Problem 77
Question
Find the vertical asymptotes of each function. $$ y=\frac{x-3}{x+3} $$
Step-by-Step Solution
Verified Answer
The vertical asymptote of the function \(y=\frac{x-3}{x+3}\) is \(x = -3\).
1Step 1: Write out the function
We have the function \( y=\frac{x-3}{x+3} \).
2Step 2: Identify the denominator of the function
The denominator of the function is \(x+3\). This equation is what we will set equal to zero.
3Step 3: Solve for x
Setting the denominator equal to zero, we get: \( x + 3 = 0 \). Solving for x, we find \( x = -3 \).
4Step 4: Determine if x value makes the numerator zero
Next, we substitute this value, \(x=-3\), into the numerator of the function. Substituting -3 in for x in \(x-3\) doesn't result in zero; thus, there is no limitation on x = -3 being a vertical asymptote.
5Step 5: Write out the asymptote
So, the vertical asymptote of the function is \(x = -3\).
Key Concepts
Rational FunctionsDenominator ZeroAsymptote IdentificationSolving Equations
Rational Functions
Rational functions are expressions that are the ratio of two polynomials. They are often given in the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials. These functions can exhibit interesting behavior, such as asymptotes and discontinuities.
Rational functions can model a variety of real-world situations, making them important in both mathematics and applied sciences. Given their polynomial nature, the x-values for which the denominator \( Q(x) \) equals zero are of particular interest, since these points can lead to vertical asymptotes.
To analyze a rational function like \( y=\frac{x-3}{x+3} \), you examine the polynomial in the numerator (\( x-3 \)) and the polynomial in the denominator (\( x+3 \)). Understanding how these polynomials interact helps us determine the behavior of the function.
Rational functions can model a variety of real-world situations, making them important in both mathematics and applied sciences. Given their polynomial nature, the x-values for which the denominator \( Q(x) \) equals zero are of particular interest, since these points can lead to vertical asymptotes.
To analyze a rational function like \( y=\frac{x-3}{x+3} \), you examine the polynomial in the numerator (\( x-3 \)) and the polynomial in the denominator (\( x+3 \)). Understanding how these polynomials interact helps us determine the behavior of the function.
Denominator Zero
The denominator in a rational function is essential for understanding where the function may become undefined. Places where the denominator equals zero are critical because a function cannot have a division by zero.
For the function \( y=\frac{x-3}{x+3} \), the denominator is \( x+3 \). Setting the denominator to zero gives us the equation \( x+3 = 0 \). Solving this equation results in \( x = -3 \).
At \( x = -3 \), the function is not defined, which is a signal that this x-value might correspond to a vertical asymptote. Finding where the denominator is zero helps in identifying important features of a rational function, such as discontinuities and vertical asymptotes.
For the function \( y=\frac{x-3}{x+3} \), the denominator is \( x+3 \). Setting the denominator to zero gives us the equation \( x+3 = 0 \). Solving this equation results in \( x = -3 \).
At \( x = -3 \), the function is not defined, which is a signal that this x-value might correspond to a vertical asymptote. Finding where the denominator is zero helps in identifying important features of a rational function, such as discontinuities and vertical asymptotes.
Asymptote Identification
Vertical asymptotes occur in rational functions when an x-value makes the denominator zero but does not make the numerator zero. They are vertical lines on a graph where the function approaches infinity or negative infinity.
In our function, \( y=\frac{x-3}{x+3} \), setting the denominator \( x+3 \) to zero results in \( x = -3 \). To confirm if this x-value is indeed a vertical asymptote, substitute \( x = -3 \) into the numerator \( x-3 \).
Calculating \( -3 - 3 \) does not yield zero, confirming that the x-value does not cancel out both the numerator and denominator. Thus, the vertical line \( x = -3 \) is identified as a vertical asymptote, indicating that the function will approach infinity as it nears this line on its graph.
In our function, \( y=\frac{x-3}{x+3} \), setting the denominator \( x+3 \) to zero results in \( x = -3 \). To confirm if this x-value is indeed a vertical asymptote, substitute \( x = -3 \) into the numerator \( x-3 \).
Calculating \( -3 - 3 \) does not yield zero, confirming that the x-value does not cancel out both the numerator and denominator. Thus, the vertical line \( x = -3 \) is identified as a vertical asymptote, indicating that the function will approach infinity as it nears this line on its graph.
Solving Equations
Solving the equation set by the denominator is key to uncovering vertical asymptotes. We simplify this process by setting the denominator equal to zero and solving for x.
In the example \( y=\frac{x-3}{x+3} \), solve \( x+3 = 0 \) by subtracting 3 from both sides. This gives \( x = -3 \).
This step is crucial because it finds potential asymptotes or undefined points. Always remember to check this x-value in the numerator to ensure it doesn't also reduce the overall expression to zero. If it does not, then you've correctly identified a vertical asymptote.
In the example \( y=\frac{x-3}{x+3} \), solve \( x+3 = 0 \) by subtracting 3 from both sides. This gives \( x = -3 \).
This step is crucial because it finds potential asymptotes or undefined points. Always remember to check this x-value in the numerator to ensure it doesn't also reduce the overall expression to zero. If it does not, then you've correctly identified a vertical asymptote.
Other exercises in this chapter
Problem 76
Write an equation of the circle with the given center and radius. Graph the circle. center \((1,1),\) radius 2
View solution Problem 76
The arithmetic mean of two terms in an arithmetic sequence is \(-6 .\) One term is \(-20 .\) Find the other term.
View solution Problem 78
Find the vertical asymptotes of each function. $$ y=\frac{x-3}{x+1} $$
View solution Problem 78
In an arithmetic sequence with \(a_{1}=2\) and \(d=-2,\) which term is \(-82 ?\)
View solution