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Chapter 2: Problem 92

The equation for \(f\) is given by the simplified expression that results after performing the indicated operation. Write the equation for \(f\) and then graph the function. $$\frac{2}{x^{2}+3 x+2}-\frac{4}{x^{2}+4 x+3}$$

Short Answer

Expert verified
The simplified function is \(f(x) = -\frac{2}{(x+2)(x+3)}\). You can graph this function by sketching two vertical asymptotes at \(x = -2\) and \(x = -3\), and the function is negative for all other x values.

Step by step solution

01

Simplifying the fraction

Firstly, factorize the two denominators \(x^{2}+3x+2\) and \(x^{2}+4x+3\) to make the subtraction easier. As per factorization rules, both would be \((x+1)(x+2)\) and \((x+1)(x+3)\) respectively. The common denominator will be \((x+1)(x+2)(x+3)\). Now, for both fractions convert into equivalent fractions with common denominator. The first fraction will become \(\frac{2(x+3)}{(x+1)(x+2)(x+3)}\) and the second fraction will be \(\frac{4(x+2)}{(x+1)(x+2)(x+3)}\). Now we can subtract the two fractions easily: \(f(x) = \frac{2x+6-4x-8}{(x+1)(x+2)(x+3)} = \frac{-2x-2}{(x+1)(x+2)(x+3)}\).
02

Further simplification

Based on the previous step, you can simplify \(f(x)\) to be \(f(x) = -\frac{2(x+1)}{(x+1)(x+2)(x+3)}\) which equals to \(-\frac{2}{(x+2)(x+3)}\) after cancelling out the \((x+1)\) terms. This is the simplified equation for \(f\).
03

Determining Characteristics of the Function

Before graphing, it is useful to consider key qualities of the function. This function f(x) have vertical asymptotes at \(x = -2\) and \(x = -3\) because at these points the function is undefined. Moreover, the function is negative for all x due to the negative numerator.
04

Graphing the Function

Use the following points to make a sketch of the graph: Choose points for different x values that are not equal to -2 and -3 (the asymptotes). Then use these point to draw the curve. The curve should approach (but never touch) the vertical lines \(x = -2\) and \(x = -3\) since these are the asymptotes. The function is negative everywhere else and approaches 0 as \(x \to \pm \infty\).

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