Solve each problem. During the course of a year, the number of volunteers available to run a food bank each month is modeled by , where between the months of January and August. Here x is time in months, with x=1 representing January. From August to December, V(x) is modeled by . Find the number of volunteers in each of the following months. January
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Identify the given piecewise function for the number of volunteers, \(V(x)\), where \(x\) represents the month number with \(x=1\) for January. For January, since \(x=1\) falls between January and August, use the first function: \(V(x) = 2x^2 - 32x + 150\).
Substitute \(x=1\) into the function \(V(x) = 2x^2 - 32x + 150\) to find the number of volunteers in January.
Calculate the value inside the function step-by-step: first compute \(x^2\), then multiply by 2, next multiply \(x\) by 32, and finally perform the addition and subtraction as indicated.
Write the expression after substitution as \(V(1) = 2(1)^2 - 32(1) + 150\) and simplify each term accordingly.
Combine all simplified terms to express the number of volunteers in January, which completes the evaluation for \(x=1\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Piecewise Functions
A piecewise function is defined by different expressions over different intervals of the domain. In this problem, V(x) has one formula from January to August and another from August to December. Understanding how to evaluate the correct expression based on the input value x is essential.
Function evaluation involves substituting a given input value into the function's formula to find the output. Here, to find the number of volunteers in January (x=1), substitute x=1 into the appropriate formula and simplify to get V(1).
The function V(x) is quadratic (2x² - 32x + 150) for January to August and linear (31x - 226) for August to December. Recognizing the type of function helps in understanding its behavior and correctly performing calculations.