Problem 18 - Algebra and Trigonometry

Question

Find all horizontal and vertical asymptotes (if any). \((x)=\frac{2 x-4}{x^{2}+2 x+1}\)

Step-by-Step Solution

Answer

Horizontal: \(y = 0\); Vertical: \(x = -1\).

Step 1: Identifying Horizontal Asymptotes

To find horizontal asymptotes, we compare the degrees of the polynomial in the numerator and the denominator. The highest degree is 1 in the numerator and 2 in the denominator. Since the degree of the denominator is greater, the horizontal asymptote is at \(y = 0\).

Step 2: Finding Vertical Asymptotes

To find vertical asymptotes, we need to determine when the denominator is zero, as long as these values do not also make the numerator zero. Solve \(x^2+2x+1=0\).\((x+1)^2=0\) The solution is \(x = -1\). Therefore, there is a vertical asymptote at \(x = -1\).

Step 3: Double-checking for Removable Discontinuities

Verify if the value \( x = -1 \) makes the numerator zero as well, which could mean a removable discontinuity rather than a vertical asymptote. The numerator \(2x-4=2(-1)-4=-6\), which is non-zero at \(x=-1\). Hence, \(x=-1\) is indeed a vertical asymptote.

Key Concepts

Horizontal Asymptote

In rational functions, a horizontal asymptote is a horizontal line that the graph approaches as the independent variable goes to positive or negative infinity. These asymptotes give us insight into the end behavior of the function. For the function \( f(x)=\frac{2x-4}{x^{2}+2x+1} \), we determine the horizontal asymptote by comparing the degrees of the polynomials in the numerator and denominator.
To find a horizontal asymptote:

• If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at \( y = 0 \).
• If the degrees are equal, the horizontal asymptote is \( y = \frac{a}{b} \), where \( a \) and \( b \) are the leading coefficients of the numerator and denominator, respectively.
• If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

Here, the degree of the numerator is 1 and the degree of the denominator is 2. Since the denominator's degree is higher, we have a horizontal asymptote at \( y = 0 \). This informs us that as \( x \) moves towards infinity, the function values get closer and closer to zero.

Vertical Asymptote

Vertical asymptotes occur in rational functions where the function approaches positive or negative infinity around specific \( x \)-values. These happen when the denominator equals zero, provided this doesn't also simultaneously zero the numerator or result in a factor cancelation.
To find a vertical asymptote for \( f(x)=\frac{2x-4}{x^{2}+2x+1} \), set the denominator to zero and solve:

• \( x^2 + 2x + 1 = 0 \) simplifies to \( (x+1)^2 = 0 \).
• Solving this equation gives \( x = -1 \).

So, there is a potential vertical asymptote at \( x = -1 \). However, to confirm, ensure this value doesn’t zero the numerator, \( 2x - 4 \). By checking, \( 2(-1) - 4 = -6 \), which is non-zero. Therefore, \( x = -1 \) is indeed a confirmed vertical asymptote, as the graph shoots to infinity at this \( x \)-value.

Removable Discontinuity

A removable discontinuity in a rational function occurs when both the numerator and the denominator equal zero at the same point. This typically indicates a hole in the graph rather than an asymptote.
To identify if there's a removable discontinuity for \( f(x)=\frac{2x-4}{x^{2}+2x+1} \), check if the \( x = -1 \) not only zeros the denominator but also zeros the numerator:

• For the numerator \( 2x - 4 \), substituting \( x = -1 \) results in \( -2 - 4 = -6 \), which is not zero.

Since the numerator does not equal zero at \( x = -1 \), there is no removable discontinuity at this point. This confirms that the zero denominator without a zero numerator results in a vertical asymptote.