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.