Let $f ( x ) = \frac { 3 x ^ { 2 } - 4 } { 2 x ^ { 2 } - 1 }$ The set of all horizontal asymptotes of ƒ is

A) $\left\{ x = \frac { 3 } { 2 } \right\}$
B) $\left\{ x = - \frac { 3 } { 2 } \right\}$
C) $\left\{ y = \frac { 3 } { 2 } \right\}$
D) $\left\{ y = - \frac { 3 } { 2 } \right\}$
E) $\varnothing$

Question

Let $f ( x ) = \frac { 3 x ^ { 2 } - 4 } { 2 x ^ { 2 } - 1 }$ The set of all horizontal asymptotes of ƒ is

A) $\left\{ x = \frac { 3 } { 2 } \right\}$
B) $\left\{ x = - \frac { 3 } { 2 } \right\}$
C) $\left\{ y = \frac { 3 } { 2 } \right\}$
D) $\left\{ y = - \frac { 3 } { 2 } \right\}$ 
E) $\varnothing$

Quizplus · Accepted Answer

Horizontal asymptotes are found by dividing the leading coefficients of the highest degree terms in the numerator and denominator as $x$ approaches infinity. Here, $\frac{3}{2}$ is the ratio of the leading coefficients, indicating a horizontal asymptote at $y = \frac{3}{2}$.

Let f(x)=3x2−42x2−1f ( x ) = \frac { 3 x ^ { 2 } - 4 } { 2 x ^ { 2 } - 1 }f(x)=2x2−13x2−4​

Let $f ( x ) = \frac { 3 x ^ { 2 } - 4 } { 2 x ^ { 2 } - 1 }$