The problem asks us to graph the quadratic function $y = x^2 + 2x + 4$ and to find the y-intercept.

Step 1: Find the y-intercept

The y-intercept occurs where $x = 0$. So, we substitute $x = 0$ into the equation:

$y = (0)^2 + 2(0) + 4 = 4$

Thus, the y-intercept is $y = 4$.

Step 2: Graph the quadratic function

The quadratic function $y = x^2 + 2x + 4$ is in standard form. We know this is a parabola that opens upwards because the coefficient of $x^2$ is positive.

Key points:

1. Vertex: The vertex can be found using the formula $x = -\frac{b}{2a}$ for a quadratic in the form $y = ax^2 + bx + c$. Here, $a = 1$, $b = 2$, and $c = 4$. $x = -\frac{2}{2(1)} = -1$ Substituting $x = -1$ 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. Axis of symmetry: This is the vertical line passing through the vertex, so $x = -1$.

3. Direction: Since the parabola opens upwards, it has a minimum point at the vertex.

Basic shape of the graph:

• The vertex is at $(-1, 3)$.
• The parabola opens upwards.
• The y-intercept is at $(0, 4)$.

Would you like me to generate a graph for this function or explain anything further? Let me know if you have any questions!

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Follow-up Questions:

1. How do you find the vertex of a quadratic function in standard form?
2. Can you explain the relationship between the coefficients and the direction of the parabola?
3. What is the significance of the axis of symmetry in a quadratic function?
4. How does the vertex relate to the maximum or minimum value of the quadratic function?
5. How would you find the x-intercepts of the quadratic function $y = x^2 + 2x + 4$?

Tip:

For any quadratic equation, the y-intercept is always equal to the constant term $c$ in the equation $y = ax^2 + bx + c$.