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To find the equation of the tangent line at the point \((2, f(2))\) given \(f(2) = 13\) and \(f'(2) = 4\), we use the point-slope form of a line:

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

where:

- \((x_1, y_1)\) is the point on the curve.
- \(m\) is the slope of the tangent line, which is given by the derivative \(f'(x)\) at \(x = x_1\).

### Given Values:
- Point on the curve: \((x_1, y_1) = (2, f(2)) = (2, 13)\)
- Slope of the tangent line: \(m = f'(2) = 4\)

### Plugging in the Values:
Using the point-slope form:

\[
y - 13 = 4(x - 2)
\]

To express this equation in standard form or slope-intercept form, we can simplify further:

\[
y - 13 = 4x - 8
\]

\[
y = 4x - 8 + 13
\]

\[
y = 4x + 5
\]

### Final Equation of the Tangent Line:
\[
y = 4x + 5
\]

Would you like more details or have any questions?

Here are 5 related questions you might find helpful:
1. How do you find the slope of a tangent line at a given point?
2. What is the difference between the tangent line and the secant line?
3. How can you find the equation of a normal line to a curve?
4. What role does the derivative play in determining the slope of a tangent?
5. How does the point-slope form of a line compare to the slope-intercept form?

**Tip:** The point-slope form is particularly useful when you know a point on a line and the slope. It is often used to write the equation of tangent lines to functions.

Find the equation of the tangent line at (2, f(2)) when f(2) = 13 and f'(2) = 4.

This calculus problem involves finding the equation of a tangent line at a specific point, given f(2) = 13 and f'(2) = 4. Learn how to use the point-slope form and apply derivatives to determine the slope of the tangent line.

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