Find an equation of the tangent line to the graph of
y = g(x) at x = 3 if g(3) = −4 and g'(3) = 5.

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.

Find an equation of the tangent line to the graph of y = g(x) at x = 3 if g(3) = −4 and g'(3) = 5.

Learn how to find the equation of the tangent line to the graph of y = g(x) at x = 3 with given values g(3) = −4 and g'(3) = 5. This problem explores key calculus concepts such as differentiation and the point-slope form, suitable for students in Grades 11-12.

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