To find the point that is symmetrical to \((-1, -3)\) for the graph of the quadratic equation \(y = x^2 + 12x + 8\), we first need to determine the vertex of this parabola.
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Identify the Vertex: The vertex of a quadratic equation in standard form \(y = ax^2 + bx + c\) can be found using the formula: \[ x = -\frac{b}{2a} \] For \(y = x^2 + 12x + 8\), we have:
- \(a = 1\)
- \(b = 12\)
Plugging in the values: \[ x = -\frac{12}{2 \cdot 1} = -6 \]
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Find the y-coordinate of the Vertex: Now, substitute \(x = -6\) back into the equation to find the y-coordinate of the vertex: \[ y = (-6)^2 + 12(-6) + 8 \] \[ y = 36 - 72 + 8 = -28 \] So, the vertex is \((-6, -28)\).
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Determine the axis of symmetry: The axis of symmetry for a parabola is a vertical line that passes through its vertex, which in this case is \(x = -6\).
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Find the symmetrical point: The point symmetric to \((-1, -3)\) relative to the axis of symmetry, which is \(x = -6\), means we need to find the reflection across the line \(x = -6\).
The distance from \((-1)\) to \(-6\) is: \[ -6 - (-1) = -6 + 1 = -5 \] Therefore, the symmetrical x-coordinate is: \[ -6 - 5 = -11 \]
The y-coordinate remains the same, so the symmetrical point is \((-11, -3)\).
In conclusion, the point that is symmetrical to \((-1, -3)\) for the graph of \(y = x^2 + 12x + 8\) is \((-11, -3)\).
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