Problem 22 - Algebra 2

Question

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

Step-by-Step Solution

Answer

\(f(-5) = -7, f(-3) = -3, f\left(\frac{1}{2}\right) = 4, f(4) = 11\)

Step 1: Calculate \(f(-5)\)

To find \(f(-5)\), substitute \(a\) with \(-5\) in the function \(f(a) = 2a + 3\). Calculate the result of the substitution: \(f(-5) = 2(-5) + 3 = -10 + 3 = -7\).

Step 2: Calculate \(f(-3)\)

To find \(f(-3)\), substitute \(a\) with \(-3\) in the function \(f(a) = 2a + 3\). Calculate the result of the substitution: \(f(-3) = 2(-3) + 3 = -6 + 3 = -3\).

Step 3: Calculate \(f\left(\frac{1}{2}\right)\)

To find \(f\left(\frac{1}{2}\right)\), substitute \(a\) with \(\frac{1}{2}\) in the function \(f(a) = 2a + 3\). Calculate the result of the substitution: \(f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right) + 3 = 1 + 3 = 4\).

Step 4: Calculate \(f(4)\)

To find \(f(4)\), substitute \(a\) with \(4\) in the function \(f(a) = 2a + 3\). Calculate the result of the substitution: \(f(4) = 2(4) + 3 = 8 + 3 = 11\).

Key Concepts

Linear Functions

Linear functions are foundational in algebra and form the basis of equations that describe a straight line on a graph. They are typically written in the form:

\( y = mx + b \), where \( m \) is the slope of the line and \( b \) is the y-intercept, where the line crosses the y-axis. In our exercise, the linear function given is \( f(a) = 2a + 3 \), where \( 2 \) is the coefficient of \( a \) (our slope) and \( 3 \) is the constant term (our y-intercept). Understanding linear functions is essential as they are used to model real-world situations with a constant rate of change.

Substitution Method

The substitution method is an invaluable technique used in algebra to solve systems of equations, evaluate functions, and simplify expressions. It involves replacing a variable with a specific value or another expression. When we evaluate the function \( f(a) = 2a + 3 \) for different values of \( a \), we are using the substitution method. Through this method, complex problems are often broken down into simpler steps, enabling a step-by-step approach to finding solutions, which is critical to learning and understanding algebra.

Evaluating Functions

Evaluating functions involves finding the output of a function for a particular input. To do this, one substitutes the input value into the function in place of the variable and computes the result. This is precisely what we did in our exercise when we calculated \( f(-5) \), \( f(-3) \), \( f\left(\frac{1}{2}\right) \), and \( f(4) \). Understanding how to evaluate functions is crucial for students as it is a recurring task in mathematics, which reinforces the relationship between variables and helps in grasping the concept of functions as mathematical models.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operation symbols that represent a certain quantity but do not have an equality sign (unlike equations). In our problem, \( 2a + 3 \) is an algebraic expression that represents the function \( f(a) \). Manipulating algebraic expressions through addition, subtraction, multiplication, and substitution is a key skill in algebra, as it allows students to understand and transform mathematical statements to solve a variety of problems.