To find the turning point of the function $P = -4Q^2 - 6Q + 7$, we can complete the square or use calculus. Here, I’ll use calculus for clarity.

The turning point of a quadratic function occurs where the derivative of $P$ with respect to $Q$ is zero:

1. Find the first derivative with respect to $Q$: $\frac{dP}{dQ} = -8Q - 6$

2. Set the derivative to zero to find the critical point: $-8Q - 6 = 0$ Solving for $Q$: $Q = -\frac{6}{-8} = \frac{3}{4}$

3. Substitute $Q = \frac{3}{4}$ back into the original function to find the corresponding $P$ value: $P = -4\left(\frac{3}{4}\right)^2 - 6\left(\frac{3}{4}\right) + 7$ $= -4 \cdot \frac{9}{16} - \frac{18}{4} + 7$ $= -\frac{36}{16} - \frac{18}{4} + 7$ $= -2.25$

So, the turning point of the function is at $\left( Q, P \right) = \left( \frac{3}{4}, -2.25 \right)$. Since the coefficient of $Q^2$ is negative, this point represents a maximum.

Would you like more details on any step?

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1. How do you determine if a turning point is a maximum or minimum?
2. Can we find the turning point by completing the square instead?
3. What are other methods to verify that the point is indeed a maximum?
4. How does changing the coefficients affect the location of the turning point?
5. How do you interpret the turning point in an economic context?

Tip: For quadratic functions $ax^2 + bx + c$, the turning point can be quickly found using $x = -\frac{b}{2a}$.