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

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UpStudy AI · Accepted Answer

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

$$
\boxed{y=7(x-7)}
$$
The equation of the tangent line at the point (7, 0) is $$ y = 7(x - 7) $$.

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

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