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

The tangent of f(x)=(x^2+5)/(x-5),\at x=3 is y=-13/2 x+25/2

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

tangent of

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

y = - \frac{13}{2} x + \frac{25}{2}

Solution steps

Find the tangent point:

(3, - 7)

Find the slope of

f (x) = \frac{x ^{2} + 5}{x - 5} : \frac{df ( x )}{d x} = \frac{x ^{2} - 10 x - 5}{( x - 5 ) ^{2}}

Compute the slope of

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

x = 3 : m = - \frac{13}{2}

Find the line with slope m=

- \frac{13}{2}

and passing through

(3, - 7) : y = - \frac{13}{2} x + \frac{25}{2}

y = - \frac{13}{2} x + \frac{25}{2}

cos (\frac{x}{2})

x \to 0 + lim ((1 + 6 x)^{\frac{3}{x}})

\frac{\partial}{\partial t} (arctan (x t))

\int cos (12 x) d x

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

What is the tangent of f(x)=(x^2+5)/(x-5),\at x=3 ?
The tangent of f(x)=(x^2+5)/(x-5),\at x=3 is y=-13/2 x+25/2