To solve this problem, let's analyze the quadratic function $h(x) = x^2 + 3$ and address the key points asked:

1. Vertex

The function $h(x) = x^2 + 3$ is in the form $h(x) = a(x - h)^2 + k$, where:

• $a = 1$,
• $h = 0$, and
• $k = 3$.

This is the standard form of a parabola with a vertex at $(h, k)$. Therefore, the vertex of $h(x) = x^2 + 3$ is at: $(0, 3)$

2. Axis of Symmetry

For any function in the form $h(x) = a(x - h)^2 + k$, the axis of symmetry is the vertical line $x = h$. Since $h = 0$ in this case, the axis of symmetry is: $x = 0$

3. Graphing

To sketch the graph:

1. Start by plotting the vertex at $(0, 3)$.
2. Because the coefficient of $x^2$ is positive (1), the parabola opens upwards.
3. Draw a solid curve representing the parabola.
4. Draw a dashed vertical line at $x = 0$ to represent the axis of symmetry.

The graph will look like a U-shaped curve centered at $(0, 3)$ with symmetry along the $y$-axis (since $x = 0$).

Would you like more details or have any questions on this?

Additional Related Questions

1. How would the graph change if the equation was $h(x) = -x^2 + 3$?
2. What happens to the vertex if we modify the equation to $h(x) = (x - 2)^2 + 3$?
3. How does changing the $+3$ term affect the graph of the function?
4. How can we determine the width of the parabola from the function?
5. How would we find the vertex and axis of symmetry for a general quadratic function $ax^2 + bx + c$?

Tip

To quickly determine the vertex of a quadratic in standard form $ax^2 + bx + c$, use the formula $x = -\frac{b}{2a}$ for the x-coordinate of the vertex.