Problem 119 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the difference quotient and simplify your answer. $$f(x)=x^{2}-2 x+9, \frac{f(3+h)-f(3)}{h}, h \neq 0$$

Step-by-Step Solution

Answer

The simplified difference quotient for the given function is \(h + 4\).

Step 1: Substitute the variable with the provided value

Substitute \(x\) with \(3 + h\) in the equation \(f(x)\), you will get \(f(3+h) = (3+h)^{2}-2(3+h)+9\).

Step 2: Simplify f(3+h)

Simplify the result from step 1: \(f(3+h) = 9 + 6h + h^{2} - 6 - 2h + 9 = h^2 + 4h + 12\)

Step 3: Calculate f(3)

Substitute \(x\) with \(3\) in the equation \(f(x)\), it gives us: \(f(3) = (3)^{2}-2(3)+9 = 9 - 6 + 9 = 12\)

Step 4: Substitute into Difference Quotient

The expression for the difference quotient is \(\frac{f(3+h)-f(3)}{h}\). Substitute \(f(3+h) = h^2 + 4h + 12\) and \(f(3) = 12\) into the difference quotient expression: \(\frac{h^2 + 4h + 12 - 12}{h} = \frac{h^2 + 4h}{h}\)

Step 5: Simplify the Difference Quotient

Simplify the difference quotient to get the final answer: \(\frac{h^2 + 4h}{h} = h + 4\), because \(h ≠ 0\).

Key Concepts

Quadratic Function

A **quadratic function** is a type of polynomial function where the highest degree of the variable is 2. The general form of a quadratic function is written as:

• \( f(x) = ax^2 + bx + c \)

Here, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). The quadratic function forms a parabolic graph that opens upward if \( a > 0 \) or downward if \( a < 0 \).

In the exercise, we are given the quadratic function \( f(x) = x^2 - 2x + 9 \). This indicates:

• \( a = 1 \)
• \( b = -2 \)
• \( c = 9 \)

Understanding the characteristics of a quadratic function is important for accurately solving and simplifying expressions involving this type of function. Quadratic functions are fundamental in algebra and appear in various real-world applications, such as physics and economics.

Function Evaluation

**Function evaluation** is the process of finding the output of a function for a given input value. To evaluate a function, simply substitute the input value into the function's formula, replacing any variables.

In the context of this exercise, we need to evaluate the quadratic function \( f(x) = x^2 - 2x + 9 \) at two specific points: \( x = 3 + h \) and \( x = 3 \).

• For \( f(3+h) \): substitute \( x \) with \( 3+h \) leading to \( (3+h)^2 - 2(3+h) + 9 \).
• For \( f(3) \): substitute \( x \) with \( 3 \) leading to \( 3^2 - 2 \cdot 3 + 9 \).

These evaluations are essential for calculating the difference quotient, especially when we need to find the rate at which the function changes near a specific point. Function evaluation is a key skill that allows you to determine how a function behaves and reacts based on different input values.

Simplification

**Simplification** involves rewriting expressions in a simpler or more concise form while maintaining their equivalence. This step is vital for making mathematical expressions easier to work with, especially when further calculations are necessary.

In this exercise, after finding \( f(3+h) \), the result:

• \( (3+h)^2 - 2(3+h) + 9 \) simplifies to \( h^2 + 4h + 12 \).

Similarly, if we calculate \( f(3) \), it simplifies directly to \( 12 \).

Simplification is crucial for calculating the difference quotient, as it allows us to clearly see and cancel terms, making it possible to derive a straightforward result of \( h + 4 \). Reducing terms and combined like elements, simplify expressions, and eliminate unnecessary complexity in your mathematical calculations.

Algebraic Expression

An **algebraic expression** is a combination of numbers, variables, and arithmetic operations (addition, subtraction, multiplication, and division) representing a particular quantity. Learning to read and manipulate algebraic expressions is a foundation of algebra.

In this exercise, we encounter multiple algebraic expressions as part of evaluating the given quadratic function.

One key expression is the difference quotient itself \( \frac{f(3+h)-f(3)}{h} \), which looks complex initially. However, after substituting the respective function evaluations, we have:

• \( f(3+h) = h^2 + 4h + 12 \)
• \( f(3) = 12 \)

Substituting these, the difference quotient simplifies to \( \frac{h^2 + 4h}{h} \), which after cancelling the common \( h \) factor results in \( h + 4 \). Understanding how to handle these pieces is essential for problem-solving in algebra, as it empowers you to transform and simplify expressions for various applications.