Problem 120 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the difference quotient and simplify your answer. $$f(x)=5+6 x-x^{2}, \frac{f(6+h)-f(6)}{h}, h \neq 0$$

Step-by-Step Solution

Answer

The difference quotient of the given function is \(-h - 6\).

Step 1: Substitute \(x = 6\) in \(f(x)\)

First, plug \(6\) into the function to get \(f(6)\). This will result in \(f(6) = 5 + 6*6 - (6)^2 = 5 + 36 - 36 = 5\).

Step 2: Substitute \(x = 6 + h\) in \(f(x)\)

Next, substitute \(6 + h\) into the function to yield \(f(6 + h)\). Simplifying this will result in \(f(6 + h) = 5 + 6*(6 + h) - (6 + h)^2 = 5 + 36 + 6h - (36 + 12h + h^2) = 5 + 36 + 6h - 36 - 12h - h^2 = 5 - h^2 - 6h\).

Step 3: Compute the difference quotient

Now, calculate the difference quotient \(\frac{f(6 + h) - f(6)}{h}\). This becomes \( \frac{(5 - h^2 - 6h) - 5}{h} = \frac{-h^2 - 6h}{h}\).

Step 4: Simplify the expression

Finally, simplify the expression by canceling \(h\) out of the equation. This gives \(-h - 6\).

Key Concepts

Quadratic Functions

Quadratic functions are a key part of algebra, often taking the form of \(f(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. In the function \(f(x) = 5 + 6x - x^2\), each component plays a pivotal role in its behavior. The term \(-x^2\) suggests the function will open downward, making it a concave parabola. This parabola has a unique symmetry and a specific vertex. The vertex determines the highest point of the curve since the parabola opens downward. Quadratic functions enable us to describe a wide array of real-world phenomena, from the trajectory of a thrown ball to economic models. Understanding how these functions operate can provide insights into their overall graph and help solve problems involving maximums and minimums.

Function Evaluation

Function evaluation is a process through which we determine the value of a function for a specific input. It involves substituting different values into the function's equation. In this exercise, we evaluate \(f(x) = 5 + 6x - x^2\) at specific points, namely, at \(x = 6\) and \(x = 6 + h\). This is done in steps:

• First, we replace \(x\) with 6 to find \(f(6)\).
• Then do the calculation: it results in \(f(6) = 5\).
• Next, \(x = 6 + h\) is substituted into \(f(x)\), leading to the expression \(5 - h^2 - 6h\).

Each step involves basic arithmetic and helps us understand how functions respond to different inputs. This is vital for solving calculus problems where precise function values are needed.

Simplifying Expressions

Simplifying expressions is an essential skill in algebra that makes complex equations easier to work with. Once you've calculated the function's outputs at particular points, as seen with \(f(6+h)\) and \(f(6)\), you find the difference and simplify it. To simplify the difference quotient \(\frac{f(6+h) - f(6)}{h}\), you need to carefully subtract one function from the other.

• Subtract \(f(6) = 5\) from \(f(6+h) = 5 - h^2 - 6h\).
• When performing the subtraction, terms cancel simplified forms like \(-h^2 - 6h\).
• Lastly, divide by \(h\) to complete the difference quotient: \(\frac{-h^2 - 6h}{h} = -h - 6\)

This final step illustrates another simplification process, removing the common factor of \(h\) to reach a concise result. Simplifying expressions helps highlight key function features and prepares them for further analysis or calculus operations.