What is the tangent of h(x)=(x^2)/(x-5) ?

The tangent of h(x)=(x^2)/(x-5) is y=(a_{0}^2-10a_{0})/((a_{0)-5)^2}x+(5a_{0}^2)/((a_{0)-5)^2}

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tangent of h(x)=(x^2)/(x-5)

tangent of

h (x) = \frac{x ^{2}}{x - 5}

y = \frac{a _{0}^{2} - 10 a _{0}}{( a _{0} - 5 ) ^{2}} x + \frac{5 a _{0}^{2}}{( a _{0} - 5 ) ^{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} - 5})

Find the slope of

h (x) = \frac{x ^{2}}{x - 5} : \frac{d h ( x )}{d x} = \frac{x ^{2} - 10 x}{( x - 5 ) ^{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{a _{0}^{2} - 10 a _{0}}{( a _{0} - 5 ) ^{2}}

Find the line with slope m=

\frac{a _{0}^{2} - 10 a _{0}}{( a _{0} - 5 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}^{2}}{a _{0} - 5}) : y = \frac{a _{0}^{2} - 10 a _{0}}{( a _{0} - 5 ) ^{2}} x + \frac{5 a _{0}^{2}}{( a _{0} - 5 ) ^{2}}

y = \frac{a _{0}^{2} - 10 a _{0}}{( a _{0} - 5 ) ^{2}} x + \frac{5 a _{0}^{2}}{( a _{0} - 5 ) ^{2}}

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What is the tangent of h(x)=(x^2)/(x-5) ?
The tangent of h(x)=(x^2)/(x-5) is y=(a_{0}^2-10a_{0})/((a_{0)-5)^2}x+(5a_{0}^2)/((a_{0)-5)^2}