Find an equation of the tangent line to the graph of the function at the given point.
f(x) = (5x + 3)/(x − 1),    (2, 13)

To find the equation of the tangent line to the graph of the function \( f(x) = \frac{5x + 3}{x - 1} \) at the point \( (2, 13) \), follow these steps:

### Step 1: Verify that the point lies on the curve.
First, we check if the point \( (2, 13) \) is indeed on the graph of the function by substituting \( x = 2 \) into \( f(x) \):

\[
f(2) = \frac{5(2) + 3}{2 - 1} = \frac{10 + 3}{1} = 13.
\]

Since \( f(2) = 13 \), the point \( (2, 13) \) is on the curve.

### Step 2: Find the derivative \( f'(x) \).
The derivative of \( f(x) \) will give us the slope of the tangent line at any point. To differentiate \( f(x) = \frac{5x + 3}{x - 1} \), we apply the quotient rule. The quotient rule states that if \( f(x) = \frac{g(x)}{h(x)} \), then

\[
f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}.
\]

Here, \( g(x) = 5x + 3 \) and \( h(x) = x - 1 \). We find the derivatives of \( g(x) \) and \( h(x) \):

\[
g'(x) = 5 \quad \text{and} \quad h'(x) = 1.
\]

Now, applying the quotient rule:

\[
f'(x) = \frac{(5)(x - 1) - (5x + 3)(1)}{(x - 1)^2} = \frac{5(x - 1) - (5x + 3)}{(x - 1)^2}.
\]

Simplifying the numerator:

\[
f'(x) = \frac{5x - 5 - 5x - 3}{(x - 1)^2} = \frac{-8}{(x - 1)^2}.
\]

Thus, the derivative is:

\[
f'(x) = \frac{-8}{(x - 1)^2}.
\]

### Step 3: Find the slope of the tangent line at \( x = 2 \).
To find the slope of the tangent line at the point \( (2, 13) \), substitute \( x = 2 \) into \( f'(x) \):

\[
f'(2) = \frac{-8}{(2 - 1)^2} = \frac{-8}{1^2} = -8.
\]

The slope of the tangent line is \( -8 \).

### Step 4: Write the equation of the tangent line.
The equation of a line in point-slope form is:

\[
y - y_1 = m(x - x_1),
\]

where \( m \) is the slope, and \( (x_1, y_1) \) is the given point. Substituting \( m = -8 \), \( x_1 = 2 \), and \( y_1 = 13 \) into the equation:

\[
y - 13 = -8(x - 2).
\]

Simplifying:

\[
y - 13 = -8x + 16,
\]

\[
y = -8x + 29.
\]

Thus, the equation of the tangent line is:

\[
y = -8x + 29.
\]

---

Would you like more details or have any other questions?

Here are 5 related questions:
1. How do you apply the product and chain rules in differentiation?
2. Can you find the tangent line at a different point on the same curve?
3. What is the second derivative of \( f(x) = \frac{5x + 3}{x - 1} \), and what does it represent?
4. How does the tangent line equation change if the point is not on the curve?
5. How do you interpret the meaning of the slope of a tangent line in real-world contexts?

**Tip:** When using the quotient rule, carefully simplify the numerator after applying the rule to avoid common errors in signs.

Find an equation of the tangent line to the graph of the function at the given point. f(x) = (5x + 3)/(x − 1), (2, 13)

Learn how to find the equation of the tangent line to the function f(x) = (5x + 3)/(x - 1) at the point (2, 13). This step-by-step guide walks through verifying the point, using the quotient rule to differentiate, and applying the point-slope form for the tangent line. Ideal for students in Grades 11-12 and Calculus 1.

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