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The image contains a series of calculus problems (7 to 18) asking to find the equation of the tangent line to the graph of a function \( f(x) \) at a given value of \( x \). To solve each of these problems, we will follow these steps:

1. **Find the derivative** of the function, \( f'(x) \), to determine the slope of the tangent line.
2. **Evaluate the derivative at the given point**, i.e., plug the \( x \)-value into \( f'(x) \), to find the slope \( m \) of the tangent line.
3. **Find the point on the curve** by calculating \( f(x) \) at the given value of \( x \).
4. **Use the point-slope form** of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is the point on the graph.

Let's work through problem 7 as an example:

**Problem 7:**
\[ f(x) = x^2 - 6, \ x = 3 \]

### Step 1: Find the derivative
\[ f'(x) = \frac{d}{dx}(x^2 - 6) = 2x \]

### Step 2: Evaluate the derivative at \( x = 3 \)
\[ f'(3) = 2(3) = 6 \]

So, the slope \( m \) of the tangent line at \( x = 3 \) is 6.

### Step 3: Find the point on the curve
We plug \( x = 3 \) into the original function to find \( y_1 \):
\[ f(3) = 3^2 - 6 = 9 - 6 = 3 \]

So, the point on the curve is \( (3, 3) \).

### Step 4: Use the point-slope form
The point-slope form is:
\[ y - y_1 = m(x - x_1) \]

Substituting \( m = 6 \), \( x_1 = 3 \), and \( y_1 = 3 \):
\[ y - 3 = 6(x - 3) \]

Simplifying:
\[ y = 6x - 18 + 3 = 6x - 15 \]

Thus, the equation of the tangent line is:
\[ y = 6x - 15 \]

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Would you like me to proceed with more problems or specific ones from the list? Feel free to ask for details or clarifications!

Here are five related questions to expand on this topic:

1. How is the derivative used to find the slope of the tangent line?
2. What is the point-slope form of a line, and when is it useful?
3. How can you verify if the tangent line equation is correct?
4. How would the tangent line change if we shift the point of tangency?
5. What happens to the tangent line if the function has a horizontal tangent?

**Tip:** The derivative of a function gives the rate of change or slope of the function at any point, which is why it's key to finding tangent lines!

Find the equation of the tangent line to the graph of the function f(x) at the given value of x.

Learn how to find the equation of the tangent line for a given function f(x) at a specific point x using derivatives. This guide explains step-by-step how to compute the slope and apply the point-slope formula. Suitable for advanced high school students in Grades 11-12.

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