f(x)=(x^2)/(x-3)

tangent of

f (x) = \frac{x ^{2}}{x - 3}

y = \frac{a _{0}^{2} - 6 a _{0}}{( a _{0} - 3 ) ^{2}} x + \frac{3 a _{0}^{2}}{( a _{0} - 3 ) ^{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} - 3})

Find the slope of

f (x) = \frac{x ^{2}}{x - 3} : \frac{df ( x )}{d x} = \frac{x ^{2} - 6 x}{( x - 3 ) ^{2}}

Compute the slope of

f (x) = \frac{x ^{2}}{x - 3}

x = a_{0} : m = \frac{a _{0}^{2} - 6 a _{0}}{( a _{0} - 3 ) ^{2}}

Find the line with slope m=

\frac{a _{0}^{2} - 6 a _{0}}{( a _{0} - 3 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}^{2}}{a _{0} - 3}) : y = \frac{a _{0}^{2} - 6 a _{0}}{( a _{0} - 3 ) ^{2}} x + \frac{3 a _{0}^{2}}{( a _{0} - 3 ) ^{2}}

y = \frac{a _{0}^{2} - 6 a _{0}}{( a _{0} - 3 ) ^{2}} x + \frac{3 a _{0}^{2}}{( a _{0} - 3 ) ^{2}}

e^{2 x} cos (3 x)

f (x) = 4 tan (x), at x = \frac{π}{3}

\int_{0}^{3} π (9 - x^{2})^{2} d x

x \to 0 + lim (\frac{x ^{3} + 3}{x})

x \to 4 lim (6 - x)

What is the tangent of f(x)=(x^2)/(x-3) ?
The tangent of f(x)=(x^2)/(x-3) is y=(a_{0}^2-6a_{0})/((a_{0)-3)^2}x+(3a_{0}^2)/((a_{0)-3)^2}