To graph the function $h(x) = (x + 3)^2 + 6$ using transformations of the basic function $f(x) = x^2$, let's analyze the transformations:

Step-by-Step Transformation:

1. Start with $f(x) = x^2$: This is the basic parabola that opens upwards with its vertex at the origin $(0, 0)$.

2. Horizontal Shift:

   • The term $(x + 3)$ inside the squared function represents a horizontal shift.
   • Specifically, $x + 3$ means that the graph is shifted 3 units to the left.

3. Vertical Shift:

   • The term $+ 6$ outside the squared function represents a vertical shift.
   • Specifically, adding 6 shifts the graph 6 units up.

Conclusion:

Based on the analysis above, the transformations needed are:

• Shift 3 units to the left (Option I).
• Shift 6 units up (Option A).

So, the correct selections are:

• A. Shift the graph 6 units up.
• I. Shift the graph 3 units to the left.

Incorrect Options:

• B and E (Reflect about the x-axis or y-axis): There is no reflection in this function.
• C (Shift 6 units down): The graph is shifted up, not down.
• D and G (Shrink/stretch horizontally): There is no horizontal stretching or shrinking involved.
• F and H (Stretch/shrink vertically): There is no vertical stretching or shrinking.

Would you like details on the individual steps or have any further questions?

Here are 5 related questions that expand on this problem:

1. How would the graph change if the function was $h(x) = -(x + 3)^2 + 6$?
2. What happens if we add a vertical stretch by a factor of 2 to $h(x) = (x + 3)^2 + 6$?
3. How would you graph $h(x) = (x - 3)^2 - 4$ using transformations?
4. What is the effect of a horizontal stretch by a factor of 2 on the graph of $f(x) = x^2$?
5. How would you find the vertex of the transformed function $h(x) = (x + 3)^2 + 6$ without graphing?

Tip: Remember that transformations inside the function $f(x + c)$ affect the graph horizontally, while transformations outside $f(x) + c$ affect the graph vertically.