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Thank you for uploading the image. It contains a math problem from a Thai textbook (KUMON K55a, Part 2), asking to find the value of a such that the function f(x) = (x - 5)² + 3 has a minimum value in the interval [1, 5], and then to find that minimum value.

The problem is broken into three parts based on different cases for a, and each case gives a graph and an inequality condition. Let’s solve it step by step.

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## 🔍 Understanding the Function

We are given:
> f(x) = (x - 5)² + 3

This is a quadratic function in vertex form. Its graph is a parabola opening upwards with vertex at (5, 3).

- Since the parabola opens upward, the minimum value occurs at the vertex.
- But we are restricted to the interval [1, 5] — so we must consider where the vertex lies relative to this interval.

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## 📌 Key Insight

The vertex is at x = 5, which is the right endpoint of the interval [1, 5].

Therefore, within the interval [1, 5], the function is decreasing as x approaches 5 from the left (since the parabola opens up and vertex is at x=5).

So, the minimum value on [1,5] occurs at x = 5, and f(5) = 3.

But the problem is structured around finding a such that the minimum occurs at a specific point depending on the value of a.

Looking at the graphs and conditions:

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## ✔ Part (1): When a + 4 < 5

This implies:
> a < 1

Graph shows the interval [a, a+4] is entirely to the left of x=5, and since the parabola opens upward and vertex is at x=5, the function is decreasing on [a, a+4].

→ So the minimum occurs at the right endpoint: x = a + 4

Therefore:
> f(a+4) = ((a+4) - 5)² + 3 = (a - 1)² + 3

✔ Answer for (1):
- x = a + 4
- f(x) = (a - 1)² + 3

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## ✔ Part (2): When 1 ≤ a ≤ 3

This means the interval [a, a+4] includes x=5, because:
- If a = 1 → interval = [1, 5]
- If a = 3 → interval = [3, 7] → but we are only considering up to x=5? Wait — let’s check.

Actually, looking at the graph in part (2), the shaded region is from a to 5, not to a+4. That suggests that although the domain is defined as [a, a+4], we are restricting our attention to [a, 5] because the problem says “ในช่วง [1,5]” — meaning we care about behavior within [1,5].

Wait — rereading the problem statement:

> "ให้ฟังก์ชัน f(x) = (x−5)² + 3 (a ≤ x ≤ a+4) จงหาค่าของ a ที่ทำให้ฟังก์ชันนี้มีค่าต่ำสุดที่ x=5 ในช่วง [1,5]"

Translation:
> Given the function f(x) = (x−5)² + 3, with domain [a, a+4]. Find the value of a such that the function has its minimum at x=5 in the interval [1,5].

So the domain is [a, a+4], but we are evaluating the minimum in [1,5] — meaning we’re looking at the intersection: [a, a+4] ∩ [1,5].

That explains why in part (2), the shaded region goes from a to 5 — because a+4 may be >5, but we only care up to 5.

So for part (2):

> Condition: 1 ≤ a ≤ 3

Why 3? Because if a = 3, then a+4 = 7 → so the interval [3,7] intersects [1,5] to give [3,5].

And since the vertex x=5 is inside [a, a+4] ∩ [1,5] = [a,5] (because a+4 ≥ 5 when a≥1), the minimum occurs at x=5.

✔ So for part (2):
- x = 5
- f(x) = f(5) = (5-5)² + 3 = 0 + 3 = 3

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## ✔ Part (3): When a > 3

Then a + 4 > 7, so the interval [a, a+4] starts after 3, and since a > 3, the intersection with [1,5] is [a,5] (if a < 5) or empty (if a ≥ 5). But since we need the function to have a minimum in [1,5], we assume a < 5.

Actually, the condition in part (3) is written as:

> "เมื่อ a > 3" — and the graph shows interval from a to 5.

So again, we are looking at [a,5] ⊆ [1,5].

Since the function is decreasing on (-∞, 5], then on [a,5] (with a > 3), the function is still decreasing → so minimum is at x=5.

Wait — but the problem says in part (3):

> "เมื่อ a > 3" — and then says "ค่าต่ำสุดเกิดที่ x = ?"

Actually, let’s look at the text in the box for part (3):

> "(ข้อ) ถ้า f(a) = f(a+4), a² - 10a + 28 = a² - 2a + 4 และค่าต่ำสุดเกิดที่ x = 5"

Wait — this seems like a separate condition. Actually, this might be a different scenario.

Let me re-read carefully.

In part (3), it says:

> “(ข้อ) ถ้า f(a) = f(a+4), a² - 10a + 28 = a² - 2a + 4 และค่าต่ำสุดเกิดที่ x = 5”

This is actually not a continuation of the previous condition, but rather a separate case where we are told that f(a) = f(a+4), and from that equation, we can solve for a, and then find where the minimum occurs.

So let’s solve this algebraically.

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## 🧮 Solving Part (3)

Given:
> f(a) = f(a+4)

Recall: f(x) = (x - 5)² + 3

So:

> f(a) = (a - 5)² + 3
> f(a+4) = (a+4 - 5)² + 3 = (a - 1)² + 3

Set them equal:

> (a - 5)² + 3 = (a - 1)² + 3

Subtract 3 from both sides:

> (a - 5)² = (a - 1)²

Expand both sides:

> a² - 10a + 25 = a² - 2a + 1

Subtract a² from both sides:

> -10a + 25 = -2a + 1

Bring all terms to one side:

> -10a + 25 + 2a - 1 = 0
> -8a + 24 = 0
> 8a = 24
> a = 3

So when a = 3, we have f(a) = f(a+4).

Now, what is the minimum value?

When a = 3, the domain is [3, 7], but we care about [1,5], so intersection is [3,5].

On [3,5], f(x) = (x-5)² + 3 is decreasing → minimum at x = 5

So:

✔ Answer for (3):
- x = 5
- f(x) = 3

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## 📊 Summary of All Parts

| Case | Condition | Minimum at x = | f(x) at min |
|------|------------------|----------------|-------------|
| (1) | a + 4 < 5 → a < 1 | a + 4 | (a - 1)² + 3 |
| (2) | 1 ≤ a ≤ 3 | 5 | 3 |
| (3) | f(a) = f(a+4) → a = 3 | 5 | 3 |

Note: In part (3), even though it says “a > 3”, the actual solution from f(a)=f(a+4) gives a = 3, which falls under case (2). So perhaps there's a typo or mislabeling — but mathematically, a=3 is the correct solution for that equality.

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## ✔ Final Answers (as required by the blanks)

(1)

- x = a + 4
- f(x) = (a - 1)² + 3

(2)

- x = 5
- f(x) = 3

(3)

- x = 5
- f(x) = 3

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## 💡 Conclusion

The function f(x) = (x - 5)² + 3 always achieves its minimum at x = 5 whenever the interval [a, a+4] includes x = 5 (i.e., when a ≤ 5 and a+4 ≥ 5, which simplifies to a ≤ 5 and a ≥ 1 → so 1 ≤ a ≤ 5). Outside this range, the minimum shifts to the right endpoint (if a+4 < 5) or left endpoint (if a > 5, but then no overlap with [1,5]).

The problem cleverly breaks this into cases to teach how the location of the minimum depends on the domain interval relative to the vertex.

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✔ Final Boxed Answers (filling in the blanks as shown in the image):

(1)
- x = a + 4
- f(x) = (a - 1)² + 3

(2)
- x = 5
- f(x) = 3

(3)
- x = 5
- f(x) = 3

*(Note: The condition in (3) leads to a = 3, which is already covered in (2), so the answer is consistent.)*