To determine the properties of the function \( f(x) = x^2 + 6x + 11 \), we can analyze it using the standard form of a quadratic function, \( ax^2 + bx + c \).
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Determine the orientation of the parabola:
- The coefficient of \( x^2 \) is positive (1), which means the parabola opens upward.
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Check for x-intercepts:
- To find the x-intercepts, we set \( f(x) = 0 \): \[ x^2 + 6x + 11 = 0 \]
- We can use the discriminant to determine the nature of the roots: \[ D = b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 11 = 36 - 44 = -8 \]
- Since the discriminant is negative (\( D < 0 \)), there are no real roots. This implies that there are no x-intercepts.
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Determine the vertex of the parabola:
- The x-coordinate of the vertex can be found using the formula \( x = -\frac{b}{2a} \): \[ x = -\frac{6}{2 \cdot 1} = -3 \]
- To find the y-coordinate of the vertex, we substitute \( x = -3 \) back into the function: \[ f(-3) = (-3)^2 + 6(-3) + 11 = 9 - 18 + 11 = 2 \]
- Therefore, the vertex is at the point (-3, 2), which is in Quadrant II.
Based on the analysis:
- The parabola opens upward (since \( a > 0 \)),
- There are no x-intercepts (since the discriminant is negative),
- The vertex is located in Quadrant II.
With these considerations, the correct match among the options provided is:
C. On a coordinate plane, an upward opening parabola is in quadrant 2 with no x-intercepts.
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