What is the tangent of y=ln(x^2-4x+1),(4,0) ?

The tangent of y=ln(x^2-4x+1),(4,0) is y=4x-16

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tangent of y=ln(x^2-4x+1),(4,0)

tangent of

y = ln (x^{2} - 4 x + 1), (4, 0)

y = 4 x - 16

Solution steps

Find the slope of

y = ln (x^{2} - 4 x + 1) : \frac{d y}{d x} = \frac{2 x - 4}{x ^{2} - 4 x + 1}

Compute the slope of

y = ln (x^{2} - 4 x + 1)

(4, 0) : m = 4

Find the line with slope m=

4

and passing through

(4, 0) : y = 4 x - 16

y = 4 x - 16

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What is the tangent of y=ln(x^2-4x+1),(4,0) ?
The tangent of y=ln(x^2-4x+1),(4,0) is y=4x-16