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

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

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

tangent of

f (x) = - x, at x = - 4

y = - \frac{1}{4} x + 1

Solution steps

Find the tangent point:

(- 4, 2)

Find the slope of

f (x) = - x : \frac{df ( x )}{d x} = - \frac{1}{2 - 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{1}{4}

Find the line with slope m=

- \frac{1}{4}

and passing through

(- 4, 2) : y = - \frac{1}{4} x + 1

y = - \frac{1}{4} x + 1

y^{^{''}} - 8 y^{^{'}} = 4 t

\int \frac{2 x}{x ^{2} - 16} d x

\frac{d y}{d x}, x^{2} + y^{2} = 0

R (t) = (t + e^{t}) (3 - t)

\frac{\partial}{\partial x} (\frac{x}{x ^{3} - y ^{5}})

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