## Calculus: Early Transcendentals 8th Edition

$y=3x-9$
$y=\ln \left(x^{2}-3 x+1\right)\\ y^{\prime}=\frac{1}{x^{2}-3 x+1} \times 2 x-3\\ y^{\prime}=\frac{2 x-3}{x^{2}-3 x+1}$ Gradient at point $(3,0)$ $y^{\prime}=\frac{2(3)-3}{(3)^{2}-3(3)+1}\\ y^{\prime}=3$ Tangent line at point $(3,0)$ $y-0 =3(x-3)\\ y=3x-9$