To find the equation of the tangent line to the graph of $y = g(x)$ at $x = 3$, we can use the point-slope form of a line:

$y - y_1 = m(x - x_1)$

where:

• $(x_1, y_1)$ is the point of tangency,
• $m$ is the slope of the tangent line.

Given:

• $x_1 = 3$,
• $y_1 = g(3) = -4$,
• $m = g'(3) = 5$ (since the derivative gives the slope of the tangent line).

Now, substituting these values into the point-slope form:

$y - (-4) = 5(x - 3)$

Simplifying:

$y + 4 = 5(x - 3)$

$y = 5(x - 3) - 4$

$y = 5x - 15 - 4$

$y = 5x - 19$

Thus, the equation of the tangent line is:

$y = 5x - 19$

Would you like more details or have any questions about this process?

Here are 5 related questions:

1. How would you find the tangent line if $g'(3)$ was not provided?
2. What is the geometric meaning of the derivative $g'(x)$?
3. How would the equation change if the slope $g'(3)$ were negative?
4. How does the point-slope form compare to the slope-intercept form for tangent lines?
5. Can you find the equation of the normal line to the curve at $x = 3$?

Tip: The slope of a tangent line at a point is the value of the derivative at that point.