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Let's break down the question step by step:

### Given:
- The concentration of the drug in the patient's body is modeled by the quadratic function:
  \[
  C(t) = -5t^2 + 10t
  \]
  where \( C(t) \) is the concentration in mg/L, and \( t \) is the time in hours after injection.

### Part A: Domain and Range
We need to find the domain and range of the function based on the context of the problem.

1. **Domain:**
   - The domain represents the possible values of \( t \), which is the time after the medication is injected.
   - Time cannot be negative, so \( t \geq 0 \).
   - The function is a quadratic opening downwards (since the coefficient of \( t^2 \) is negative), meaning there is a maximum point. After a certain point, the concentration decreases back to 0.
   - We need to find when \( C(t) = 0 \) to determine when the concentration reaches 0 again.
   - Solve \( -5t^2 + 10t = 0 \):
     \[
     t(-5t + 10) = 0
     \]
     Solutions: \( t = 0 \) or \( t = 2 \).

   Therefore, the domain is \( t \in [0, 2] \).

2. **Range:**
   - The range depends on the maximum value of \( C(t) \). Since it's a downward-opening parabola, we need to find the vertex to determine the maximum value.
   - The vertex occurs at \( t = \frac{-b}{2a} \), where \( a = -5 \) and \( b = 10 \):
     \[
     t = \frac{-10}{2(-5)} = 1.
     \]
   - Now, find \( C(1) \):
     \[
     C(1) = -5(1)^2 + 10(1) = -5 + 10 = 5 \, \text{mg/L}.
     \]
   - Therefore, the maximum concentration is 5 mg/L, and since \( C(t) \) starts at 0 and reaches 0 again at \( t = 2 \), the range is \( [0, 5] \).

   **Summary for Part A:**
   - **Domain:** \( [0, 2] \) hours
   - **Range:** \( [0, 5] \) mg/L

### Part B: Graphing and Maximum Concentration
The graph of \( C(t) = -5t^2 + 10t \) is a parabola that opens downwards. Based on Part A, the vertex is at \( t = 1 \), where the maximum concentration is 5 mg/L. The graph starts at \( t = 0 \) with \( C(0) = 0 \), peaks at \( t = 1 \) with \( C(1) = 5 \), and returns to 0 at \( t = 2 \).

**Maximum concentration:** 5 mg/L at \( t = 1 \) hour.

Would you like a more detailed explanation of any step or the graph plotted for better understanding?

Here are 5 related questions to deepen your understanding:
1. How do you calculate the vertex of a parabola algebraically?
2. Why is the domain limited to \( [0, 2] \) hours?
3. How can you determine the maximum point of any quadratic function?
4. What real-world scenarios might exhibit a quadratic concentration pattern like this?
5. How does the sign of the quadratic term affect the shape of the parabola?

**Tip:** For any quadratic function \( ax^2 + bx + c \), the maximum or minimum occurs at the vertex, and you can find it using \( x = -\frac{b}{2a} \).

Many doctors rely on the use of intravenous medication administration in order to achieve an immediate response of a particular drug's effects. The concentration, C, in mg/L, of a particular medication after being injected into a patient can be given by the function C(t) = -5t^2 + 10t, where the time, t, is hours after injection.
Part A: What are the domain and range of the function C(t) based on the context of the problem? Show all necessary calculations.
Part B: Graph the function to determine the greatest concentration of the medication that a patient will have in their body, and the time when that occurs.

Analyze a quadratic function C(t) = -5t^2 + 10t modeling the concentration of medication in a patient's body. Learn how to find the domain, range, and graph to determine the maximum concentration and the time it occurs. Suitable for high school students studying quadratic functions.

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