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

The tangent of sqrt(1-6x),\at x=-1 is y=-3/(sqrt(7))x+sqrt(7)-3/(sqrt(7))

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

tangent of

1 - 6 x, at x = - 1

y = - \frac{3}{7} x + 7 - \frac{3}{7}

Solution steps

Find the tangent point:

(- 1, 7)

Find the slope of

y = 1 - 6 x : \frac{d y}{d x} = - \frac{3}{1 - 6 x}

Compute the slope of

y = 1 - 6 x

x = - 1 : m = - \frac{3}{7}

Find the line with slope m=

- \frac{3}{7}

and passing through

(- 1, 7) : y = - \frac{3}{7} x + 7 - \frac{3}{7}

y = - \frac{3}{7} x + 7 - \frac{3}{7}

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What is the tangent of sqrt(1-6x),\at x=-1 ?
The tangent of sqrt(1-6x),\at x=-1 is y=-3/(sqrt(7))x+sqrt(7)-3/(sqrt(7))