Identify the asymptotes.
-$$f ( x ) = \frac { x ^ { 3 } - 4 x ^ { 2 } - 9 x + 8 } { x ^ { 2 } - 6 }$$
A) Vertical asymptotes: $x = \sqrt { 6 }$ and $x = - \sqrt { 6 }$
Slant asymptote: $y = x - 4$

B) Vertical asymptotes: $x = \sqrt { 6 }$ and $x = - \sqrt { 6 }$
Slant asymptote: $y = - 3 x - 16$

C) Vertical asymptotes: $x = \sqrt { 6 }$ and $x = - \sqrt { 6 }$
Horizontal asymptote: $y = 1$

D) Vertical asymptotes: $x = 6$ and $x = - 6$
Horizontal asymptote: $y = 1$

Question

Identify the asymptotes.
-$$f ( x ) = \frac { x ^ { 3 } - 4 x ^ { 2 } - 9 x + 8 } { x ^ { 2 } - 6 }$$
A) Vertical asymptotes: $x = \sqrt { 6 }$ and $x = - \sqrt { 6 }$
Slant asymptote: $y = x - 4$

B) Vertical asymptotes: $x = \sqrt { 6 }$ and $x = - \sqrt { 6 }$
Slant asymptote: $y = - 3 x - 16$

C) Vertical asymptotes: $x = \sqrt { 6 }$ and $x = - \sqrt { 6 }$
Horizontal asymptote: $y = 1$

D) Vertical asymptotes: $x = 6$ and $x = - 6$
Horizontal asymptote: $y = 1$

Quizplus · Accepted Answer

Vertical asymptotes occur where the denominator equals zero, so setting $x^2 - 6 = 0$ gives $x = \pm \sqrt{6}$. A slant (or oblique) asymptote occurs when the degree of the numerator is exactly one more than the degree of the denominator. By dividing the numerator by the denominator, we find the slant asymptote to be $y = x - 4$.

Identify the Asymptotes f(x)=x3−4x2−9x+8x2−6f ( x ) = \frac { x ^ { 3 } - 4 x ^ { 2 } - 9 x + 8 } { x ^ { 2 } - 6 }f(x)=x2−6x3−4x2−9x+8​

Identify the Asymptotes $f ( x ) = \frac { x ^ { 3 } - 4 x ^ { 2 } - 9 x + 8 } { x ^ { 2 } - 6 }$