To find the point that is symmetrical to \((-1, -3)\) for the graph of the function \(y = x^2 + 12x + 8\), we need to identify the axis of symmetry of the parabola defined by this quadratic equation.
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Find the vertex of the parabola using the formula \(x = -\frac{b}{2a}\). Here, \(a = 1\) and \(b = 12\):
\[ x = -\frac{12}{2 \times 1} = -6 \]
Plugging \(x = -6\) back into the equation to find the \(y\)-coordinate:
\[ y = (-6)^2 + 12(-6) + 8 = 36 - 72 + 8 = -28 \]
So, the vertex of the parabola is at \((-6, -28)\).
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Find the symmetrical point: Since the axis of symmetry is a vertical line at \(x = -6\), the symmetrical point to \((-1, -3)\) will have the same \(y\)-coordinate but will be equidistant from \(-6\).
- Calculate the distance from \(-1\) to the axis of symmetry \(-6\):
\[ |(-1) - (-6)| = |-1 + 6| = |5| = 5 \]
Then, move to the left from -6 by 5 units to find the symmetrical point:
\[ -6 - 5 = -11 \]
Thus, the symmetrical point \( (-11, -3) \) retains the same \(y\)-coordinate of \(-3\).
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Final Result: The point symmetrical to \((-1, -3)\) with respect to the graph of \(y = x^2 + 12x + 8\) is
\[ \boxed{(-11, -3)}. \]
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