What is the tangent of f(x)=(6x)/(x+3),(-1,-3) ?

The tangent of f(x)=(6x)/(x+3),(-1,-3) is y= 9/2 x+3/2

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tangent of f(x)=(6x)/(x+3),(-1,-3)

tangent of

f (x) = \frac{6 x}{x + 3}, (- 1, - 3)

y = \frac{9}{2} x + \frac{3}{2}

Solution steps

Find the slope of

f (x) = \frac{6 x}{x + 3} : \frac{df ( x )}{d x} = \frac{18}{( x + 3 ) ^{2}}

Compute the slope of

f (x) = \frac{6 x}{x + 3}

(- 1, - 3) : m = \frac{9}{2}

Find the line with slope m=

\frac{9}{2}

and passing through

(- 1, - 3) : y = \frac{9}{2} x + \frac{3}{2}

y = \frac{9}{2} x + \frac{3}{2}

\int e^{x} sin (3 x) d x

\frac{\partial}{\partial z} (x^{2} z^{2} sin (4 y))

\frac{\partial}{\partial x} (sin (\frac{2 x}{y}))

- 5 w^{- 7} + 3 w

x \to 1 lim (\frac{x}{x - 1})

What is the tangent of f(x)=(6x)/(x+3),(-1,-3) ?
The tangent of f(x)=(6x)/(x+3),(-1,-3) is y= 9/2 x+3/2