What is the tangent of y=sqrt(9-4x),\at x=-1 ?

The tangent of y=sqrt(9-4x),\at x=-1 is y=-2/(sqrt(13))x+sqrt(13)-2/(sqrt(13))

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tangent of y=sqrt(9-4x),\at x=-1

tangent of

y = 9 - 4 x, at x = - 1

y = - \frac{2}{13} x + 13 - \frac{2}{13}

Solution steps

Find the tangent point:

(- 1, 13)

Find the slope of

y = 9 - 4 x : \frac{d y}{d x} = - \frac{2}{9 - 4 x}

Compute the slope of

y = 9 - 4 x

x = - 1 : m = - \frac{2}{13}

Find the line with slope m=

- \frac{2}{13}

and passing through

(- 1, 13) : y = - \frac{2}{13} x + 13 - \frac{2}{13}

y = - \frac{2}{13} x + 13 - \frac{2}{13}

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What is the tangent of y=sqrt(9-4x),\at x=-1 ?
The tangent of y=sqrt(9-4x),\at x=-1 is y=-2/(sqrt(13))x+sqrt(13)-2/(sqrt(13))