To find the slope of the function's graph at the given point $(-3, 3)$ for the function $f(x) = 2x^2 + 5x$, we need to calculate the derivative of $f(x)$ and then evaluate it at $x = -3$.

Step 1: Find the derivative

The derivative of $f(x) = 2x^2 + 5x$ with respect to $x$ is: $f'(x) = \frac{d}{dx}(2x^2 + 5x)$ Using the power rule for differentiation: $f'(x) = 4x + 5$

Step 2: Evaluate the derivative at $x = -3$

Now, substitute $x = -3$ into $f'(x)$: $f'(-3) = 4(-3) + 5 = -12 + 5 = -7$

Conclusion

The slope of the function at the point $(-3, 3)$ is $\boxed{-7}$.

Would you like further details or have any questions?

Here are 5 related questions for further exploration:

1. What is the geometric interpretation of a negative slope?
2. How would you find the equation of the tangent line at $x = -3$?
3. What is the second derivative of $f(x) = 2x^2 + 5x$ and what does it tell us?
4. How does the slope of a quadratic function change as $x$ increases or decreases?
5. What are the critical points of $f(x) = 2x^2 + 5x$, and how do we find them?

Tip: When differentiating polynomial functions, the power rule $\frac{d}{dx} x^n = nx^{n-1}$ is your go-to method for quick and accurate results.