Problem 77 - A Graphical Approach to Precalculus with Limits

Question

Determine the difference quotient \(\frac{f(x+h)-f(x)}{h}\) (where \(h \neq 0\) j for each function \(f\). Simplify completely. $$f(x)=-6 x^{2}-x+4$$

Step-by-Step Solution

Answer

The simplified difference quotient is \(-12x - 6h - 1\).

Step 1: Identify the function

Given the function is \(f(x) = -6x^2 - x + 4\). This is a quadratic function with coefficients \(-6\), \(-1\), and constant term \(4\).

Step 2: Calculate \(f(x+h)\)

Substitute \(x+h\) into the function: \[ f(x+h) = -6(x+h)^2 - (x+h) + 4 \]Expand \((x+h)^2\) to \(x^2 + 2xh + h^2\):\[ f(x+h) = -6(x^2 + 2xh + h^2) - x - h + 4 \] Simplify to obtain:\[ f(x+h) = -6x^2 - 12xh - 6h^2 - x - h + 4 \].

Step 3: Calculate \(f(x+h) - f(x)\)

Subtract \(f(x)\) from \(f(x+h)\): \[ f(x+h) - f(x) = (-6x^2 - 12xh - 6h^2 - x - h + 4) - (-6x^2 - x + 4) \]Distribute the negative sign:\[ = -6x^2 - 12xh - 6h^2 - x - h + 4 + 6x^2 + x - 4 \]Combine like terms to get:\[ = -12xh - 6h^2 - h \].

Step 4: Form the difference quotient

The difference quotient is:\[ \frac{f(x+h) - f(x)}{h} = \frac{-12xh - 6h^2 - h}{h} \].

Step 5: Simplify the difference quotient

Factor \(h\) out of the numerator:\[ \frac{h(-12x - 6h - 1)}{h} \]Since \(h eq 0\), cancel \(h\) to get:\[ -12x - 6h - 1 \].

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial that is characterized by an expression of the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are coefficients and \( a eq 0 \). These functions create a parabolic graph, which can open upwards or downwards depending on the sign of the coefficient \( a \).
In our exercise, the quadratic function is \( f(x) = -6x^2 - x + 4 \), with coefficients \( -6 \), \( -1 \), and a constant \( 4 \).
Given its negative leading coefficient, this particular parabola opens downwards. Understanding how to manipulate and extract valuable information from such functions is crucial for calculus and algebra.

Function Simplification

Function simplification is a process used to make algebraic expressions easier to work with. In this exercise, we simplified the expression \( f(x+h) \).

• First, substituted \( x+h \) for \( x \) in the original function.
• Expanded the squared term \((x+h)^2 = x^2 + 2xh + h^2\).
• Then, we distributed and combined like terms to simplify further.

This simplification step is essential as it prepares the expression for use in more complex operations, like forming the difference quotient. By obtaining the simplified form \( -6x^2 - 12xh - 6h^2 - x - h + 4 \), you ensure that the operations that follow are more straightforward and less prone to error.

Algebraic Expressions

Algebraic expressions consist of numbers, variables, and operations. These expressions can be manipulated to achieve desired forms or to simplify calculations.
In the context of our exercise, you dealt with expressions like \( -12xh - 6h^2 - h \) during the subtraction of \( f(x) \) from \( f(x+h) \).
Breaking these expressions down involves:

• Identifying like terms, which can be combined.
• Factoring where possible to simplify further.
• Understanding the rules of negative signs and distribution to ensure accuracy.

This foundation in understanding how to work with algebraic expressions allows you to tackle problems involving complex operations, like difference quotients, easily.