What is the tangent of f(x)= 6/(5x+2),(-1,-2) ?

The tangent of f(x)= 6/(5x+2),(-1,-2) is y=-10/3 x-16/3

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tangent of f(x)= 6/(5x+2),(-1,-2)

tangent of

f (x) = \frac{6}{5 x + 2}, (- 1, - 2)

y = - \frac{10}{3} x - \frac{16}{3}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = \frac{6}{5 x + 2}

(- 1, - 2) : m = - \frac{10}{3}

Find the line with slope m=

- \frac{10}{3}

and passing through

(- 1, - 2) : y = - \frac{10}{3} x - \frac{16}{3}

y = - \frac{10}{3} x - \frac{16}{3}

\int (\frac{sin ( 4 x )}{4}) - (\frac{sin ^{3} ( x )}{4}) d x

n = 0 \sum \infty n (n + 1) - 1

\int cos (x + π) d x

\frac{e ^{- s}}{s ^{2} ( s + 2 )}

\int 1 + e^{2 x} d x

What is the tangent of f(x)= 6/(5x+2),(-1,-2) ?
The tangent of f(x)= 6/(5x+2),(-1,-2) is y=-10/3 x-16/3