To find the \( y \)-intercept of the quadratic equation \( y = x^2 + 2x + 4 \), we set \( x = 0 \):
\[ y = (0)^2 + 2(0) + 4 = 0 + 0 + 4 = 4 \]
Thus, the \( y \)-value of the \( y \)-intercept is \( 4 \).
To visualize the graph, we can summarize the key points:
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Vertex: The vertex of the quadratic can be found using the formula \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = 2 \):
\[ x = -\frac{2}{2 \cdot 1} = -1 \]
Plugging this back into the equation to find the \( y \)-coordinate of the vertex:
\[ y = (-1)^2 + 2(-1) + 4 = 1 - 2 + 4 = 3 \]
So, the vertex is at \( (-1, 3) \).
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Y-Intercept: As calculated, the \( y \)-intercept is \( (0, 4) \).
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X-Intercepts: To find the \( x \)-intercepts, we set the equation to zero:
\[ x^2 + 2x + 4 = 0 \]
The discriminant \( b^2 - 4ac \) will tell us if there are real roots:
\[ 2^2 - 4(1)(4) = 4 - 16 = -12 \]
Since the discriminant is negative, there are no real \( x \)-intercepts.
This quadratic opens upwards (as \( a > 0 \)), has a vertex at \( (-1, 3) \), and a \( y \)-intercept at \( (0, 4) \).
Summary
- Y-Intercept: \( 4 \)
- Vertex: \( (-1, 3) \)
- No Real X-Intercepts
You can graph this information accordingly, showing the vertex and the intercepts. The graph will be a parabola opening upwards, crossing the \( y \)-axis at \( (0, 4) \).