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The problem shows the curve \( y = x^2 + 6x - 14 \), and we are asked to find the coordinates of the turning point \( P \). To do this, we need to find the vertex of the parabola, which represents the turning point.

### Steps:
1. The vertex of a parabola in the form \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex:
   \[
   x = -\frac{b}{2a}
   \]
   Here, \( a = 1 \), \( b = 6 \), and \( c = -14 \).

   Substituting the values:
   \[
   x = -\frac{6}{2(1)} = -3
   \]

2. To find the y-coordinate of the turning point, substitute \( x = -3 \) into the original equation:
   \[
   y = (-3)^2 + 6(-3) - 14
   \]
   \[
   y = 9 - 18 - 14 = -23
   \]

Thus, the coordinates of the turning point \( P \) are \( (-3, -23) \).

Would you like a detailed explanation on any specific step?

### Related questions:
1. How do you complete the square for the equation \( y = x^2 + 6x - 14 \)?
2. What is the axis of symmetry for the curve \( y = x^2 + 6x - 14 \)?
3. How can we determine if the turning point is a maximum or a minimum?
4. How do we find the roots of the equation \( y = x^2 + 6x - 14 = 0 \)?
5. How does changing the coefficients in a quadratic equation affect the graph?

### Tip:
The x-coordinate of the turning point of a parabola can always be found using \( x = -\frac{b}{2a} \), regardless of the complexity of the quadratic.

The diagram shows the curve y = x^2 + 6x - 14. P is the turning point of the curve. Work out the coordinates of P.

Learn how to find the turning point of the quadratic curve y = x^2 + 6x - 14. Follow a step-by-step solution to calculate the vertex using the formula x = -b/(2a) and substitute the value back into the equation to find the y-coordinate. Suitable for Grades 9-11.

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