What is the tangent of f(x)=ln(x^2-9x+1),(9,0) ?

The tangent of f(x)=ln(x^2-9x+1),(9,0) is y=9x-81

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tangent of f(x)=ln(x^2-9x+1),(9,0)

tangent of

f (x) = ln (x^{2} - 9 x + 1), (9, 0)

y = 9 x - 81

Solution steps

Find the slope of

f (x) = ln (x^{2} - 9 x + 1) : \frac{df ( x )}{d x} = \frac{2 x - 9}{x ^{2} - 9 x + 1}

Compute the slope of

f (x) = ln (x^{2} - 9 x + 1)

(9, 0) : m = 9

Find the line with slope m=

9

and passing through

(9, 0) : y = 9 x - 81

y = 9 x - 81

(- 2, 2), (1, - 2)

6 t^{2}

\int \frac{1}{484 - x ^{2}} d x

\frac{d}{d x} (I n^{2} x)

n = 1 \sum \infty \frac{e ^{2 n}}{n}

What is the tangent of f(x)=ln(x^2-9x+1),(9,0) ?
The tangent of f(x)=ln(x^2-9x+1),(9,0) is y=9x-81