To find the vertex of the quadratic function given the x-intercepts, we can use the fact that the vertex lies on the axis of symmetry of the parabola, which is halfway between the two x-intercepts.
Given the x-intercepts at \( x = -6 \) and \( x = 2 \), we can calculate the x-coordinate of the vertex as follows:
\[ \text{Vertex } x = \frac{-6 + 2}{2} = \frac{-4}{2} = -2 \]
Next, we need to find the y-coordinate of the vertex. We can express the quadratic function in vertex form or standard form. Since we have the x-intercepts, we can initially write the quadratic in factored form as:
\[ f(x) = a(x + 6)(x - 2) \]
We need to determine the value of \( a \) using the additional point provided: (-5, -14).
Substituting \( x = -5 \) and \( f(x) = -14 \):
\[ -14 = a(-5 + 6)(-5 - 2) \]
This simplifies to:
\[ -14 = a(1)(-7) \implies -14 = -7a \implies a = 2 \]
Now that we have \( a \), the quadratic function becomes:
\[ f(x) = 2(x + 6)(x - 2) \]
Now we'll expand this to get it into standard form:
\[ f(x) = 2(x^2 + 4x - 12) \] \[ = 2x^2 + 8x - 24 \]
In standard form, the quadratic is \( f(x) = 2x^2 + 8x - 24 \).
To find the vertex, we can use the vertex formula for the x-coordinate:
\[ x = -\frac{b}{2a} = -\frac{8}{2 \cdot 2} = -\frac{8}{4} = -2 \]
Now substituting \( x = -2 \) back into the quadratic function to find the y-coordinate of the vertex:
\[ f(-2) = 2(-2)^2 + 8(-2) - 24 \] \[ = 2(4) - 16 - 24 \] \[ = 8 - 16 - 24 = -32 \]
Thus, the vertex of the quadratic function is:
\[ \boxed{(-2, -32)} \]
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