The image appears to contain mathematical problems, asking for the determination of points on the graph where the derivative is equal to zero, suggesting that the task is to find the critical points of the functions provided. This generally involves finding where the slope of the tangent line to the graph (i.e., the derivative) equals zero, which can indicate maxima, minima, or saddle points.

Let's work through a few examples from the list:

1. $y = x^2 - 6x$ (Exercise 21)

   • Step 1: Find the derivative $y'$. $y' = \frac{d}{dx}(x^2 - 6x) = 2x - 6$
   • Step 2: Set the derivative equal to zero to find the critical points. $2x - 6 = 0 \implies x = 3$
   • Step 3: Substitute $x = 3$ back into the original function to find the y-coordinate. $y = 3^2 - 6(3) = 9 - 18 = -9$
   • Critical Point: $(3, -9)$

2. $y = -3x^2 + 3x + 2$ (Exercise 22)

   • Derivative: $y' = \frac{d}{dx}(-3x^2 + 3x + 2) = -6x + 3$
   • Set to zero: $-6x + 3 = 0 \implies x = \frac{1}{2}$
   • Substitute $x = \frac{1}{2}$ back: $y = -3\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2}\right) + 2 = -\frac{3}{4} + \frac{3}{2} + 2 = \frac{13}{4}$
   • Critical Point: $\left(\frac{1}{2}, \frac{13}{4}\right)$

If you would like, I can continue solving more of these exercises or provide a detailed step-by-step approach for each function.

Would you like to see further solutions or explanations?

Relative Questions:

1. How do you determine if a critical point is a maximum, minimum, or saddle point?
2. Can you explain how the second derivative test helps in identifying the nature of the critical points?
3. What are some practical applications of finding critical points in real-world scenarios?
4. Why is it important to understand the behavior of a function's graph through its critical points?
5. What is the difference between local maxima/minima and global maxima/minima?

Tip: When finding critical points, always check if the points are within the domain of the function to ensure they are valid solutions.