What is the tangent of f(x)=((x^2))/(x-5),\at x=-3 ?

The tangent of f(x)=((x^2))/(x-5),\at x=-3 is y= 39/64 x+45/64

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tangent of f(x)=((x^2))/(x-5),\at x=-3

tangent of

f (x) = \frac{( x ^{2} )}{x - 5}, at x = - 3

y = \frac{39}{64} x + \frac{45}{64}

Solution steps

Find the tangent point:

(- 3, - \frac{9}{8})

Find the slope of

f (x) = \frac{x ^{2}}{x - 5} : \frac{df ( 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{39}{64}

Find the line with slope m=

\frac{39}{64}

and passing through

(- 3, - \frac{9}{8}) : y = \frac{39}{64} x + \frac{45}{64}

y = \frac{39}{64} x + \frac{45}{64}

0.001 t

t y^{^{'}} = 4 y

(x^{2} + 1) y^{^{'}} + 4 x (y - 1) = 0, y (0) = 3

\frac{d y}{d x} + y^{3} x + 8 y = 0

\int sin^{5} (9 x) d x

What is the tangent of f(x)=((x^2))/(x-5),\at x=-3 ?
The tangent of f(x)=((x^2))/(x-5),\at x=-3 is y= 39/64 x+45/64