What is the tangent of f(x)=sqrt(x+4),(12,4) ?

The tangent of f(x)=sqrt(x+4),(12,4) is y= 1/8 x+5/2

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tangent of f(x)=sqrt(x+4),(12,4)

tangent of

f (x) = x + 4, (12, 4)

y = \frac{1}{8} x + \frac{5}{2}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x + 4

(12, 4) : m = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(12, 4) : y = \frac{1}{8} x + \frac{5}{2}

y = \frac{1}{8} x + \frac{5}{2}

4 y^{^{''}} + 6 y^{^{'}} - 4 y = 0

\int 1 - cos (x) d x

f (x) = \frac{x ^{3}}{6 x - 3}

x \to 3 + lim (\frac{2 + x}{3 - x})

\frac{d y}{d t} = \frac{t ^{2} + t y + y ^{2}}{t ^{2}}

What is the tangent of f(x)=sqrt(x+4),(12,4) ?
The tangent of f(x)=sqrt(x+4),(12,4) is y= 1/8 x+5/2