Problem 7 - Precalculus

Question

Evaluate the algebraic expressions. If \(f(x)=x^{2}+3 x+5,\) evaluate \(f(2+i)\)

Step-by-Step Solution

Answer

\(f(2+i) = 14 + 7i\)

Step 1: Substitute the Input Value

We start by substituting the expression \(2+i\) into the function \(f(x) = x^2 + 3x + 5\). This gives us \(f(2+i) = (2+i)^2 + 3(2+i) + 5\).

Step 2: Expand the Squared Term

We need to expand \((2+i)^2\). Recall that \((a+b)^2 = a^2 + 2ab + b^2\). Using this formula, we get \((2+i)^2 = 2^2 + 2 \cdot 2 \cdot i + i^2 = 4 + 4i + i^2\). Here, \(i^2 = -1\), so \(4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i\).

Step 3: Simplify the Linear Term

Calculate \(3(2+i)\). Distribute the 3 to each term: \(3 \times 2 + 3 \times i = 6 + 3i\).

Step 4: Combine All Parts

Add all the components from the previous steps: \(f(2+i) = (3 + 4i) + (6 + 3i) + 5\).

Step 5: Simplify the Expression

Combine like terms: the real parts \(3 + 6 + 5 = 14\), and the imaginary parts \(4i + 3i = 7i\). Thus, \(f(2+i) = 14 + 7i\).

Key Concepts

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and mathematical operations. In our problem, we have the function \(f(x) = x^2 + 3x + 5\). This expression includes:

• \(x^2\) (a quadratic term)
• \(3x\) (a linear term)
• \(5\) (a constant term)

To evaluate this function at a specific point, such as \(2+i\), we substitute \(2+i\) in place of \(x\). Remember to follow order of operations (parentheses, exponents, multiplication and division, addition and subtraction) to correctly simplify the expression after substitution.
By substituting, we transform the task into handling complex numbers within our algebraic expression.

Function Evaluation

Function evaluation involves determining the output of a function for a specific input. It's like asking, "What result do we get if we plug \(2+i\) into \(f(x)\)?"
For the function \(f(x) = x^2 + 3x + 5\), substituting \(2+i\) gives:

• Substitute \((2+i)\) into the function: \(f(2+i) = (2+i)^2 + 3(2+i) + 5\).
• This step requires carefully expanding and simplifying each part of the expression, taking care of complex number interaction.

Function evaluation often involves plugging in values, performing polynomial expansions, and simplifying results while following algebraic rules.

Imaginary Unit i

The imaginary unit \(i\) is a fundamental part of complex numbers, where \(i^2 = -1\). It is used to extend the real number system to the complex number system when solving algebraic expressions involving roots of negative numbers.
In our problem:

• When expanding \((2+i)^2\), we eventually use \(i^2\) as \(i^2 = -1\). This transforms the term \(4 + 4i + i^2\) into \(4 + 4i - 1\), resulting in \(3 + 4i\).

The imaginary unit allows for all real polynomials to have roots, providing a deeper understanding of numbers and operations.

Polynomial Functions

Polynomial functions, like \(f(x) = x^2 + 3x + 5\), are essential tools in algebra. They are sums of terms consisting of variables, exponents, and coefficients.

• Polynomials can be evaluated by plug-in points, leveraging distribution and power expansion. This helps map mathematical problems involving real and complex numbers.
• When evaluating polynomials, we manage both the real and complex parts. By breaking expressions into manageable parts, such as separately dealing with real and imaginary parts after substitution, polynomial functions retain their importance in handling complex operations.

Understanding polynomial functions equips students to solve complex problems with components like quadratic terms and linear additions, leading to comprehensive solutions when dealing with any form of numerical inputs.