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

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

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

tangent of

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

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

Solution steps

Find the slope of

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

Compute the slope of

y = \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}

f (x) = \frac{2 x}{x ^{2} - 8}

\int \frac{1}{t ^{2} - 2 t + 5} d t

(2 x - 7) (5 - 3 x)

x \to - 1 + lim (x + 12)

\frac{\partial}{\partial x} (2 e^{x} y)

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