To find the coordinates of the vertex of the parabola defined by the equation \( y = x^2 + 8x + 11 \), we can use the vertex formula. The x-coordinate of the vertex of a parabola given by \( y = ax^2 + bx + c \) can be found using the formula:
\[ x = -\frac{b}{2a} \]
In this equation:
- \( a = 1 \) (the coefficient of \( x^2 \))
- \( b = 8 \) (the coefficient of \( x \))
- \( c = 11 \) (the constant term)
Substituting \( a \) and \( b \) into the vertex formula:
\[ x = -\frac{8}{2 \cdot 1} = -\frac{8}{2} = -4 \]
Now that we have the x-coordinate of the vertex, we can find the y-coordinate by substituting \( x = -4 \) back into the original equation:
\[ y = (-4)^2 + 8(-4) + 11 \] \[ y = 16 - 32 + 11 \] \[ y = 16 - 32 + 11 = -5 \]
Thus, the coordinates of the vertex of the parabola are:
\[ \boxed{(-4, -5)} \]
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