https://file.aiquickdraw.com/mathbot/file/1727042324353jrdbrj8p.jpg

The task in question involves finding the slope of the tangent line to the graph of a function at a given point. We'll address problem number 11 from the provided image:

**Problem 11:**
\[ f(x) = 2x^2 - 3 \]
At the point \( (2, 5) \).

To find the slope of the tangent line at any given point on a function's graph, we need to compute the derivative of the function, which gives us the slope of the tangent line at any point \( x \).

### Step 1: Compute the derivative of \( f(x) \)
The function given is:
\[ f(x) = 2x^2 - 3 \]

The derivative \( f'(x) \) is computed as follows:
\[ f'(x) = d/dx[2x^2 - 3] \]
\[ f'(x) = 4x \]

### Step 2: Evaluate the derivative at \( x = 2 \)
\[ f'(2) = 4 \times 2 = 8 \]

Thus, the slope of the tangent line to the graph of \( f(x) \) at the point \( (2, 5) \) is 8.

This means that the tangent line is rising at a rate of 8 units vertically for every 1 unit horizontally at the point \( (2, 5) \).

Would you like more details on any of the steps or have any questions?

#### Follow-up Questions:
1. Would you like to solve another problem from the image?
2. How does the derivative relate to the concavity of the function?
3. What is the significance of finding the slope of the tangent line in real-world applications?
4. Would you like to learn how to find the equation of the tangent line itself?
5. Can we explore the behavior of \( f(x) \) at other points?

#### Tip:
When calculating derivatives, remember to apply the power rule which states that \( d/dx[x^n] = nx^{n-1} \). This is a fundamental rule in calculus and is often the first step in solving problems involving rates of change.

Find the slope of the tangent line to the function f(x) = 2x^2 - 3 at the point (2, 5).

This problem demonstrates how to calculate the slope of the tangent line to the quadratic function f(x) = 2x^2 - 3 at the point (2, 5). It covers the derivative using the power rule and evaluates the slope at the specific point.

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