What is the tangent of f(x)=sqrt(x+3),\at x=-2 ?

The tangent of f(x)=sqrt(x+3),\at x=-2 is y= 1/2 x+2

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tangent of f(x)=sqrt(x+3),\at x=-2

tangent of

f (x) = x + 3, at x = - 2

y = \frac{1}{2} x + 2

Solution steps

Find the tangent point:

(- 2, 1)

Find the slope of

f (x) = x + 3 : \frac{df ( x )}{d x} = \frac{1}{2 x + 3}

Compute the slope of

f (x) = x + 3

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

Find the line with slope m=

\frac{1}{2}

and passing through

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

y = \frac{1}{2} x + 2

s^{2}

n lo g_{2} (n)

\int \frac{1}{sin ( u ) + 1} d u

y^{^{'}} = \frac{x ^{2} y ^{4} + 3 x ^{2}}{y ^{3}}

n = 1 \sum \infty \frac{1}{n} (\frac{3}{2})^{n}

What is the tangent of f(x)=sqrt(x+3),\at x=-2 ?
The tangent of f(x)=sqrt(x+3),\at x=-2 is y= 1/2 x+2