What is the tangent of (6x)/(x+5),\at x=-3 ?

The tangent of (6x)/(x+5),\at x=-3 is y= 15/2 x+27/2

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tangent of (6x)/(x+5),\at x=-3

tangent of

\frac{6 x}{x + 5}, at x = - 3

y = \frac{15}{2} x + \frac{27}{2}

Solution steps

Find the tangent point:

(- 3, - 9)

Find the slope of

y = \frac{6 x}{x + 5} : \frac{d y}{d x} = \frac{30}{( x + 5 ) ^{2}}

Compute the slope of

y = \frac{6 x}{x + 5}

x = - 3 : m = \frac{15}{2}

Find the line with slope m=

\frac{15}{2}

and passing through

(- 3, - 9) : y = \frac{15}{2} x + \frac{27}{2}

y = \frac{15}{2} x + \frac{27}{2}

\int_{- 1}^{1} 0 d x

n = 0 \sum \infty \frac{8}{5 ^{n}}

(x^{3} + 9) e^{x}

\int \frac{( ( x - 2 ) ^{4} )}{4 x - x ^{2}} d x

\frac{( - 2 x - 3 ) ^{7}}{( 4 x + 2 ) ^{4}}

What is the tangent of (6x)/(x+5),\at x=-3 ?
The tangent of (6x)/(x+5),\at x=-3 is y= 15/2 x+27/2