To find the vertex, axis of symmetry, maximum or minimum value, and range of the parabola given by the equation \( y = x^2 + 2x + 5 \), we can follow these steps:
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Finding the vertex: The vertex form of a quadratic equation can be determined using the formula: \[ x = -\frac{b}{2a} \] where \( a \) is the coefficient of \( x^2 \) and \( b \) is the coefficient of \( x \).
In the equation \( y = x^2 + 2x + 5 \):
- \( a = 1 \)
- \( b = 2 \)
Now, substituting in the values: \[ x = -\frac{2}{2 \cdot 1} = -1 \]
Next, substitute \( x = -1 \) back into the equation to find the \( y \)-coordinate of the vertex: \[ y = (-1)^2 + 2(-1) + 5 = 1 - 2 + 5 = 4 \]
Thus, the vertex is: \[ \text{Vertex} = (-1, 4) \]
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Axis of symmetry: The axis of symmetry is the vertical line that passes through the vertex. This can be given by the equation: \[ x = -1 \]
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Maximum or minimum value: Since the coefficient of \( x^2 \) (which is \( a = 1 \)) is positive, the parabola opens upwards. Therefore, the vertex represents the minimum point of the parabola.
Minimum value of \( y \): \[ \text{Minimum value} = 4 \]
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Range of the parabola: Since the parabola opens upwards and the minimum value is 4, the range of the function is: \[ \text{Range} = [4, \infty) \]
Final Results:
- Vertex: \( (-1, 4) \)
- Axis of symmetry: \( x = -1 \)
- Minimum value: \( 4 \)
- Range: \( [4, \infty) \)