Find the turning point of P = -4Q2 - 6Q + 7

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 \( x = -\frac{b}{2a} \).

Find the turning point of P = -4Q^2 - 6Q + 7

Discover how to find the turning point of the quadratic function P = -4Q^2 - 6Q + 7 using calculus methods. This solution includes differentiation techniques to identify critical points and explains how to confirm the maximum point. Suitable for students in Grades 10-12.

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