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.