To graph the quadratic equation \( y = x^2 + 2x + 4 \), we can start by identifying key features of the parabola.
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Y-intercept: The y-intercept occurs where \( x = 0 \). \[ y = 0^2 + 2(0) + 4 = 4 \] Thus, the y-value of the y-intercept is \( 4 \).
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Vertex: The vertex of a quadratic equation in the standard form \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = 2 \). \[ x = -\frac{2}{2(1)} = -1 \] Now, substitute \( x = -1 \) back into the equation to find the y-value of the vertex. \[ y = (-1)^2 + 2(-1) + 4 = 1 - 2 + 4 = 3 \] So, the vertex is at the point \( (-1, 3) \).
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Direction: Since \( a = 1 \) (which is positive), the parabola opens upwards.
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Finding Additional Points: To get a better sense of the graph, we can calculate some additional points. Let's plug in \( x = -2 \) and \( x = 0 \) (already done) and \( x = 1 \):
- For \( x = -2 \): \[ y = (-2)^2 + 2(-2) + 4 = 4 - 4 + 4 = 4 \quad \text{(point: } (-2, 4) \text{)} \]
- For \( x = 1 \): \[ y = (1)^2 + 2(1) + 4 = 1 + 2 + 4 = 7 \quad \text{(point: } (1, 7) \text{)} \]
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Plotting Points: The points we have identified are:
- Vertex: \( (-1, 3) \)
- Y-intercept: \( (0, 4) \)
- Another point: \( (-2, 4) \)
- Another point: \( (1, 7) \)
Putting these points together, you would plot them on a graph. The parabola will open upwards, with the vertex as the lowest point.
In conclusion, the y-value of the y-intercept is 4.