What is the tangent of f(x)=sqrt(x-8),\at x=12 ?

The tangent of f(x)=sqrt(x-8),\at x=12 is y= 1/4 x-1

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tangent of f(x)=sqrt(x-8),\at x=12

tangent of

f (x) = x - 8, at x = 12

y = \frac{1}{4} x - 1

Solution steps

Find the tangent point:

(12, 2)

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

x = 12 : m = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(12, 2) : y = \frac{1}{4} x - 1

y = \frac{1}{4} x - 1

\int x^{2} + 4 e^{x} + C d x

\frac{\partial}{\partial x} (2 \cdot x^{3} \cdot y^{2})

\int_{0}^{1} cos (nπ x) d x

\frac{d}{d x} (ln (6 x^{4} - 9 x - 3))

y^{^{''}} + 8 y = 0

What is the tangent of f(x)=sqrt(x-8),\at x=12 ?
The tangent of f(x)=sqrt(x-8),\at x=12 is y= 1/4 x-1