What is the tangent of f(x)=sqrt(x^2+18x+86),\at x=0 ?

The tangent of f(x)=sqrt(x^2+18x+86),\at x=0 is y= 9/(sqrt(86))x+sqrt(86)

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tangent of f(x)=sqrt(x^2+18x+86),\at x=0

tangent of

f (x) = x^{2} + 18 x + 86, at x = 0

y = \frac{9}{86} x + 86

Solution steps

Find the tangent point:

(0, 86)

Find the slope of

f (x) = x^{2} + 18 x + 86 : \frac{df ( x )}{d x} = \frac{x + 9}{x ^{2} + 18 x + 86}

Compute the slope of

f (x) = x^{2} + 18 x + 86

x = 0 : m = \frac{9}{86}

Find the line with slope m=

\frac{9}{86}

and passing through

(0, 86) : y = \frac{9}{86} x + 86

y = \frac{9}{86} x + 86

\int - \frac{1}{2} e^{- 2 x} d x

\int_{0}^{x} \frac{1}{1 + t ^{2}} d t

\frac{\partial}{\partial x} ((x e^{x + 2 y}) (1 + 2 y))

\int \frac{- x ^{3} - x ^{2} - 4 x + 28}{x ^{3} + 4 x} d x

What is the tangent of f(x)=sqrt(x^2+18x+86),\at x=0 ?
The tangent of f(x)=sqrt(x^2+18x+86),\at x=0 is y= 9/(sqrt(86))x+sqrt(86)