Problem 69 - College Algebra Essentials

Question

find and simplify the difference quotient $$ \frac{f(x+h)-f(x)}{h}, h \neq 0 $$ for the given function. $$ f(x)=-2 x^{2}-x+3 $$

Step-by-Step Solution

Answer

The simplified difference quotient of the function \(f(x)=-2x^{2}-x+3\) is \(-4x-2h-1\).

Step 1: Calculating f(x + h)

First, replace each x in f(x) with (x + h). So, the \(f(x)=-2x^{2}-x+3\) becomes \(f(x+h)= -2(x+h)^{2}-(x+h)+3 = -2(x^{2}+2xh+h^{2})-(x+h)+3 = -2x^{2}-4xh-2h^{2}-x-h+3\).

Step 2: Substituting into the difference quotient formula

Now substitute the expressions for f(x + h) and f(x) into the difference quotient formula. This gives us: \[\frac{(-2x^{2}-4xh-2h^{2}-x-h+3)-(-2x^{2}-x+3)}{h}= \frac{-2x^{2}-4xh-2h^{2}-x-h+3+2x^{2}+x-3}{h} = \frac{-4xh-2h^{2}-h}{h}\]

Step 3: Simplifying the difference quotient

The next step is to simplify the numerator by factorizing. The numerator -4xh-2h^{2}-h can be factorized to h(-4x-2h-1). This gives us: \[\frac{h(-4x-2h-1)}{h}\]. We then cancel out h from the numerator and the denominator, leaving us with: \[-4x-2h-1\]

Key Concepts

Simplifying Expressions

To simplify expressions, you want to make them as simple as possible while keeping the same value. Simplification involves a series of steps that help transform complex mathematical expressions into more straightforward forms. In the context of the difference quotient from our exercise, the key aim is to make the numerator simpler so that the quotient itself becomes easier to evaluate.
In this particular case, the expression \(-4xh - 2h^2 - h\) was simplified using factorization. By factoring out the common term \(h\), we reduced the expression to \(h(-4x - 2h - 1)\). This simplification step is crucial in algebra because it often leads to expressions that can be easily manipulated or solved.
Remember that not all terms are always factorable, but identifying common factors or using distributive laws can greatly simplify the process.

• Combine like terms whenever possible.
• Look for common factors in each term.
• Apply arithmetic operations like addition or subtraction carefully.

With practice, simplifying expressions becomes a skill that greatly facilitates solving more complex mathematical problems.

Polynomial Functions

Polynomial functions are expressions that consist of variables raised to whole number powers and have coefficients. The polynomial in our exercise is \(-2x^2 - x + 3\), with degrees of these terms specified by the power of \(x\). Each term in this polynomial function is either a constant, a variable, or a product of a constant and a variable raised to a power.
There are several key characteristics of polynomial functions:

• **Degree:** The highest power of the variable in the polynomial. In our polynomial, the degree is 2 because of the term \(-2x^2\).
• **Coefficients:** The numbers in front of the variables. For the polynomial \(-2x^2 - x + 3\), the coefficients are \(-2, -1,\) and \(3\).
• **Constant Term:** A term without a variable, which is \(3\) in the given polynomial function.

Understanding these parts helps in performing operations such as addition, subtraction, and especially division like in the difference quotient, where recognizing how terms interact is crucial.
Polynomials are foundational in algebra and calculus, serving as a stepping stone in understanding more complex functions and mathematical concepts.

Factorization

Factorization involves breaking down a complex expression or number into simpler parts, or 'factors', that when multiplied together give the original expression or number. It's a critical skill in algebra that deals with simplifying equations and understanding relationships between terms.
In our exercise, factorization was used to simplify the numerator \(-4xh - 2h^2 - h\). By identifying the common factor \(h\), the expression was rewritten as \(h(-4x - 2h - 1)\). This step is essential for simplifying the difference quotient and allows for terms to be canceled out, ultimately simplifying the expression to \(-4x - 2h - 1\).
Key factorization techniques include:

• **Common Factoring:** Finding common factors across terms, as we did by extracting \(h\) from the expression.
• **Grouping:** When expressions have more than two terms, sometimes grouping terms can reveal a common factor.
• **Differences of Squares:** Recognizing and factoring expressions like \(a^2 - b^2 = (a + b)(a - b)\).

Having a strong grasp of these techniques helps in tackling polynomials and complex algebraic expressions, making them manageable and easier to work with.