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Determine the vertical asymptote(s) of each function. If none exists, state that fact. $$ f(x)=\frac{5 x}{x^{2}-25} $$

Short Answer

Expert verified
The vertical asymptotes of the function are at \( x = 5 \) and \( x = -5 \).

Step by step solution

01

Identify the Denominator

The vertical asymptotes occur where the function is undefined, which typically happens where the denominator is equal to zero. Begin by setting the denominator equal to zero: \( x^2 - 25 = 0 \).
02

Solve for Zero in the Denominator

Solve the equation \( x^2 - 25 = 0 \) by moving 25 to the other side to get \( x^2 = 25 \).
03

Take Square Root of Both Sides

Take the square root of both sides to solve for \( x \): \( x = ext{±} 5 \). This gives the potential locations for vertical asymptotes.
04

Verify Asymptotes Availability

Verify that both \( x = 5 \) and \( x = -5 \) make the denominator zero without being canceled by the numerator. Since \( 5x \) does not have any factor to cancel \( x = 5 \) or \( x = -5 \), they are indeed vertical asymptotes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Denominator of a Function
In mathematics, a polynomial or rational function's structure often involves a numerator (the top part) and a denominator (the bottom part). The denominator of a function can reveal important characteristics about the function's behavior. Specifically, it defines the points where the function may be undefined if it equals zero. This is crucial when finding vertical asymptotes. Vertical asymptotes typically occur at the values for which the denominator is zero, as the function approaches infinity or negative infinity at these points.

To find these points, take the expression in the denominator and set it equal to zero. For example, in the function \( f(x) = \frac{5x}{x^2 - 25} \), the denominator is \( x^2 - 25 \). Setting \( x^2 - 25 = 0 \) helps identify the inputs (or \( x \)-values) that will make the function yield undefined or infinite outputs, indicating potential vertical asymptotes.

This process helps in analyzing the function's graph and understanding how the function behaves across different parts of the domain.
Solving Equations
Solving equations is a fundamental skill in algebra. It involves finding the values of variables that make the equation true. In the exercise, the equation \( x^2 - 25 = 0 \) needs to be solved to determine the vertical asymptotes for the function.

Once you have your equation from setting the denominator to zero, like \( x^2 - 25 = 0 \), the task is to isolate \( x \). This often means rearranging terms and applying mathematical operations to both sides of the equation. Here’s how this particular equation can be approached:
  • Move 25 to the other side of the equation: \( x^2 = 25 \)
  • Apply the square root operation, recognizing it has both a positive and negative solution: \( x = \pm 5 \)
These solutions \( x = 5 \) and \( x = -5 \) are points where the denominator equals zero, making the function undefined at these \( x \)-values. Thus, they indicate the positions of vertical asymptotes on the graph.
Square Roots
Square roots are essential in solving equations, especially when dealing with quadratic terms like \( x^2 \). Understanding square roots helps in solving equations of the form \( x^2 = n \), where \( n \) is a non-negative number. The square root operation provides values that satisfy the original squared term.

To solve \( x^2 = 25 \), take the square root of both sides:
  • The square root of 25 is \( 5 \), since \( 5 \times 5 = 25 \).
  • This means \( x \) could be \( 5 \).
  • But remember, since the square of a negative number is also positive, \( x \) could also be \( -5 \).
When you square a number, you lose the information about its original sign (positive or negative). That's why taking the square root often gives two possible solutions: positive and negative. These solutions highlight critical points on the graph where vertical asymptotes can occur if they don't appear in the numerator, too.

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