Problem 51 - Intermediate Algebra

Question

If \(f(x)=2 x^{2}-x-1\), find \(f(-a), f(a+1)\), and \(f(a+h)\).

Step-by-Step Solution

Answer

\(f(-a) = 2a^2 + a - 1\), \(f(a+1) = 2a^2 + 3a\), \(f(a+h) = 2a^2 + 4ah + 2h^2 - a - h - 1\).

Step 1: Understand the function

The function given is \( f(x) = 2x^2 - x - 1 \). This is a quadratic function in the standard form, where the coefficients are specifically 2 for \(x^2\), -1 for \(x\), and the constant is -1.

Step 2: Substitute \\(-a\\) into the Function

Find \( f(-a) \) by substituting \(-a\) into the function:\[ f(-a) = 2(-a)^2 - (-a) - 1 \]\[ = 2a^2 + a - 1 \].

Step 3: Substitute \\(a+1\\) into the Function

Find \( f(a+1) \) by substituting \(a+1\) into the function:\[ f(a+1) = 2(a+1)^2 - (a+1) - 1 \]Expand to solve:\[ = 2(a^2 + 2a + 1) - a - 1 - 1 \]\[ = 2a^2 + 4a + 2 - a - 2 \]Combine like terms:\[ = 2a^2 + 3a \].

Step 4: Substitute \\(a+h\\) into the Function

Find \( f(a+h) \) by substituting \(a+h\) into the function:\[ f(a+h) = 2(a+h)^2 - (a+h) - 1 \]Expand to solve:\[ = 2(a^2 + 2ah + h^2) - a - h - 1 \]\[ = 2a^2 + 4ah + 2h^2 - a - h - 1 \].

Key Concepts

Function Evaluation

Evaluating a function means finding the result of a function when specific input values are provided. Consider a quadratic function, like the one given in the exercise. The formula is typically expressed as \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. To evaluate the function, we need to substitute the variable \( x \) with a specific value, such as \(-a\), \(a+1\), or \(a+h\), as shown in the exercise.
This process allows us to find the output of the function at different points. Remember, different values of \( x \) will yield different unique outputs from the function, helping us understand the behavior of quadratic functions at various points.

Substitution in Functions

Substitution is a key process in evaluating functions. By replacing the variable in a function with a specific value or expression, you can calculate the output of a function for any given input. In the exercise, the focus was on substituting sets like \(-a\), \(a+1\), and \(a+h\) into the function.

• **Substitute \(-a\):** Replace \( x \) with \(-a\) and follow the algebraic rules, paying careful attention to signs (negative and positive values).


• **Substitute \(a+1\):** Insert \( x = a + 1 \) into the function, expanding and simplifying as needed to find the result.


• **Substitute \(a+h\):** Similar to the others, replacing \( x \) with \( a + h \) involves expansion and combination of like terms.

Through substitution, students learn how each part of the function interacts with different inputs, contributing to a deeper understanding of quadratic functions.

Algebraic Expansion

Algebraic expansion is a technique used to simplify expressions, particularly when dealing with powers and products. When you're substituting values into a quadratic function, expansion becomes crucial to solve for the function's value.
Take, for example, substituting \((a+1)\) or \((a+h)\) into the quadratic function. The binomial expression \((a+1)^2\) expands to \(a^2 + 2a + 1\), while \((a+h)^2\) expands to \(a^2 + 2ah + h^2\).
After substitution, each term gets multiplied by the corresponding coefficient from the function, and like terms are combined:

• **For \(f(a+1)\):** The steps result in \(2a^2 + 3a\) after expansion and simplification.


• **For \(f(a+h)\):** More terms appear, combining into \(2a^2 + 4ah + 2h^2 - a - h - 1\).
  

Algebraic expansion helps clarify the structure of quadratic expressions, turning them into more manageable, simplified components. It allows students to derive results with clear, logical steps.