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We are tasked with finding the equation of the tangent line to the function \( f(x) = 2x^2 + 5x \) at the point \( (-3, 3) \).

### Steps to find the equation of the tangent line:

1. **Find the derivative \( f'(x) \)**: The derivative will give us the slope of the tangent line at any point \( x \).
   
   \( f(x) = 2x^2 + 5x \)
   
   To differentiate:
   \[
   f'(x) = \frac{d}{dx}(2x^2) + \frac{d}{dx}(5x) = 4x + 5
   \]

2. **Evaluate the derivative at \( x = -3 \)** to get the slope of the tangent line:
   
   \[
   f'(-3) = 4(-3) + 5 = -12 + 5 = -7
   \]
   So, the slope of the tangent line is \( m = -7 \).

3. **Use the point-slope form of the equation of a line**: The point-slope form is given by:
   \[
   y - y_1 = m(x - x_1)
   \]
   where \( (x_1, y_1) \) is the point \( (-3, 3) \) and \( m = -7 \).
   
   Substituting the values:
   \[
   y - 3 = -7(x + 3)
   \]
   Expanding this:
   \[
   y - 3 = -7x - 21
   \]
   \[
   y = -7x - 21 + 3
   \]
   \[
   y = -7x - 18
   \]

Thus, the equation of the tangent line is \( y = -7x - 18 \).

Let me know if you'd like more details or have questions!

Here are 5 related questions you might find interesting:
1. How do you compute the second derivative of \( f(x) = 2x^2 + 5x \)?
2. What does the second derivative tell us about the concavity of the function at \( x = -3 \)?
3. Can you find the equation of the normal line to \( f(x) \) at \( (-3, 3) \)?
4. How would the tangent line change if we evaluated it at a different point on the curve?
5. How do we interpret the slope of the tangent line in terms of the graph's behavior?

**Tip**: The derivative represents the slope of the tangent line, which is the rate of change of the function at a specific point!

find an equation for the line tangent f(x) = 2x^2 + 5x, (-3, 3)

Learn how to find the equation of the tangent line to the function f(x) = 2x^2 + 5x at the point (-3, 3). This solution covers key concepts in differentiation and algebra, ideal for high school math students.

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