Problem 31
Question
Find all horizontal and vertical asymptotes (if any). $$ t(x)=\frac{x^{2}+2}{x-1} $$
Step-by-Step Solution
Verified Answer
Vertical asymptote at \( x = 1 \), no horizontal asymptotes, oblique asymptote at \( y = x + 1 \).
1Step 1: Find Vertical Asymptotes
Vertical asymptotes occur where the denominator of the function is equal to zero, provided the numerator does not also equal zero at these points. For the given function \( t(x)= \frac{x^{2}+2}{x-1} \), set the denominator to zero: \( x - 1 = 0 \). Solve for \( x \): \( x = 1 \). Since the numerator \( x^2 + 2 \) does not equal zero at \( x = 1 \), there is a vertical asymptote at \( x = 1 \).
2Step 2: Find Horizontal Asymptotes
Horizontal asymptotes are determined by the degrees of the polynomials in the numerator and denominator. For \( t(x)= \frac{x^{2}+2}{x-1} \), the degree of the numerator (2) is greater than the degree of the denominator (1). When the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes. However, because the degrees differ by 1, there may be an oblique (slant) asymptote, which can be found by polynomial long division.
3Step 3: Perform Polynomial Long Division for Oblique Asymptote
Divide \( x^2 + 2 \) by \( x - 1 \) using polynomial long division. This process involves dividing the leading term of the numerator by the leading term of the denominator: \( x^2 \div x = x \). Multiply \( x \) by \( x-1 \) to get \( x^2 - x \), and subtract it from the original numerator \( x^2 + 2 \) to get the new numerator \( x + 2 \). Then, divide \( x \) by \( x \) to get 1, and multiply by \( x-1 \), subtracting it from \( x + 2 \) to yield a remainder. The quotient \( x + 1 \) represents the equation of the oblique asymptote: \( y = x + 1 \).
Key Concepts
Vertical AsymptoteHorizontal AsymptoteOblique AsymptotePolynomial Long Division
Vertical Asymptote
Vertical asymptotes are like invisible lines that the graph of a function approaches but never crosses. They are an essential concept in understanding the behavior of rational functions, such as the one given in the exercise. For the function \( t(x) = \frac{x^2 + 2}{x - 1} \), a vertical asymptote occurs where the denominator equals zero because division by zero is undefined.
To find the vertical asymptote, you set the denominator of the function \( x - 1 = 0 \) and solve for \( x \). This gives you \( x = 1 \). An important step is to verify that this value does not make the numerator zero as well, as that would indicate a hole rather than an asymptote. In this case, \( x^2 + 2 \) is not zero when \( x = 1 \). Thus, the function has a vertical asymptote at \( x = 1 \).
Key Points:
To find the vertical asymptote, you set the denominator of the function \( x - 1 = 0 \) and solve for \( x \). This gives you \( x = 1 \). An important step is to verify that this value does not make the numerator zero as well, as that would indicate a hole rather than an asymptote. In this case, \( x^2 + 2 \) is not zero when \( x = 1 \). Thus, the function has a vertical asymptote at \( x = 1 \).
Key Points:
- Set the denominator to zero.
- Ensure the numerator is not zero at that point.
- The vertical asymptote is where the denominator zeros out alone.
Horizontal Asymptote
Horizontal asymptotes describe the behavior of a function as \( x \) approaches positive or negative infinity. They give an idea of what the graph will look like as you move far to the left or right. Generally, they depend on the degrees of the polynomials in the numerator and denominator.
For \( t(x) = \frac{x^2 + 2}{x - 1} \), the degree of the numerator is 2, while the denominator's degree is 1.
When the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes. Instead, if the numerator's degree is exactly one higher, an oblique asymptote might exist. This will involve checking for a slant asymptote via polynomial long division.
Remember:
For \( t(x) = \frac{x^2 + 2}{x - 1} \), the degree of the numerator is 2, while the denominator's degree is 1.
When the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes. Instead, if the numerator's degree is exactly one higher, an oblique asymptote might exist. This will involve checking for a slant asymptote via polynomial long division.
Remember:
- Compare the degrees of numerator and denominator.
- Higher numerator degree means no horizontal asymptote.
Oblique Asymptote
An oblique asymptote, also known as a slant asymptote, occurs when the degree of the numerator is exactly one higher than the degree of the denominator in a rational function. The presence of an oblique asymptote indicates a slanted line that the graph will follow at extreme values of \( x \).
For the function \( t(x) = \frac{x^2 + 2}{x - 1} \), polynomial long division must be used to find the oblique asymptote. By dividing \( x^2 + 2 \) by \( x - 1 \), the result is \( x + 1 \), which is the equation for the oblique asymptote. This means the graph approaches the line \( y = x + 1 \) as \( x \) goes to infinity or negative infinity.
Remember:
For the function \( t(x) = \frac{x^2 + 2}{x - 1} \), polynomial long division must be used to find the oblique asymptote. By dividing \( x^2 + 2 \) by \( x - 1 \), the result is \( x + 1 \), which is the equation for the oblique asymptote. This means the graph approaches the line \( y = x + 1 \) as \( x \) goes to infinity or negative infinity.
Remember:
- Exists when numerator's degree is one more than the denominator's.
- Found through polynomial long division.
- Represents a linear trend the function approaches.
Polynomial Long Division
Polynomial long division is a method used to divide one polynomial by another. It's similar to long division with numbers and is particularly useful in finding oblique asymptotes of rational functions.
Using our function \( t(x) = \frac{x^2 + 2}{x - 1} \), we need to divide \( x^2 + 2 \) by \( x - 1 \). Start by dividing the first term of the numerator \( x^2 \) by the first term of the denominator \( x \), which gives \( x \).
Next, multiply \( x \) by the entire divisor \( x - 1 \), resulting in \( x^2 - x \). Subtract this from the original numerator to find \( x + 2 \). Then, continue the process: divide \( x \) by \( x \) to get 1 and multiply \( 1 \) by \( x - 1 \), giving \( x - 1 \). Subtract this result, leaving a remainder. The quotient \( x + 1 \) becomes the oblique asymptote \( y = x + 1 \).
Key Steps:
Using our function \( t(x) = \frac{x^2 + 2}{x - 1} \), we need to divide \( x^2 + 2 \) by \( x - 1 \). Start by dividing the first term of the numerator \( x^2 \) by the first term of the denominator \( x \), which gives \( x \).
Next, multiply \( x \) by the entire divisor \( x - 1 \), resulting in \( x^2 - x \). Subtract this from the original numerator to find \( x + 2 \). Then, continue the process: divide \( x \) by \( x \) to get 1 and multiply \( 1 \) by \( x - 1 \), giving \( x - 1 \). Subtract this result, leaving a remainder. The quotient \( x + 1 \) becomes the oblique asymptote \( y = x + 1 \).
Key Steps:
- Divide leading terms of numerator and denominator.
- Multiply and subtract like in basic long division.
- The quotient reveals the oblique asymptote equation.
Other exercises in this chapter
Problem 30
A quadratic function is given. (a) Express the quadratic function in standard form. (b) Sketch its graph. (c) Find its maximum or minimum value. $$ g(x)=2 x^{2}
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Find all rational zeros of the polynomial, and write the polynomial in factored form. $$ P(x)=3 x^{4}-10 x^{3}-9 x^{2}+40 x-12 $$
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25-38 . Find the quotient and remainder using synthetic division. $$ \frac{x^{3}-8 x+2}{x+3} $$
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Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero. \(P(x)=x^{4}+3 x^{2}-4\)
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