Math Problem Statement

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.

Solution

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)=5t2+10tC(t) = -5t^2 + 10t where C(t)C(t) is the concentration in mg/L, and tt 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 tt, which is the time after the medication is injected.
    • Time cannot be negative, so t0t \geq 0.
    • The function is a quadratic opening downwards (since the coefficient of t2t^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)=0C(t) = 0 to determine when the concentration reaches 0 again.
    • Solve 5t2+10t=0-5t^2 + 10t = 0: t(5t+10)=0t(-5t + 10) = 0 Solutions: t=0t = 0 or t=2t = 2.

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

  2. Range:

    • The range depends on the maximum value of C(t)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=b2at = \frac{-b}{2a}, where a=5a = -5 and b=10b = 10: t=102(5)=1.t = \frac{-10}{2(-5)} = 1.
    • Now, find C(1)C(1): C(1)=5(1)2+10(1)=5+10=5mg/L.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)C(t) starts at 0 and reaches 0 again at t=2t = 2, the range is [0,5][0, 5].

    Summary for Part A:

    • Domain: [0,2][0, 2] hours
    • Range: [0,5][0, 5] mg/L

Part B: Graphing and Maximum Concentration

The graph of C(t)=5t2+10tC(t) = -5t^2 + 10t is a parabola that opens downwards. Based on Part A, the vertex is at t=1t = 1, where the maximum concentration is 5 mg/L. The graph starts at t=0t = 0 with C(0)=0C(0) = 0, peaks at t=1t = 1 with C(1)=5C(1) = 5, and returns to 0 at t=2t = 2.

Maximum concentration: 5 mg/L at t=1t = 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][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 ax2+bx+cax^2 + bx + c, the maximum or minimum occurs at the vertex, and you can find it using x=b2ax = -\frac{b}{2a}.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Graphing Parabolas

Formulas

C(t) = -5t^2 + 10t
Vertex formula: t = -b / (2a)
Quadratic roots
Solving C(t) = 0

Theorems

Quadratic function and its maximum point theorem
Domain and range for real-world contexts

Suitable Grade Level

Grades 9-12