Question

Graph the quadratic y=x^2+2x+4
. What is the y-value of the y-intercept?

Answer 1

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:

1. **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 \( (-1, 3) \).

2. **Y-Intercept**: As calculated, the \( y \)-intercept is \( (0, 4) \).

3. **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 \( (0, 4) \).