What is the tangent of y=ln(x^2-5x+1),(5,0) ?

The tangent of y=ln(x^2-5x+1),(5,0) is y=5x-25

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tangent of y=ln(x^2-5x+1),(5,0)

tangent of

y = ln (x^{2} - 5 x + 1), (5, 0)

y = 5 x - 25

Solution steps

Find the slope of

y = ln (x^{2} - 5 x + 1) : \frac{d y}{d x} = \frac{2 x - 5}{x ^{2} - 5 x + 1}

Compute the slope of

y = ln (x^{2} - 5 x + 1)

(5, 0) : m = 5

Find the line with slope m=

5

and passing through

(5, 0) : y = 5 x - 25

y = 5 x - 25

\int_{0}^{\frac{π}{3}} x sin (6 x) d x

\int (1 + x)^{0} d x

\frac{d}{d x} (x - (\frac{1}{2})^{x})

\int cos^{3} (θ) + cos (θ) sin^{2} (θ) d θ

n = 1 \sum \infty \frac{5}{7 ^{n}}

What is the tangent of y=ln(x^2-5x+1),(5,0) ?
The tangent of y=ln(x^2-5x+1),(5,0) is y=5x-25