What is the tangent of y= 3/x ,(-2,-3/2) ?

The tangent of y= 3/x ,(-2,-3/2) is y=-3/4 x-3

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tangent of y= 3/x ,(-2,-3/2)

tangent of

y = \frac{3}{x}, (- 2, - \frac{3}{2})

y = - \frac{3}{4} x - 3

Solution steps

Find the slope of

y = \frac{3}{x} : \frac{d y}{d x} = - \frac{3}{x ^{2}}

Compute the slope of

y = \frac{3}{x}

(- 2, - \frac{3}{2}) : m = - \frac{3}{4}

Find the line with slope m=

- \frac{3}{4}

and passing through

(- 2, - \frac{3}{2}) : y = - \frac{3}{4} x - 3

y = - \frac{3}{4} x - 3

\int x^{\frac{1}{9}} d x

\int (x + 1) sec (x^{2} + 2 x) tan (x^{2} + 2 x) d x

sin^{3} (5 x)

y \to - 3 lim (4 y^{2} + 4 y + 1)

y = (2)^{x}, at x = 8

What is the tangent of y= 3/x ,(-2,-3/2) ?
The tangent of y= 3/x ,(-2,-3/2) is y=-3/4 x-3