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

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

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

tangent of

y = x + 8, (8, 4)

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

Solution steps

Find the slope of

y = x + 8 : \frac{d y}{d x} = \frac{1}{2 x + 8}

Compute the slope of

y = 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 8 e^{t} - 5 d t

\frac{8 x ^{2} + 18 x}{( 9 + 8 x ) ^{2}}

x^{\frac{1}{6}}

\frac{d}{d x} (- x^{2} + 13 x - 22)

\int e^{t - s t}

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