To analyze the function $h(x) = \frac{1}{2}(x - 3)^2 - 1$ on the interval $[-1, 5]$ based on the graph, here's the kind of information we can determine:

1. Vertex of the parabola:

The given function is in vertex form $h(x) = a(x - h)^2 + k$, where the vertex is at $(h, k)$.

For $h(x) = \frac{1}{2}(x - 3)^2 - 1$, the vertex is:

• $h = 3$
• $k = -1$

Thus, the vertex is $(3, -1)$.

2. Axis of symmetry:

The axis of symmetry of a parabola in this form is given by the line $x = h$, where $h$ is the x-coordinate of the vertex. So, the axis of symmetry is:

• $x = 3$

3. Direction of opening:

The coefficient of $(x - 3)^2$ is $\frac{1}{2}$, which is positive. This means the parabola opens upwards.

4. Y-intercept:

To find the y-intercept, substitute $x = 0$ into the function: $h(0) = \frac{1}{2}(0 - 3)^2 - 1 = \frac{1}{2}(9) - 1 = 4.5 - 1 = 3.5$ Thus, the y-intercept is at $(0, 3.5)$.

5. X-intercepts (if any):

To find the x-intercepts, we set $h(x) = 0$ and solve for $x$: $0 = \frac{1}{2}(x - 3)^2 - 1$ $1 = \frac{1}{2}(x - 3)^2$ $2 = (x - 3)^2$ $x - 3 = \pm \sqrt{2}$ $x = 3 \pm \sqrt{2}$

Thus, the x-intercepts are $x = 3 + \sqrt{2} \approx 4.41$ and $x = 3 - \sqrt{2} \approx 1.59$.

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

---

Here are 5 questions that build upon this:

1. How does changing the coefficient of $\frac{1}{2}$ affect the graph's shape?
2. What would happen to the graph if the vertex was moved to $(2, -3)$?
3. How can you calculate the domain and range of this function?
4. How would the graph change if it opened downwards instead?
5. What are the minimum and maximum values of $h(x)$ on the interval $[-1, 5]$?

Tip: For vertex form, the vertex $(h, k)$ can be directly read from $y = a(x - h)^2 + k$, simplifying graphing!