What is the tangent of f(x)=sqrt(x^2+5),\at x=-2 ?

The tangent of f(x)=sqrt(x^2+5),\at x=-2 is y=-2/3 x+5/3

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tangent of f(x)=sqrt(x^2+5),\at x=-2

tangent of

f (x) = x^{2} + 5, at x = - 2

y = - \frac{2}{3} x + \frac{5}{3}

Solution steps

Find the tangent point:

(- 2, 3)

Find the slope of

f (x) = x^{2} + 5 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 5}

Compute the slope of

f (x) = x^{2} + 5

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

Find the line with slope m=

- \frac{2}{3}

and passing through

(- 2, 3) : y = - \frac{2}{3} x + \frac{5}{3}

y = - \frac{2}{3} x + \frac{5}{3}

\frac{16}{4 + x ^{2}}, y = 1

y = \frac{1}{4 - 9 x ^{2}}, (0, \frac{1}{3})

\frac{d}{d x} (- 7 sin (cos (x)))

\frac{\partial}{\partial x} (\frac{4 x - 4 z}{y + 4 z})

(2 x + 2 y)^{3} = 8 x^{3} + 8 y^{3}, (- 1, 1)

What is the tangent of f(x)=sqrt(x^2+5),\at x=-2 ?
The tangent of f(x)=sqrt(x^2+5),\at x=-2 is y=-2/3 x+5/3