https://file.aiquickdraw.com/mathbot/file/1727228717930wc9u0i23.jpg

To solve the problem:

We are tasked with finding the equation of the tangent line to the graph of \( y = g(x) \) at \( x = 5 \), given that \( g(5) = -2 \) and \( g'(5) = 6 \). The tangent line equation takes the form:

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

Where:
- \( m \) is the slope of the tangent line, which is the derivative \( g'(x) \) at \( x = 5 \), so \( m = 6 \).
- \( (x_1, y_1) \) is the point where the tangent line touches the curve, which is \( (5, g(5)) = (5, -2) \).

Now, using the point-slope form:

\[
y - (-2) = 6(x - 5)
\]

Simplify the equation:

\[
y + 2 = 6(x - 5)
\]
\[
y + 2 = 6x - 30
\]
\[
y = 6x - 32
\]

Thus, the equation of the tangent line is:

\[
y = 6x - 32
\]

Would you like further details or have any questions?

Here are five related questions to explore this concept:
1. How do you find the derivative of a given function at a point?
2. Why is the point-slope form used for tangent lines?
3. How can you determine if a function has a horizontal tangent line?
4. What is the geometric interpretation of the derivative at a point?
5. How do you interpret the behavior of a function using the slope of the tangent line?

**Tip**: When given \( g'(x) \), the value represents the slope of the tangent line at that specific point on the function's graph.

Find an equation of the tangent line to the graph of y = g(x) at x = 5 if g(5) = -2 and g'(5) = 6.

Learn how to find the equation of the tangent line to the curve y = g(x) at x = 5. The problem involves applying the point-slope formula with the derivative g'(x) to determine the tangent line. Given g(5) = -2 and g'(5) = 6, this solution is suitable for Grade 12 and early college calculus students.

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