Problem 97 - Algebra 2

Question

PREREQUISITE SKILL. Evaluate each function for the given value. $$ f(x)=\frac{2}{3} x^{2}+2 x-1, x=-3 $$

Step-by-Step Solution

Answer

The value of the function at \( x = -3 \) is \( -1 \).

Step 1: Substitute the Value into the Function

The given function is \( f(x)=\frac{2}{3}x^{2}+2x-1 \), and we need to evaluate this function at \( x = -3 \). Begin by substituting \( x = -3 \) into the function: \[ f(-3) = \frac{2}{3}(-3)^{2} + 2(-3) - 1 \]

Step 2: Evaluate the Exponentiation

Calculate \((-3)^{2}\). Since squared means multiplying the number by itself: \[ (-3)^{2} = 9 \]

Step 3: Multiply the Square by the Coefficient

Take the squared value and multiply by the coefficient \( \frac{2}{3} \): \[ \frac{2}{3} \times 9 = 6 \]

Step 4: Evaluate the Linear Term

Calculate the expression \(2(-3)\), which simplifies to: \[ 2 \times -3 = -6 \]

Step 5: Combine the Results

Substitute the results from the previous steps into the simplified expression: \[ f(-3) = 6 - 6 - 1 \]

Step 6: Simplify the Expression

Combine the terms to find the final result: \[ 6 - 6 - 1 = -1 \]

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial function where the highest power of the variable is 2. Quadratic functions are commonly represented in the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. The notable characteristic of quadratic functions is the presence of the \( x^2 \) term. This term creates a parabolic graph, which can either be

• concave up (like a U shape) if \( a > 0 \), or
• concave down (like an upside-down U) if \( a < 0 \).

In our exercise, the quadratic function given is \( f(x)=\frac{2}{3}x^{2}+2x-1 \), where:

• \( \frac{2}{3} \) is the coefficient of the quadratic term \( x^2 \)
• \( 2 \) is the coefficient of the linear term \( x \)
• \( -1 \) is the constant term.

This specific function is known to represent a parabola that opens upward because the coefficient of \( x^2 \) is positive.

Substitution

Substitution is a fundamental process in algebra used to evaluate functions at specific values. It involves replacing the variable in a function with a given number. For instance, when the exercise asks us to evaluate \( f(x)=\frac{2}{3}x^{2}+2x-1 \) for \( x = -3 \), we replace every occurrence of \( x \) with \( -3 \).
This is the first step in solving the problem,

• resulting in \( f(-3) = \frac{2}{3}(-3)^{2} + 2(-3) - 1 \).

This substitution allows us to transform an abstract function into a numeric expression, which can then be simplified. Correct substitution is crucial, as any mistake in this initial step can lead to incorrect results.
The key here is to ensure that each substitution is performed accurately and consistently throughout the entire function.

Exponentiation

Exponentiation is the mathematical operation of raising a number to a power, which means multiplying the number by itself a certain number of times. In our case, we raise \( -3 \) to the power of 2, represented by \((-3)^2\).
This operation involves two key steps:

• First, we multiply \(-3\) by itself, resulting in \(9\).
• This is because \((-3) \times (-3) = 9\), considering that two negative numbers multiply to give a positive number.

Understanding the exponentiation step is vital as it lays the foundation for further calculations in the function evaluation. It is crucial to pay careful attention to the signs when dealing with negative numbers, as errors here directly impact the final result. This step is a building block for simplifying expressions involving quadratic terms.

Linear Term Simplification

Linear term simplification involves reducing the linear part of a function, which typically involves the first-degree variable term. In our exercise, the linear term is represented by \( 2x \), and when substituted, becomes \( 2(-3) \).
To simplify this term, we perform basic multiplication:

• Multiply \( 2 \) by \( -3 \) to yield \(-6\).

This process is straightforward since linear terms are not exponentiated. Instead, it's primarily about straightforward multiplication and carrying over the correct sign. Simplifying the linear term correctly contributes to accurately arriving at the solution. The simplicity of this step highlights its importance, ensuring no overlooked opportunities for simplification, which might otherwise complicate the resolution of the function.