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Magazine Chapter 9: Problem 20
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
Expert verified
The equation is \( y = \frac{2x^2 - 8x + 6}{4x^2 - 12x - 112} \)
Step by step solution
01
- 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 \( y = \frac{1}{4} f[-(x-3)] \).
02
- Substitute the argument in the given function
Start by replacing the argument \( x \) in \( f(x) \) with \( -(x-3) \):\[ f(-(x-3)) = \frac{2[(-(x-3))]^2 - 4[(-(x-3))]}{[-(x-3)]^2 + 3[-(x-3)] - 28} \]
03
- 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:\[= \frac{2(x^2 - 6x + 9) + 4(x-3)}{(x-3)^2 + 3(x-3) - 28} \]\[= \frac{2x^2 - 12x + 18 + 4x - 12}{x^2 - 6x + 9 + 3x - 9 - 28} \]\[= \frac{2x^2 - 8x + 6}{x^2 - 3x - 28} \]
04
- Multiply by the scalar
Next, multiply the simplified function by \( \frac{1}{4} \):\[ y = \frac{1}{4} f(-(x-3)) = \frac{1}{4} \cdot \frac{2x^2 - 8x + 6}{x^2 - 3x - 28} \]\[ = \frac{2x^2 - 8x + 6}{4(x^2 - 3x - 28)} \]
05
- Simplify the final expression
Simplify the numerator and the denominator, if possible. However, in this case, we cannot further simplify:\[ y = \frac{2x^2 - 8x + 6}{4x^2 - 12x - 112} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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 \( y = \frac{2x^2 - 8x + 6}{4x^2 - 12x - 112} \).
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:
\( y = \frac{1}{4} \cdot \frac{2x^2 - 8x + 6}{x^2 - 3x - 28} \).
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.
- Calculate the value of the function you want to scale.
- Multiply each term in the function's output by the scalar.
- Simplify, if possible.
\( y = \frac{1}{4} \cdot \frac{2x^2 - 8x + 6}{x^2 - 3x - 28} \).
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.
