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

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

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

tangent of

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}

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{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}

\int_{1}^{4} - 4 e^{x} d x

\frac{d}{d x} (\frac{x - 8}{7 x ^{2} e ^{x}})

f (x) = (x - 2) (x - 1) (x + 1) (x + 2)

\frac{d}{d x} (x^{4} + \frac{1}{x ^{4}})

\frac{s - 4}{s + 2}

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