What is the tangent of y=sqrt(9-6x),\at x=-1 ?

The tangent of y=sqrt(9-6x),\at x=-1 is y=-sqrt(3/5)x+sqrt(15)-sqrt(3/5)

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tangent of y=sqrt(9-6x),\at x=-1

tangent of

y = 9 - 6 x, at x = - 1

y = - \frac{3}{5} x + 15 - \frac{3}{5}

Solution steps

Find the tangent point:

(- 1, 15)

Find the slope of

y = 9 - 6 x : \frac{d y}{d x} = - \frac{3}{9 - 6 x}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{3}{5}

Find the line with slope m=

- \frac{3}{5}

and passing through

(- 1, 15) : y = - \frac{3}{5} x + 15 - \frac{3}{5}

y = - \frac{3}{5} x + 15 - \frac{3}{5}

\int \frac{x ^{6}}{( x - 2 ) ^{2} ( 1 - x ) ^{5}} d x

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

x \to 4 - lim (\frac{- 1}{x - 4})

\frac{\partial}{\partial x} (\frac{2 x y ^{2}}{2 + x ^{2} y ^{2}})

x \to - 4 lim (25)

What is the tangent of y=sqrt(9-6x),\at x=-1 ?
The tangent of y=sqrt(9-6x),\at x=-1 is y=-sqrt(3/5)x+sqrt(15)-sqrt(3/5)