Find an equation of the tangent line to the curve at the given poin \[ y=\ln \left(x^{2}-7 x+1\right), \quad(7,0) \] \( y=\square \)

We begin with the function \[ y=\ln\left(x^2-7x+1\right) \] and the point \[ (7, 0). \] ### Step 1: Compute the derivative of \( y \) Using the chain rule, if \[ y=\ln(u(x)), \] then \[ y'=\frac{u'(x)}{u(x)}. \] Here, \( u(x)=x^2-7x+1 \), so its derivative is \[ u'(x)=2x-7. \] Thus, the derivative of \( y \) is \[ y'=\frac{2x-7}{x^2-7x+1}. \] ### Step 2: Evaluate the derivative at \( x=7 \) Substitute \( x = 7 \) into the derivative: \[ y'(7)=\frac{2(7)-7}{7^2-7(7)+1} = \frac{14-7}{49-49+1} = \frac{7}{1}=7. \] So, the slope of the tangent line at \( x=7 \) is \( 7 \). ### Step 3: Form the equation of the tangent line Using the point-slope form: \[ y-y_1=m(x-x_1), \] where \( m=7 \) and the point is \( (x_1, y_1)=(7,0) \), we have \[ y-0=7(x-7). \] ### Final Answer \[ \boxed{y=7(x-7)} \]