Problem 30 - Algebra 2

Question

For each function, find \(f(-5), f(-3), f\left(\frac{1}{2}\right),\) and \(f(4)\) $$ f(t)=\frac{1}{2} t-2 $$

Step-by-Step Solution

Answer

f(-5) = -4.5, f(-3) = -3.5, f(1/2) = -1.75, f(4) = 0

Step 1: Calculate f(-5)

To find the value of the function when t is -5, substitute t with -5 in the function f(t): \( f(-5) = \frac{1}{2}(-5) - 2 \). Multiply \frac{1}{2}(-5) to get -\frac{5}{2} or -2.5, then subtract 2: \( f(-5) = -2.5 - 2 = -4.5 \).

Step 2: Calculate f(-3)

To calculate the value of the function when t is -3, substitute t with -3: \( f(-3) = \frac{1}{2}(-3) - 2 \). Multiply \frac{1}{2}(-3) to get -\frac{3}{2} or -1.5, then subtract 2: \( f(-3) = -1.5 - 2 = -3.5 \).

Step 3: Calculate f(1/2)

To find the value of the function when t is \frac{1}{2}, substitute t with \frac{1}{2}: \( f\left(\frac{1}{2}\right) = \frac{1}{2}\left(\frac{1}{2}\right) - 2 \). Multiply \frac{1}{2}\left(\frac{1}{2}\right) to get \frac{1}{4}, then subtract 2: \( f\left(\frac{1}{2}\right) = \frac{1}{4} - 2 = -\frac{7}{4} \) or -1.75.

Step 4: Calculate f(4)

To determine the value of the function when t is 4, substitute t with 4: \( f(4) = \frac{1}{2}(4) - 2 \). Multiply \frac{1}{2}(4) to get 2, then subtract 2: \( f(4) = 2 - 2 = 0 \).

Key Concepts

Function Evaluation

Function evaluation is the process of determining the output of a function given a specific input. In algebra, it involves substituting the input value for the variable in the function’s equation and performing the necessary calculations to find the result. For example, if we have a function defined as \( f(t)=\frac{1}{2}t-2 \), evaluating this function for a specific value involves finding the result when we replace \( t \) with the value of interest.

When asked to evaluate \( f(-5) \), we substitute -5 into the equation, resulting in \( f(-5)=\frac{1}{2}(-5)-2 \), which simplifies to -4.5. The process is repeated for other values, applying arithmetic operations to reach the final results. Students often visualize function evaluation as plugging in numbers into a machine where the function itself is the machinery, and the output is the result you get after the machine processes the input number.

Linear Functions

Linear functions are among the simplest and most commonly studied types of functions in algebra. They can be recognized by their straight-line graphs and are typically written in the form \( f(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.

The function \( f(t)=\frac{1}{2}t-2 \) is an example of a linear function. Its graph would be a straight line with a slope of \( \frac{1}{2} \), representing a moderate ascending slope, and a y-intercept at -2, meaning the line crosses the y-axis at -2. Linear functions are predictable and have a constant rate of change, which makes evaluating them at different points straightforward.

Substitution Method

The substitution method is a fundamental technique used in algebra to evaluate functions, solve systems of equations, and simplify expressions. The key idea is replacing a variable with a given number or another expression.

To apply this method to evaluate our function \( f(t) \) for different values of \( t \), we 'substitute' the values of \( t \) directly into the equation. For instance, to find \( f(-3) \), we substitute -3 for \( t \) to get \( f(-3)=\frac{1}{2}(-3)-2 \), then calculate the arithmetic to obtain -3.5 as the output. This process makes evaluating functions at specific points efficient and precise.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operation symbols but do not contain an equals sign (as opposed to equations). They are used to represent relationships and can be as simple as a single term or as complex as multiple terms connected by addition, subtraction, multiplication, and division.

The function \( f(t)=\frac{1}{2}t-2 \) itself is an algebraic expression, specifically a binomial with two terms: \( \frac{1}{2}t \) and \( -2 \). Evaluating algebraic expressions requires understanding the order of operations and combining like terms where applicable. Mastery of manipulating these expressions is crucial for success in algebra and higher-level mathematics.