What is the tangent of y=(27)/(x^2+9),(-3, 3/2) ?

The tangent of y=(27)/(x^2+9),(-3, 3/2) is y= 1/2 x+3

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tangent of y=(27)/(x^2+9),(-3, 3/2)

tangent of

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

y = \frac{1}{2} x + 3

Solution steps

Find the slope of

y = \frac{27}{x ^{2} + 9} : \frac{d y}{d x} = - \frac{54 x}{( x ^{2} + 9 ) ^{2}}

Compute the slope of

y = \frac{27}{x ^{2} + 9}

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

Find the line with slope m=

\frac{1}{2}

and passing through

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

y = \frac{1}{2} x + 3

x \to \infty + lim (3 x^{2} - 8)

f (x) = (\frac{1}{4} x)^{3}

\frac{d}{d x} ((tan (x))^{2})

y^{^{'}} = - y - 2 x, y (0) = 1

\frac{d y}{d x} + 4 x y^{3} = 0

What is the tangent of y=(27)/(x^2+9),(-3, 3/2) ?
The tangent of y=(27)/(x^2+9),(-3, 3/2) is y= 1/2 x+3