What is the tangent of f(x)=sqrt(x-8),\at x=9 ?

The tangent of f(x)=sqrt(x-8),\at x=9 is y= 1/2 x-7/2

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tangent of f(x)=sqrt(x-8),\at x=9

tangent of

f (x) = x - 8, at x = 9

y = \frac{1}{2} x - \frac{7}{2}

Solution steps

Find the tangent point:

(9, 1)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(9, 1) : y = \frac{1}{2} x - \frac{7}{2}

y = \frac{1}{2} x - \frac{7}{2}

\int 0.06 e^{0.04 x} d x

\int 3 x^{2} cos (x) d x

n = 0 \sum \infty \frac{n !}{10 7 ^{n}}

\frac{d}{d x} (\frac{x ^{2}}{5 x + 2})

3 t^{2} - 30 t + 72

What is the tangent of f(x)=sqrt(x-8),\at x=9 ?
The tangent of f(x)=sqrt(x-8),\at x=9 is y= 1/2 x-7/2