Question

find an equation of the tangent line to the curve at the given point. y = ln(x^2 - 3x + 1), (3, 0) y =

Explanation:

Step1: Differentiate the function

Use the chain - rule. If $y = \ln(u)$ and $u=x^{2}-3x + 1$, then $\frac{dy}{du}=\frac{1}{u}$ and $\frac{du}{dx}=2x - 3$. By the chain - rule $\frac{dy}{dx}=\frac{2x - 3}{x^{2}-3x + 1}$.

Step2: Find the slope of the tangent line

Substitute $x = 3$ into $\frac{dy}{dx}$. When $x = 3$, $\frac{dy}{dx}=\frac{2\times3-3}{3^{2}-3\times3 + 1}=\frac{6 - 3}{9-9 + 1}=3$.

Step3: Use the point - slope form of a line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(3,0)$ and $m = 3$. Substituting these values gives $y-0 = 3(x - 3)$.

Step4: Simplify the equation

$y=3x-9$.

Answer:

$y = 3x-9$