Identify the vertical asymptote of the function.
[tex]f(x)=\frac{x^2+1}{3(x-8)}[/tex]
The vertical asymptote is at [tex]x=[/tex] $\square$.
Answer (2)
The function is f ( x ) = 3 ( x − 8 ) x 2 + 1 .
Vertical asymptotes occur when the denominator equals zero, so set 3 ( x − 8 ) = 0 .
Solve for x : x − 8 = 0 ⟹ x = 8 .
The vertical asymptote is at 8 .
Explanation
Problem Analysis We are given the function f ( x ) = 3 ( x − 8 ) x 2 + 1 and asked to find its vertical asymptote.
Finding the Zeros of the Denominator Vertical asymptotes occur at values of x where the denominator of the rational function is equal to zero. So, we need to find the values of x such that 3 ( x − 8 ) = 0 .
Solving for x To solve for x , we set the denominator equal to zero: 3 ( x − 8 ) = 0
Divide both sides by 3: x − 8 = 0
Add 8 to both sides: x = 8
Conclusion Therefore, the vertical asymptote is at x = 8 .
Examples
Vertical asymptotes are useful in various fields, such as physics and engineering, when modeling real-world phenomena. For example, when analyzing the behavior of electrical circuits, a function might have a vertical asymptote that represents a point where the current or voltage becomes infinitely large, indicating a critical limit or instability in the circuit. Understanding and identifying these asymptotes helps engineers design more stable and reliable systems.
The vertical asymptote of the function f ( x ) = 3 ( x − 8 ) x 2 + 1 is found by setting the denominator equal to zero. This leads to x = 8 . Thus, the vertical asymptote is at x = 8 .
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