What is the tangent of y=sqrt(-3x-9),(-9,sqrt(18)) ?

The tangent of y=sqrt(-3x-9),(-9,sqrt(18)) is y=-1/(2sqrt(2))x+3sqrt(2)-9/(2sqrt(2))

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tangent of y=sqrt(-3x-9),(-9,sqrt(18))

tangent of

y = - 3 x - 9, (- 9, 18)

y = - \frac{1}{2 2} x + 32 - \frac{9}{2 2}

Solution steps

Find the slope of

y = - 3 x - 9 : \frac{d y}{d x} = - \frac{3}{2 - 3 x - 9}

Compute the slope of

y = - 3 x - 9

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

Find the line with slope m=

- \frac{1}{2 2}

and passing through

(- 9, 18) : y = - \frac{1}{2 2} x + 32 - \frac{9}{2 2}

y = - \frac{1}{2 2} x + 32 - \frac{9}{2 2}

g (x) = 2 x^{0.8} + 4 x^{- 0.8}

\int 3 x^{2} + 2 y^{2} d x

f (x) = cot^{2} (cos (x))

y = \frac{2}{x - 6}

y = - 3 x

What is the tangent of y=sqrt(-3x-9),(-9,sqrt(18)) ?
The tangent of y=sqrt(-3x-9),(-9,sqrt(18)) is y=-1/(2sqrt(2))x+3sqrt(2)-9/(2sqrt(2))