What is the tangent of y=sqrt(2-6x),\at x=-1 ?

The tangent of y=sqrt(2-6x),\at x=-1 is y=-3/(2sqrt(2))x+2sqrt(2)-3/(2sqrt(2))

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tangent of y=sqrt(2-6x),\at x=-1

tangent of

y = 2 - 6 x, at x = - 1

y = - \frac{3}{2 2} x + 22 - \frac{3}{2 2}

Solution steps

Find the tangent point:

(- 1, 22)

Find the slope of

y = 2 - 6 x : \frac{d y}{d x} = - \frac{3}{2 - 6 x}

Compute the slope of

y = 2 - 6 x

x = - 1 : m = - \frac{3}{2 2}

Find the line with slope m=

- \frac{3}{2 2}

and passing through

(- 1, 22) : y = - \frac{3}{2 2} x + 22 - \frac{3}{2 2}

y = - \frac{3}{2 2} x + 22 - \frac{3}{2 2}

\frac{s}{s ^{2} - 3}

\int_{0}^{3} \frac{3}{2} x x + 1 d x

\int_{2}^{8} - \frac{x}{2} d x

\int \frac{cos ( ln ( 9 t ) )}{t} d t

\frac{d}{d x} (e^{2 c o s (3 x)})

What is the tangent of y=sqrt(2-6x),\at x=-1 ?
The tangent of y=sqrt(2-6x),\at x=-1 is y=-3/(2sqrt(2))x+2sqrt(2)-3/(2sqrt(2))