The equation given is $y = x^2 + 4$, which is a quadratic equation of the form $y = ax^2 + c$ where $a = 1$ and $c = 4$.

Here’s how we can analyze this to identify the correct graph:

1. Shape of the graph: Since $a = 1$, the graph will be a parabola that opens upwards (as $a > 0$).

2. Vertex: The vertex form of a quadratic function $y = (x - h)^2 + k$ shows that the vertex of the parabola is at $(h, k)$. Here, $h = 0$ and $k = 4$, meaning the vertex of the parabola is at $(0, 4)$.

3. Y-intercept: The graph intersects the y-axis at $(0, 4)$.

4. General behavior: Since this is a simple upward-facing parabola with its vertex shifted 4 units up from the origin, the graph will rise symmetrically on both sides.

Now, let’s examine the options:

• Option b) correctly shows a parabola opening upwards with its vertex at $(0, 4)$, which corresponds to $y = x^2 + 4$.
• Other options either open downwards or have the wrong vertex location.

Thus, option b) is the correct graph for $y = x^2 + 4$.

Would you like more details or have any questions?

Here are some follow-up questions:

1. What happens to the graph of $y = x^2 + c$ as $c$ increases or decreases?
2. How does changing the coefficient $a$ affect the width of the parabola?
3. What is the equation of a parabola that opens downwards with vertex at $(0, -4)$?
4. How would you graph $y = (x - 2)^2 + 4$?
5. What are the roots of $y = x^2 + 4$?

Tip: The value of $c$ in the equation $y = ax^2 + c$ shifts the parabola vertically by $c$ units.