To find the vertex of the quadratic function \( y = x^2 + 6x + 13 \), we can use the vertex formula. The x-coordinate of the vertex for a quadratic in the standard form \( y = ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \).
Here, \( a = 1 \) and \( b = 6 \).
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Calculate the x-coordinate of the vertex: \[ x = -\frac{6}{2(1)} = -\frac{6}{2} = -3 \]
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Now substitute \( x = -3 \) back into the equation to find the y-coordinate: \[ y = (-3)^2 + 6(-3) + 13 \] \[ y = 9 - 18 + 13 \] \[ y = 4 \]
Thus, the vertex of the quadratic function \( y = x^2 + 6x + 13 \) is \( (-3, 4) \).
So the correct response is: (-3, 4)