91Ó°ÊÓ

A vertical asymptote occurs when the denominator of a rational function is equal to zero, and the numerator is not zero at those points. To test for vertical asymptotes in \( r(x) = \frac{x^6 + 10}{x^4 + 8x^2 + 15} \), consider the equation of the denominator: \( x^4 + 8x^2 + 15 = 0 \). This resembles a quadratic in terms of \( y = x^2 \): \( y^2 + 8y + 15 = 0 \). Solving for \( y \), use the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1, b = 8, c = 15 \). The discriminant is \( 8^2 - 4 \cdot 1 \cdot 15 = 64 - 60 = 4 \), which is positive, giving real roots. Hence no values of \( x \) make the denominator zero without making the numerator zero as well, thus no vertical asymptote.

To find \( x \)-intercepts, set the numerator \( x^6 + 10 = 0 \). Solving for \( x^6 = -10 \) results in a contradiction, as any real number raised to an even power is non-negative. Therefore, \( r(x) \) has no real \( x \)-intercepts.

Horizontal asymptotes for a rational function occur by the degrees of the polynomials. If the degree of the numerator equals the degree of the denominator, the asymptote is \( y = \frac{a}{b} \) where \( a \) and \( b \) are the leading coefficients. In \( r(x) = \frac{x^6 + 10}{x^4 + 8x^2 + 15} \), the numerator has degree 6 and the denominator is degree 4, exceeding the denominator. This implies no horizontal asymptote.

A slant (oblique) asymptote occurs when the degree of the numerator is exactly one more than the degree of the denominator. In this function, since the numerator is of a higher degree than exactly one more, the condition doesn’t hold. Thus, \( r(x) \) has no slant asymptote.

As \( x \to \infty \) or \( x \to -\infty \), the polynomial terms dominate rational functions. Here, the \( x^6 \) term in the numerator will significantly overshadow all others, and the same goes for \( x^4 \) in the denominator. Thus, the end behavior can be described by \( y = \frac{x^6}{x^4} = x^2 \), leading \( r(x) \to \infty \) as \( x \to \pm\infty \).