- Identify the point where the function is undefined by solving \(8x - 64 = 0\), which gives \(x = 8\).
- Factor the numerator as \((x-8)(x+8)\) and the denominator as \(8(x-8)\) to simplify the function to \(f(x) = \frac{x+8}{8}\) for \(x
eq 8\).
- Evaluate the limit as \(x \to 8\) to find it equals \(2\), indicating a removable discontinuity (hole) rather than an asymptote.
- Conclude that the graph has no vertical asymptote, so the statement is \(\boxed{\text{The graph has no vertical asymptote.}}\)
Explanation
- Identify the points where the function is undefined
To determine the vertical asymptotes of the function \(f(x) = \frac{x^2 - 64}{8x - 64}\), we must examine the points where the function is undefined. These points occur where the denominator equals zero.
Setting the denominator equal to zero, we have:
\[8x - 64 = 0\]
\[8x = 64\]
\[x = 8\]
So, the function is undefined at \(x = 8\).
- Analyze the limit at the point of discontinuity
Next, we analyze the behavior of the function near \(x = 8\). We factor both the numerator and the denominator to simplify the expression:
Numerator: \(x^2 - 64 = (x - 8)(x + 8)\)
Denominator: \(8x - 64 = 8(x - 8)\)
Substituting these factors into the function:
\[f(x) = \frac{(x - 8)(x + 8)}{8(x - 8)}\]
For \(x
eq 8\), we can cancel the common factor \((x - 8)\):
\[f(x) = \frac{x + 8}{8}\]
As \(x \to 8\), the value of the simplified function is:
\[\lim_{x \to 8} f(x) = \lim_{x \to 8} \frac{x + 8}{8} = \frac{8 + 8}{8} = \frac{16}{8} = 2\]
Since the limit as \(x \to 8\) is a finite number, there is a removable discontinuity (a hole) at \(x = 8\), not a vertical asymptote.
- Conclusion regarding vertical asymptotes
A vertical asymptote exists at a value \(x = a\) if the function approaches \(\infty\) or \(-\infty\) as \(x\) approaches \(a\). In this case, because the factor \((x - 8)\) in the denominator is canceled out by a factor in the numerator, the function does not approach infinity at \(x = 8\). There are no other values of \(x\) that make the denominator zero. Therefore, the function \(f(x) = \frac{x^2 - 64}{8x - 64}\) has no vertical asymptotes. The correct statement is that the graph has no vertical asymptote.
Examples
In fields like electrical engineering, analyzing functions helps determine the stable regions of a circuit. If a circuit's impedance is modeled by a rational function, vertical asymptotes would indicate points of infinite resistance or failure. Identifying these points, or distinguishing them from simple holes in the domain, is vital for ensuring circuit stability and reliability in electronic design.