what point is symmetrical to (-1, -3) for the graph of y = x^2 + 12x + 8?
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To find the point that is symmetrical to (-1, -3), we need to calculate the axis of symmetry and then reflect the given point across it.
The axis of symmetry for a quadratic function in the form y = ax^2 + bx + c can be found using the formula: x = -b/2a.
For the given equation y = x^2 + 12x + 8:
a = 1, b = 12
The axis of symmetry can be found as follows:
x = -b/2a
x = -12 / (2 * 1)
x = -12 / 2
x = -6
The x-coordinate of the point that is symmetrical to (-1, -3) is -6. To find the corresponding y-coordinate, we substitute x = -6 into the equation y = x^2 + 12x + 8:
y = (-6)^2 + 12(-6) + 8
y = 36 - 72 + 8
y = -28
Therefore, the point that is symmetrical to (-1, -3) for the graph of y = x^2 + 12x + 8 is (-6, -28).