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To solve this problem, we need to find the equation of the tangent line to the graph of \( y = g(x) \) at the point \( x = 2 \). 

Given:
- \( g(2) = -4 \) 
- \( g'(2) = 3 \)

The equation of the tangent line at a point \( (x_0, y_0) \) is given by:

\[
y - y_0 = m(x - x_0)
\]

Where \( m \) is the slope of the tangent line, and \( (x_0, y_0) \) is the point of tangency. 

Here:
- \( x_0 = 2 \)
- \( y_0 = g(2) = -4 \)
- \( m = g'(2) = 3 \)

Substitute these values into the equation:

\[
y - (-4) = 3(x - 2)
\]

Simplify this to get the equation in standard form:

\[
y + 4 = 3(x - 2)
\]

\[
y + 4 = 3x - 6
\]

\[
y = 3x - 10
\]

So, the equation of the tangent line is:

\[
y = 3x - 10
\]

Would you like any further details on this solution? Here are some related questions to deepen your understanding:

1. What does the derivative \( g'(x) \) represent geometrically?
2. How do you find the slope of a tangent line if given a function's graph?
3. How would the equation change if \( g'(2) \) was negative?
4. Can you derive the equation of the tangent line using a different method?
5. How does the tangent line equation relate to linear approximations?

**Tip:** Remember, the slope of the tangent line is always equal to the value of the derivative at that specific point.

Find an equation of the tangent line to the graph of y = g(x) at x = 2 if g(2) = -4 and g'(2) = 3.

Learn how to find the equation of the tangent line to the graph of y = g(x) at x = 2, with g(2) = -4 and g'(2) = 3. This problem involves the derivative, the slope of a tangent line, and linear approximation concepts, suitable for Grades 11-12.

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