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To solve this problem, we need to find the equation of the tangent line to the graph of \( f(x) = 2x - 6 \) at \( x = 1 \).

Here’s how to solve it:

### 1. Find the slope of the tangent line.
The derivative of \( f(x) \) gives the slope of the tangent line. Since \( f(x) = 2x - 6 \), the derivative is:

\[
f'(x) = 2
\]

Thus, the slope of the tangent line at any point is 2, including at \( x = 1 \).

### 2. Find the point on the graph at \( x = 1 \).
Substitute \( x = 1 \) into \( f(x) = 2x - 6 \) to find the \( y \)-coordinate of the point:

\[
f(1) = 2(1) - 6 = -4
\]

So, the point of tangency is \( (1, -4) \).

### 3. Use the point-slope form of the line equation.
The point-slope form of a line is:

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

Where \( m \) is the slope and \( (x_1, y_1) \) is the point on the line. Substituting \( m = 2 \), \( x_1 = 1 \), and \( y_1 = -4 \):

\[
y - (-4) = 2(x - 1)
\]

Simplifying:

\[
y + 4 = 2x - 2
\]

\[
y = 2x - 6
\]

### Conclusion:
The equation of the tangent line is:

\[
y = 2x - 6
\]

Would you like more details on this process? Here are some related questions to deepen your understanding:

1. How do you find the derivative of more complex functions?
2. What is the significance of the point-slope form in determining the equation of a tangent line?
3. How would this problem change if the function was non-linear, such as \( f(x) = x^2 - 6x \)?
4. How does the slope of a tangent line relate to the behavior of a function at that point?
5. What would happen if you were asked to find the normal line instead of the tangent line?

**Tip:** The derivative of a function gives the slope of the tangent line at any point on the curve. Always start by finding the derivative when solving problems involving tangent lines.

Find the equation of the line tangent to the graph of f(x) = 2x - 6 at x = 1.

Find the equation of the tangent line to the function f(x) = 2x - 6 at the point x = 1. This problem involves finding the derivative of the linear function and using the point-slope form of the line equation. Suitable for high school students learning calculus.

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