Question

Given \(f(x)=\frac{2 x^{2}-4 x}{x^{2}+3 x-28},\) what is the equation of \(y=\frac{1}{4} f[-(x-3)]\) in simplest form?

Short answer

The equation is \( y = \frac{2x^2 - 8x + 6}{4x^2 - 12x - 112} \)

Step-by-step solution

- Understand the given function

The given function is: \[ f(x) = \frac{2x^2 - 4x}{x^2 + 3x - 28} \]We need to rewrite this function for the new function \( y = \frac{1}{4} f[-(x-3)] \).

- Substitute the argument in the given function

Start by replacing the argument \( x \) in \( f(x) \) with \( -(x-3) \):\[ f(-(x-3)) = \frac{2[(-(x-3))]^2 - 4[(-(x-3))]}{[-(x-3)]^2 + 3[-(x-3)] - 28} \]

- Simplify the expression inside the function

Calculate \( (-(x-3))^2 \) and \(-(x-3)\):\[ f(-(x-3)) = \frac{2(x-3)^2 - 4(-(x-3))}{(x-3)^2 - 3(x-3) - 28} \]Expanding the expressions, we get:\[= \frac{2(x^2 - 6x + 9) + 4(x-3)}{(x-3)^2 + 3(x-3) - 28} \]\[= \frac{2x^2 - 12x + 18 + 4x - 12}{x^2 - 6x + 9 + 3x - 9 - 28} \]\[= \frac{2x^2 - 8x + 6}{x^2 - 3x - 28} \]

- Multiply by the scalar

Next, multiply the simplified function by \( \frac{1}{4} \):\[ y = \frac{1}{4} f(-(x-3)) = \frac{1}{4} \cdot \frac{2x^2 - 8x + 6}{x^2 - 3x - 28} \]\[ = \frac{2x^2 - 8x + 6}{4(x^2 - 3x - 28)} \]

- Simplify the final expression

Simplify the numerator and the denominator, if possible. However, in this case, we cannot further simplify:\[ y = \frac{2x^2 - 8x + 6}{4x^2 - 12x - 112} \]

Key concepts

rational functions

A rational function is a ratio of two polynomials. For example, the given function \( f(x) = \frac{2x^2 - 4x}{x^2 + 3x - 28} \) is a rational function because the numerator \( 2x^2 - 4x \) and the denominator \( x^2 + 3x - 28 \) are both polynomials. Rational functions can have various behaviors such as vertical and horizontal asymptotes, which are important in understanding the graph of the function. Asymptotes are lines that the graph approaches but never actually touches. To get a better understanding of rational functions:

• Identify the numerator and the denominator.
• Determine the factors of each polynomial, if possible.
• Look for common factors to simplify the function.
• Study the behavior around asymptotes by checking the values of the function as \( x \to \infty \) and as \( x \to -\infty \).

function composition

Function composition involves combining two functions such that the output of one function becomes the input of another. In the given problem, we see function composition when the argument of \( f \) is replaced: \( f(-(x-3)) \). This means we need to plug \( -(x-3) \) into the function \( f \). Here's how we approach it:

• Take the inner function or modification, which in this case is \( -(x-3) \).
• Substitute \( x \) with \( -(x-3) \) in the original function \( f(x) \).
• Simplify the resulting expression.

This method allows us to create new functions from existing ones by changing their inputs, showcasing flexibility in function manipulation.

function simplification

Function simplification aims to make a function easier to work with by reducing it to its simplest form. During simplification, factors common to both the numerator and denominator are canceled out. For the given exercise:

• We started with \( f(-(x-3)) = \frac{2(x^2-6x+9) + 4(x-3)}{(x-3)^2 + 3(x-3) - 28} \).
• We expanded and combined like terms to get \( \frac{2x^2 - 8x + 6}{x^2 - 3x - 28} \).
• Lastly, we were given a scalar multiplication \( \frac{1}{4} \), leading to the final function \( y = \frac{2x^2 - 8x + 6}{4x^2 - 12x - 112} \).

Remember, simplification does not change the value of the function but makes it more manageable. Always look for common factors and use basic algebraic techniques like factoring and combining like terms.

scalar multiplication

Scalar multiplication involves multiplying a function by a constant value. In this exercise, we multiplied the function \( f(-(x-3)) \) by \( \frac{1}{4} \), which is a scalar. The main steps include:

• Calculate the value of the function you want to scale.
• Multiply each term in the function's output by the scalar.
• Simplify, if possible.

In the case of \( y = \frac{1}{4} f(-[x-3]) \), the step of multiplying by \( \frac{1}{4} \) resulted in:
\( y = \frac{1}{4} \cdot \frac{2x^2 - 8x + 6}{x^2 - 3x - 28} \).
It eventually simplified to \( y = \frac{2x^2 - 8x + 6}{4x^2 - 12x - 112} \). Scalar multiplication simply scales the amplitude of the function's output without affecting its domain or general behavior.