Questions: Find the horizontal and vertical asymptotes of the curve. You may want to use a graph y=(x^3-x)/(x^2-9x+8) x= y=

Solution Steps

To find the horizontal and vertical asymptotes of the given rational function, we need to analyze the behavior of the function as \( x \) approaches infinity and the values that make the denominator zero. The horizontal asymptote is determined by comparing the degrees of the numerator and the denominator. The vertical asymptotes occur at the values of \( x \) that make the denominator zero, provided they do not also make the numerator zero.

The given function is: \[ y = \frac{x^3 - x}{x^2 - 9x + 8} \]

Vertical asymptotes occur where the denominator is zero, provided the numerator is not zero at those points. Set the denominator equal to zero and solve for \(x\): \[ x^2 - 9x + 8 = 0 \]

Factor the quadratic equation: \[ x^2 - 9x + 8 = (x - 1)(x - 8) = 0 \]

Thus, the potential vertical asymptotes are at \(x = 1\) and \(x = 8\).

Check if the numerator is also zero at \(x = 1\) and \(x = 8\) to determine if these are removable discontinuities.

Evaluate the numerator at \(x = 1\): \[ x^3 - x = 1^3 - 1 = 0 \]

Evaluate the numerator at \(x = 8\): \[ x^3 - x = 8^3 - 8 = 512 - 8 = 504 \neq 0 \]

Since the numerator is zero at \(x = 1\), \(x = 1\) is a removable discontinuity, not a vertical asymptote. Therefore, the vertical asymptote is only at \(x = 8\).

Horizontal asymptotes are determined by the degrees of the numerator and the denominator.

• The degree of the numerator \(x^3 - x\) is 3.
• The degree of the denominator \(x^2 - 9x + 8\) is 2.

Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

Final Answer