What is the tangent of f(x)=sqrt(x+8),(8,4) ?

The tangent of f(x)=sqrt(x+8),(8,4) is y= 1/8 x+3

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tangent of f(x)=sqrt(x+8),(8,4)

tangent of

f (x) = x + 8, (8, 4)

y = \frac{1}{8} x + 3

Solution steps

Find the slope of

f (x) = x + 8 : \frac{df ( x )}{d x} = \frac{1}{2 x + 8}

Compute the slope of

f (x) = x + 8

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

Find the line with slope m=

\frac{1}{8}

and passing through

(8, 4) : y = \frac{1}{8} x + 3

y = \frac{1}{8} x + 3

\int \frac{2 x}{9 - 4 x ^{2}} d x

\frac{12}{0.5 s ^{2} + 4 s + 5}

\int (x - 3) d x

\int_{0}^{\frac{1}{2}} 2 arcsin (2 y) d y

\int \frac{1}{x ( 49 + x )} d x

What is the tangent of f(x)=sqrt(x+8),(8,4) ?
The tangent of f(x)=sqrt(x+8),(8,4) is y= 1/8 x+3