What is the tangent of y=ln(x^2-3x+1),(3,0) ?

The tangent of y=ln(x^2-3x+1),(3,0) is y=3x-9

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tangent of y=ln(x^2-3x+1),(3,0)

tangent of

y = ln (x^{2} - 3 x + 1), (3, 0)

y = 3 x - 9

Solution steps

Find the slope of

y = ln (x^{2} - 3 x + 1) : \frac{d y}{d x} = \frac{2 x - 3}{x ^{2} - 3 x + 1}

Compute the slope of

y = ln (x^{2} - 3 x + 1)

(3, 0) : m = 3

Find the line with slope m=

3

and passing through

(3, 0) : y = 3 x - 9

y = 3 x - 9

y^{^{'}} = \frac{x - y sin ( 2 x )}{sin ^{2} ( x )}

f (x) = 3 sin^{2} (x)

y^{^{''}} + 9 y = t^{2} + cos (3 t)

\frac{s - 5}{s ^{2} + 4 s + 20}

\int \frac{y ^{2}}{( y + 1 ) ^{4}} d y

What is the tangent of y=ln(x^2-3x+1),(3,0) ?
The tangent of y=ln(x^2-3x+1),(3,0) is y=3x-9